Download السلطة الفلسطينية الوطنية

‫السلطة الوطنية الفلسطينية‬
‫وزارة التربية والتعليم العالي‬
‫كلية العلوم والتكنولوجيا‬
‫قسم العلوم الهندسية والفنون التطبيقية‬
‫بكالوريوس هندسة المباني‬
‫الكتاب المنهجي لمساق الكهرباء التطبيقية‬
‫إعداد ‪ :‬م‪ .‬سمير أحمد جبر‬
‫الفصل الثاني‬
‫‪2008/2007‬‬
TABLE OF CONTENTS:
UNIT 1 : AC, DC and Electrical Signals
UNIT 2 : DC Laws and Theorems
UNIT 3 : Single Phase AC
UNIT 4 : Three Phase AC System
UNIT 5 : Electrical Transformers
UNIT 6 : Electrical Motors
UNIT 7 : Switching and Controlling of AC Motors
UNIT 8 : Introduction To Protection
2
pp 3
pp 16
pp 28
pp 50
pp 58
pp 72
pp 101
pp 113
UNIT 1
AC, DC and Electrical Signals
AC means Alternating Current and DC means Direct Current. AC and DC are also used when referring to
voltages and electrical signals which are not currents! For example: a 12V AC power supply has an
alternating voltage (which will make an alternating current flow). An electrical signal is a voltage or
current which conveys information, usually it means a voltage. The term can be used for any voltage or
current in a circuit.
1.1 Direct Current (DC)
Direct Current (DC) always flows in the same direction,
but it may increase and decrease.
A DC voltage is always positive (or always negative),
but it may increase and decrease. Electronic circuits
normally require a steady DC supply which is constant
at one value or a smooth DC supply which has a small
Steady DC
variation called ripple.
Cells, batteries and regulated power supplies provide
from a battery or regulated power supply,
this is ideal for electronic circuits.
steady DC which is ideal for electronic circuits.
Power supplies contain a transformer which converts the
mains AC supply to a safe low voltage AC. Then the AC
is converted to DC by a bridge rectifier but the output is
varying DC which is unsuitable for electronic circuits.
Smooth DC
Some power supplies include a capacitor to provide
smooth DC which is suitable for less-sensitive electronic
from a smoothed power supply,
this is suitable for some electronics.
circuits, including most of the projects on this website.
Lamps, heaters and motors will work with any DC supply.
Varying DC
from a power supply without smoothing,
this is not suitable for electronics.
Applications
Direct-current installations usually have different types of sockets, switches, and fixtures, mostly due to
the low voltages used, from those suitable for alternating current. It is usually important with a directcurrent appliance not to reverse polarity unless the device has a diode bridge to correct for this (most
battery-powered devices do not).
3
This symbol is found on many electronic devices that either require or produce direct current
DC is commonly found in many low-voltage applications, especially where these are powered by
batteries, which can produce only DC, or solar power systems, since solar cells can produce only DC.
Most automotive applications use DC, although the alternator is an AC device which uses a rectifier to
produce DC. Most electronic circuits require a DC power supply. Applications using fuel cells (mixing
hydrogen and oxygen together with a catalyst to produce electricity and water as byproducts) also
produce only DC.
Many telephones connect to a twisted pair of wires, and internally separate the AC component of the
voltage between the two wires (the audio signal) from the DC component of the voltage between the two
wires (used to power the phone).
Telephone exchange communication equipment, such as DSLAM, uses standard -48V DC power supply.
The negative polarity is achieved by grounding the positive terminal of power supply system and the
battery bank. This is done to prevent electrolysis depositions.
An electrified third rail can be used to power both underground (subway) and overground trains.
1.2 Alternating Current (AC)
Alternating Current (AC) flows one way, then the other
way, continually reversing direction.
An AC voltage is continually changing between positive
(+) and negative (-).
The rate of changing direction is called the frequency of
the AC and it is measured in hertz (Hz) which is the
number of forwards-backwards cycles per second.
AC from a power supply
This shape is called a sine wave.
Mains electricity in the USA and Canada have a
frequency 60Hz while other counties have a frequency of
50Hz.
An AC supply is suitable for powering some devices
such as lamps and heaters but almost all electronic
This triangular signal is AC because it changes
circuits require a steady DC supply (see below).
between positive (+) and negative (-).
4
1.3 Properties of electrical signals
An electrical signal is a voltage or current which
conveys information, usually it means a voltage. The
term can be used for any voltage or current in a circuit.
The voltage-time graph on the right shows various
properties of an electrical signal. In addition to the
properties labelled on the graph, there is frequency
which is the number of cycles per second.
The diagram shows a sine wave but these properties apply to any signal with a constant shape.





Amplitude is the maximum voltage reached by the signal.
It is measured in volts, V.
Peak voltage is another name for amplitude.
Peak-peak voltage is twice the peak voltage (amplitude). When reading an oscilloscope trace it is
usual to measure peak-peak voltage.
Time period is the time taken for the signal to complete one cycle.
It is measured in seconds (s), but time periods tend to be short so milliseconds (ms) and
microseconds (µs) are often used. 1ms = 0.001s and 1µs = 0.000001s.
Frequency is the number of cycles per second.
It is measured in hertz (Hz), but frequencies tend to be high so kilohertz (kHz) and megahertz
(MHz) are often used. 1kHz = 1000Hz and 1MHz = 1000000Hz.
frequency =

1
time period
and
time period =
1
frequency
Mains electricity in the Palestine has a frequency of 50Hz,
so it has a time period of 1/50 = 0.02s = 20ms
1.4 Root Mean Square (RMS) Values
The value of an AC voltage is continually changing from zero up to the
positive peak, through zero to the negative peak and back to zero again.
Clearly for most of the time it is less than the peak voltage, so this is not
a good measure of its real effect.
Instead we use the root mean square voltage (VRMS) which is 0.7 of
the peak voltage (Vpeak):
VRMS = 0.7 × Vpeak and Vpeak = 1.4 × VRMS
(1.1)
These equations also apply to current.
They are only true for sine waves (the most common type of AC) because the 0.7 and 1.4 are different
values for other shapes.
The RMS value is the effective value of a varying voltage or current. It is the equivalent steady DC
(constant) value which gives the same effect.
5
For example a lamp connected to a 6V RMS AC supply will light with the same brightness when
connected to a steady 6V DC supply. However, the lamp will be dimmer if connected to a 6V peak AC
supply because the RMS value of this is only 4.2V (it is equivalent to a steady 4.2V DC).
You may find it helps to think of the RMS value as a sort of average, but please remember that it is NOT
really the average! In fact the average voltage (or current) of an AC signal is zero because the positive
and negative parts exactly cancel out!
What do AC meters show, is it the RMS or peak voltage?
AC voltmeters and ammeters show the RMS value of the voltage or current. DC meters also show the
RMS value when connected to varying DC providing the DC is varying quickly, if the frequency is less
than about 10Hz you will see the meter reading fluctuating instead.
What does '6V AC' really mean, is it the RMS or peak voltage?
If the peak value is meant it should be clearly stated, otherwise assume it is the RMS value. In everyday
use AC voltages (and currents) are always given as RMS values because this allows a sensible
comparison to be made with steady DC voltages (and currents), such as from a battery.
For example a '6V AC supply' means 6V RMS, the peak voltage is 8.6V. The Palestinian mains supply is
220V AC, this means 220V RMS so the peak voltage of the mains is about 311V.
1.5 Insulators and Conductors
An Insulator is a material that resists the flow of electric current. It is an object intended to support or
separate electrical conductors without passing current through itself. An insulation material has atoms
with tightly bonded valence electrons. The term electrical insulation has the same meaning as the term
dielectric.
Some materials such as silicon dioxide or teflon are very good electrical insulators. A much larger class
of materials, for example rubber-like polymers and most plastics are still "good enough" to insulate
electrical wiring and cables even though they may have lower bulk resistivity. These materials can serve
as practical and safe insulators for low to moderate voltages (hundreds, or even thousands, of volts
Electrical conductor: In science and engineering, conductors, such as copper or aluminum, are materials
wAll conductors contain movable electric charges which will move when an electric potential difference
(measured in volts) is applied across separate points on a wire (etc) made from the material. This flow of
charge (measured in amperes) is what is meant by electric current. In most materials, the amount of
current is proportional to the voltage (Ohm's law) provided the temperature remains constant and the
material remains in the same shape and state. The ratio between the voltage and the current is called the
resistance (measured in ohms) of the object between the points where the voltage was applied. The
resistance across a standard mass (and shape) of a material at a given temperature is called the resistivity
of the material. The inverse of resistance and resistivity is conductance and conductivityith atoms having
loosely held valence electrons. See electrical conduction.
Since all conductors have some resistance, and all insulators will carry some current, there is no
theoretical dividing line between conductors and insulators. However, there is a large gap between the
conductance of materials that will carry a useful current at working voltages and those that will carry a
negligible current for the purpose in hand, so the categories of insulator and conductor do have practical
utility.
Thermal and electrical conductivity often go together (for instance, most metals are both electrical and
thermal conductors). However, some materials are practical electrical conductors without being a good
thermal conductor.
6
Conductor size
In many countries, conductors are measured by their cross section in square millimeters.
However, in the United States, conductors are measured by American wire gauge for smaller ones, and
circular mils for larger ones.
Conductor materials
Of the metals commonly used for conductors, copper, has a high conductivity. Silver is more conductive,
but due to cost it is not practical in most cases. However, it is used in specialized equipment, such as
satellites, and as a thin plating to mitigate skin effect losses at high frequencies. Because of its ease of
connection by soldering or clamping, copper is still the most common choice for most light-gauge wires.
Conductor voltage
The voltage on a conductor is determined by the connected circuitry and has nothing to do with the
conductor itself. Conductors are usually surrounded by and/or supported by insulators and the insulation
determines the maximum voltage that can be applied to any given conductor
1.6 Electrical Resistance
Electrical resistance is a measure of the degree to which an object opposes an electric current through it.
Its reciprocal quantity is electrical conductance (provided the electrical impedence is real) measured in
siemens. Assuming a uniform current density, an object's electrical resistance is a function of both its
physical geometry and the resistivity of the material it is made from:
R
l
(1.2)
A
where
" " is the length
"A" is the cross sectional area, and
"ρ" is the resistivity of the material
Electrical resistance shares some conceptual parallels with the mechanical notion of friction. The SI unit
of electrical resistance is the ohm, symbol Ω.
The electrical resistivity ρ (rho) of a material is given by

Where :
RA
l
(1.3)
ρ is the static resistivity (measured in ohm metres, Ω-m);
R is the electrical resistance of a uniform specimen of the material (measured in ohms, Ω);
is the length of the piece of material (measured in metres, m);
A is the cross-sectional area of the specimen (measured in square metres, m²).
Electrical resistivity can also be defined as

E
J
(1.4)
7
Where :
E is the magnitude of the electric field (measured in volts per metre, V/m);
J is the magnitude of the current density (measured in amperes per square metre, A/m²).
Electrical resistivity is also defined as the inverse of the conductivity σ (sigma), of the material, or
1
(1.5)


Table (1): Resistivity of Famous Conducting and Insulating Materials
Material
Silver
Resistivity (Ω-m) at 20 °C
1.59×10−8
Coefficient
0.0038
Copper
1.72×10−8
0.0039
2.44×10
−8
0.0034
2.82×10−8
0.0039
−8
0.0045
Gold
Aluminum
Tungsten
Nickel
Brass
5.60×10
6.99×10−8
?
−7
0.0015
−7
0.8×10
Iron
1.0×10
0.005
Tin
1.09×10−7
0.0045
Platinum
Lead
−7
1.1×10
0.00392
2.2×10−7
0.0039
−7
Manganin
4.82×10
0.000002
Constantan
4.9×10−7
0.00001
Mercury
Nichrome
Carbon
Germanium
Silicon
Glass
Hard rubber
Sulfur
Quartz (fused)
Teflon
−7
9.8×10
0.0009
−6
1.10×10
0.0004
3.5×10−5
-0.0005
−1
-0.048
6.40×102
-0.075
4.6×10
10
14
?
approx. 1013
?
10 to 10
15
10
?
17
7.5×10
?
1022 to 1024
?
Example 1.1 :What is the value of the resistance of a copper wire with length 250m and cross section
area equals 2.5 mm²
Solution :
Plug into the formula and using the value of the resistivity of the copper from the table :
R
l
A
1.72 x 10-8 x 250
R
 1.72 
2.5 x 10 6
8
1.7 Ohm's law
Ohm's law states that in an electrical circuit, the current passing through a conductor between two points
is directly proportional to the potential difference (i.e. voltage drop or voltage) across the two points, and
inversely proportional to the resistance between them, and for the same temperature.
The mathematical equation that describes this relationship is:
I
V
R
(1.6)
where I is the current in amperes, V is the potential difference between two points of interest in volts, and
R is a circuit parameter, measured in ohms (which is equivalent to volts per ampere), and is called the
resistance. The potential difference is also known as the voltage drop, and is sometimes denoted by U, E
or emf (electromotive force) instead of V.
Ohm’s law defines the relationship between current, voltage, and resistance. There are three ways to
express Ohm’s law mathematically.
1. The current in a circuit is equal to the voltage applied to the circuit divided by the resistance of the
circuit:
I
V
R
(1.7)
2. The resistance of a circuit is equal to the voltage applied to the circuit divided by the current in the
circuit:
R
V
I
(1.8)
3. The applied voltage to a circuit is equal to the product of the current and the resistance of the circuit:
V  I. R
(1.9)
where I = current, A
R = resistance, 
V = voltage, V
Example 1.2 :What is the value of the resistance if the current is 18 mA and the voltage is 229 mV?
Solution :
First, convert these values to amperes and volts. This gives I = 0.018 A and E = 0.229 V. Then plug into
the equation R= E/I = 0.229/0.018 = 13  . You’re justified in
giving your answer to two significant figures, because the current is only given to that many digits.
Example 1.3 :Suppose the ammeter reads 52 A and the voltmeter indicates 2.33 kV. What is the
resistance?
Solution :
Convert to amperes and volts, getting I = 0.000052 A and E = 2330 V. Then plug into the formula:
R = 2330/0.000052 = 45,000,000 = 45 M 
Example 1.4 :Find I when V = 120V and R = 30 
Solution :
Using Ohm's Law
I
V
R
9
I
120
4 A
30
Example 1.5 :Find V when I = 120nA and R = 25 k
Solution :
Convert to amperes and ohms, getting 120 x109 A and R = 25 x 103  . Then plug into the formula:
V  I. R
V 120 x109 x 25 x103  3x103  3 mV
Example 1.6 :Find V when I = 85A and R = 6.5 k
Solution :
Plug into the formula:
V  I. R
V  85 x 6.5x103  552500  552.5 kV
Example 1.7 :Find I when V = 22kV and R = 440 
Solution :
Using Ohm's Law Eq. (1.7)
V
R
22 x 103
I
 50 A
440
I
1.8 Electrical Power
The electric power P used in any part of a circuit is equal to the current I in that part multiplied by the
voltage V across that part of the circuit. Its formula is:
P = VI
(1.10)
where P = power, W
V = voltage, V
I = current, A
Other forms for P = VI are I = P/V and V = P/I.
If we know the current I and the resistance R but not the voltage V, we can find the power P by using
Ohm’s law for voltage, so that substituting V = IR into (1.10) we have
PI2 R
(1.11)
In the same manner, if we know the voltage V and the resistance R but not the current I, we can find the
power P by using Ohm’s law for current, so that substituting I 
P
V2
R
V
into (1.10) we have
R
(1.12)
Example 1.8 : The current through a 100Ω resistor to be used in a circuit is 0.20 A. Find the power rating
of the resistor.
Solution :
Since I and R are known, use Eq. (1.11) to find P.
P  I 2 R  0.22 x100  4W
Example 1.9 If the voltage across a 25000 Ω resistor is 500 V, what is the power dissipated in the
resistor?
Solution :
10
Since R and V are known, use Eq. (1.12) to find P.
V2
5002
P

10W
R
25000
1.9 Electrical energy
Electrical energy in any part of the circuit is that power of that part multiplied of the time of consumption,
and it can be found from the formula:
Electrical energy U = Power x time
(1.13)
U = VIt Joules
(1.14)
Although the unit of energy is the joule, when dealing with large amounts of energy, the unit used is the
kilowatt hour (kWh) where
1 kWh = 1000 watt hour
= 1000 x 3600 watt seconds or joules
= 3 600 000 J
Example 1.10 A source e.m.f. of 5 V supplies a current of 3 A for 10 minutes. How much energy is
provided in this time?
Solution :
Energy = power x time,
and power = voltage x current. Hence
U = VIt = 5 x 3 x (10 x 60) = 9000 Ws or J
= 9 kJ
Example 1.11: An electric heater consumes 1.8 MJ when connected to a 250 V supply for 30 minutes.
Find the power rating of the heater and the current taken from the supply.
Solution :
Power rating of heater = P 
Thus I 
U 1.8 x106

1000W 1 kW
t
30 x 60
P 1000

4 A
V
250
Example 1.12: An electric bulb has a power 100 W. If it works 10 hours per a day. Find the energy
consumed in 1 month.
Solution :
Since energy = power x time
Then, Energy U = P x t = 100 x 10 x 30 = 30000 Wh
= 30 kWh
11
1.10 Series and Parallel Networks
Series circuits: The Figure shows three resistors
R1, R2 and R3 connected end to end, i.e., in series,
with a battery source of V volts. Since the circuit
is closed a current I will flow and the p.d. across
each resistor may be determined from the
voltmeter readings V1, V2 and V3
In a series circuit
(a) the current I is the same in all parts of the circuit and hence the same reading is found on each of the
two ammeters shown, and
(b) the sum of the voltages V1, V2 and V3 is equal to the total applied voltage, V, i.e.
From Ohm’s law:
where R is the total circuit resistance.
Since V = V1 + V2 + V3
then IR = IR1 + IR2 + IR3
Dividing throughout by I gives
(1.15)
Thus for a series circuit, the total resistance is obtained by adding together the values of the separate
resistances.
Example 1.13: For the circuit shown in the illustrated figure, determine
(a) the battery voltage V,
(b) the total resistance of the circuit, and
(c) the values of resistance of resistors R1, R2 and R3, given
that the p.d.’s across R1, R2 and R3 are 5 V, 2 V and 6 V respectively.
Solution :
Voltage divider The voltage distribution for the circuit shown
in the Figure is given by:
(1.16)
(1.17)
12
Example 1.14: Two resistors are connected in series
across a 24 V supply and a current of 3 A flows in the
circuit. If one of the resistors has a resistance of 2 Ω
determine: (a) the value of the other resistor, and
(b) the volt. difference. across the 2 Ω resistor.
Solution :
(a) Total circuit resistance
Value of unknown resistance,
(b) Volt. Difference across 2 Ω resistor, V1 = IR1 = 3 x 2 = 6 V
Alternatively, from above,
Parallel networks The Figure shows three resistors,
R1, R2 and R3 connected across each other, i.e.,
in parallel, across a battery source of V volts.
In a parallel circuit:
(a) the sum of the currents I1, I2 and I3 is equal to the total circuit
current, I, i.e. I = I1 + I2 + I3, and
(b) the source p.d., V volts, is the same across each of the resistors.
From Ohm’s law:
where R is the total circuit resistance.
Since I = I1 + I2 + I3
then,
Dividing throughout by V gives:
(1.18)
This equation must be used when finding the total resistance R of a parallel circuit. For the special case of
two resistors in parallel
(1.19)
Example 1.15: For the circuit shown in the Figure, find
(a) the value of the supply voltage V
(b) the value of current I.
Solution :
(a) V.d. across 20 Ω resistor = I2R2 = 3 x 20 = 60 V, hence
supply voltage V = 60 V since the circuit is connected in parallel.
13
Current I = I1 + I2 + I3 and hence I = 6 + 3 + 1 = 10 A
Another solution for part (b):
Current division: For the circuit shown in the Figure,
the total circuit resistance, RT is given by:
Summarizing, with reference to the above Figure
(1.20)
(1.21)
Example 1.16: For the circuit shown in the Figure, find the current Ix
Solution :
Commencing at the right-hand side of the arrangement shown in the Figure below, the circuit is gradually
reduced in stages as shown in Figure (a)–(d).
14
From Figure (d)
From Figure (b)
From the above Figure
15
UNIT 2
DC Laws and Theorems
Introduction The laws which determine the currents and voltage drops in d.c. networks are:
(a) Ohm’s law,
(b) the laws for resistors in series and in parallel ,
(c) Kirchhoff’s laws
(d) the superposition theorem
(e) Th´evenin’s theorem
(f) Norton’s theorem
2.1 Kirchhoff’s laws
Kirchhoff’s laws state:
(a) Current Law. At any junction in an electric circuit the total
current flowing towards that junction is equal to the total current
flowing away from the junction, i.e. ∑I = 0
Thus, referring to Figure 2.1:
I1 + I2 = I3 + I4 + I5 or I1 + I2 - I3 - I4 - I5 = 0
(b) Voltage Law. In any closed loop in a network, the
algebraic sum of the voltage drops (i.e. products of
current and resistance) taken around the loop is equal
to the resultant e.m.f. acting in that loop.
Thus, referring to Figure 2.2:
E1 - E2 = IR1 + IR2 + IR3
(Note that if current flows away from the positive
terminal of a source, that source is considered by
convention to be positive. Thus moving anticlockwise
around the loop of Figure 2.2, E1 is positive and E2 is negative.)
Example 2. 1. (a) Find the unknown currents marked in Figure 2.3(a).
(b) Determine the value of e.m.f. E in Figure 2.3(b).
Figure 2.3
Figure 2.2
Solution:
(a) Applying Kirchhoff’s current law:
For junction B: 50 = 20 + I1. Hence I1 = 30 A
For junction C: 20 + 15 = I2. Hence I2 = 35 A
For junction D: I1 = I3 + 120
i.e. 30 = I3 + 120. Hence I3 = −90 A
(i.e. in the opposite direction to that shown in Figure 2.3(a))
For junction E: I4 + I3 = 15
i.e. I4 = 15 – (-90). Hence I4 = 105 A
For junction F: 120 = I5 + 40. Hence I5 = 80 A
16
Figure 2.1
Figure 2.2
(b) Applying Kirchhoff’s voltage law and moving clockwise around the loop of Figure 2.3(b) starting at
point A:
3 + 6 + E - 4 = (I)(2) + (I)(2.5) + (I)(1.5) + (I)(1)
= I(2 + 2.5 + 1.5 + 1)
i.e.
5 + E = 2(7), since I = 2 A
Hence E = 14 - 5 = 9 V
Example 2. 2. Use Kirchhoff’s laws to determine the currents flowing in each branch of the network
shown in Figure 2.4.
Figure 2.5
Figure 2.4
Solution:
Procedure
1. Use Kirchhoff’s current law and label current directions on the original circuit diagram. The directions
chosen are arbitrary, but it is usual, as a starting point, to assume that current flows from the positive
terminals of the batteries. This is shown in Figure 2.5 where the three branch currents are expressed in
terms of I1 and I2 only, since the current through R is I1 + I2.
2. Divide the circuit into two loops and apply Kirchhoff’s voltage law to each. From loop 1 of Figure 2.5,
and moving in a clockwise direction as indicated (the direction chosen does not matter), gives
(1)
From loop 2 of Figure 2.5, and moving in an anticlockwise direction as indicated (once again, the choice
of direction does not matter; it does not have to be in the same direction as that chosen for the first loop),
gives:
(2)
3. Solve equations (1) and (2) for I1 and I2.
(3)
(4)
(i.e. I2 is flowing in the opposite direction to that shown in Figure 2.5.)
From (1)
Hence
Current flowing through resistance R is
17
Note that a third loop is possible, as shown in Figure 2.6,
giving a third equation which can be used as a check:
Figure 2.6
Example 2. 3. Determine, using Kirchhoff’s laws, each
branch current for the network shown in Figure 2.7.
Solution:
1. Currents, and their directions are shown labelled in
Figure 2.8 following Kirchhoff’s current law. It is usual,
although not essential, to follow conventional current flow
with current flowing from the positive terminal of the source.
Figure 2.7
2. The network is divided into two loops as shown in
Figure 2.8.
Applying Kirchhoff’s voltage law gives:
For loop 1:
(1)
For loop 2:
Figure 2.8
Note that since loop 2 is in the opposite direction to current
the volt drop across
is by convention negative).
Thus
(2)
3. Solving equations (1) and (2) to find I1 and I2:
(3)
(2) + (3) gives :
From (1):
Current flowing in
18
Example 2. 4. For the bridge network shown in Figure 2.9
determine the currents in each of the resistors.
Solution:
Let the current in the 2Ω resistor be I1, then by Kirchhoff’s
current law, the current in the 14 Ω resistor is (I - I1). Let the
current in the 32 Ω resistor be I2 as shown in Figure 2.10.
Then the current in the 11 resistor is (I1 - I2) and that in the
3 Ω resistor is (I - I1 + I2). Applying Kirchhoff’s voltage law
to loop 1 and moving in a clockwise direction as shown in Figure 2.10 gives:
Figure 2.9
(1)
Applying Kirchhoff’s voltage law to loop 2 and moving in an
anticlockwise direction as shown in Figure 2.10 gives:
However
Figure 2.10
Hence
(2)
Equations (1) and (2) are simultaneous equations with two unknowns, I1 and I2.
(3)
(4)
(4) – (3) gives:
Substituting for I2 in (1) gives:
Hence,
the current flowing in the 2 Ω resistor = I1 = 5 A
the current flowing in the 14 Ω resistor = I - I1 = 8 - 5 = 3 A
the current flowing in the 32 Ω resistor = I2 = 1 A
the current flowing in the 11 Ω resistor = I1 - I2 = 5 - 1 = 4 A and
the current flowing in the 3 Ω resistor = I - I1 + I2 = 8 - 5 + 1 = 4 A
19
2.2 The Superposition Theorem
The superposition theorem states:
‘In any network made up of linear resistances and containing more than one source of e.m.f., the
resultant current flowing in any branch is the algebraic sum of the currents that would flow in that
branch if each source was considered separately, all other sources being replaced at that time by their
respective internal resistances.’
Example 2. 5. Figure 2.11 shows a circuit containing two sources
of e.m.f., each with their internal resistance. Determine the current
in each branch of the network by using the superposition theorem.
Solution Procedure:
1. Redraw the original circuit with source E2 removed, being
replaced by r2 only, as shown in Figure 2.12(a).
2. Label the currents in each branch and their directions as shown in
Figure 2.12(a) and determine their values. (Note that the choice of
current directions depends on the battery polarity, which, by convention
is taken as flowing from the positive battery terminal as shown.)
Figure 2.11
R in parallel with r2 gives an equivalent resistance of:
From the equivalent circuit of Figure 2.12(b)
From Figure 2.12(a)
and
Figure 2.12
3. Redraw the original circuit with source E1 removed, being replaced
by r1 only, as shown in Figure 2.13(a).
4. Label the currents in each branch and their directions as shown in
Figure 2.13(a) and determine their values.
r1 in parallel with R gives an equivalent resistance of:
From the equivalent circuit of Figure 2.13(b)
From Figure 2.13(a)
5. Superimpose Figure 2.13(a) on to Figure 2.12(a) as shown in Figure 2.14.
20
Figure 2.13
6. Determine the algebraic sum of the currents flowing in each branch.
Resultant current flowing through source 1, i.e.
Resultant current flowing through source 2, i.e.
Figure 2.14
Resultant current flowing through resistor R, i.e.
The resultant currents with their directions are
shown in Figure 13.15.
Figure 2.15
2.3 Th´evenin’s Theorem
Th´evenin’s theorem states:
‘The current in any branch of a network is that which would result if an e.m.f. equal to the p.d. across
a break made in the branch, were introduced into the branch, all other e.m.f.’s being removed and
represented by the internal resistances of the sources.’
The procedure adopted when using Th´evenin’s theorem is summarized below.
To determine the current in any branch of an active network (i.e. one containing a source of e.m.f.):
(i) remove the resistance R from that branch,
(ii) determine the open-circuit voltage, E, across the break,
(iii) remove each source of e.m.f. and replace them by their internal
resistances and then determine the resistance, r, ‘looking-in’ at the break,
(iv) determine the value of the current from the equivalent circuit shown
in Figure 2.16, i.e.
Figure 2.16
Example 2. 6. Use Th´evenin’s theorem to find the current flowing in the 10 Ω resistor for the circuit
shown in Figure 2.17
Figure 2.17
21
Solution:
Following the above procedure:
(i) The 10Ω resistance is removed from the circuit as shown in
Figure 2.18
(ii) There is no current flowing in the 5 Ω resistor and
current I1 is given by:
Figure 2.18
P.d. across R2 = I1R2 = 1 x 8 = 8 V
Hence p.d. across AB, i.e. the open-circuit voltage across the break,
E = 8 V.
(iii) Removing the source of e.m.f. gives the circuit of Figure 2.19.
Resistance
(iv) The equivalent Th´evenin’s circuit is shown in Figure 2.20
Current
Figure 2.19
Hence the current flowing in the 10 resistor of Figure 2.17 is 0.482 A
Figure 2.20
Example 2.7. A Wheatstone Bridge network is shown in
Figure 2.21(a). Calculate the current flowing in the 32Ω resistor,
and its direction, using Th´evenin’s theorem. Assume the source of
e.m.f. to have negligible resistance.
Figure 2.21
22
Solution:
Following the procedure:
(i) The 32Ω resistor is removed from the circuit as shown in
Figure 2.21(b)
(ii) The p.d. between A and C,
The p.d. between B and C,
Hence the p.d. between A and B = 44.47 - 8.31 = 36.16 V
Point C is at a potential of +54 V. Between C and A is a voltage drop of 8.31 V.
Hence the voltage at point A is 54 - 8.31 = 45.69 V.
Between C and B is a voltage drop of 44.47 V.
Hence the voltage at point B is 54 - 44.47 = 9.53 V.
Since the voltage at A is greater than at B, current must flow in the direction A to B.
(iii) Replacing the source of e.m.f. with a short-circuit (i.e. zero internal resistance) gives the circuit
shown in Figure 2.21(c). The circuit is redrawn and simplified as shown in Figure 2.21(d) and (e), from
which the resistance between terminals A and B,
(iv) The equivalent Th´evenin’s circuit is shown in Figure 13.32(f), from which,
Current
Hence the current in the 32 Ω resistor of Figure 2.21(a) is 1 A, flowing from A to B
Example 2.8. Use Th´evenin’s theorem to determine the current
flowing in the 3 Ω resistance of the network shown in
Figure 2.22. The voltage source has negligible internal
resistance.
Solution:
Following the procedure:
(i) The 3 Ω resistance is removed from the circuit as shown in
Figure 2.23
(ii) The Ω resistance now carries no current.
Figure 2.22
= 16V
Hence p.d. across AB, E = 16 V
Figure 2.23
23
(iii) Removing the source of e.m.f. and replacing it by its
internal resistance means that the 20 Ω resistance is
short-circuited as shown in Figure 2.24 since its internal
resistance is zero. The 20 Ω resistance may thus be removed
as shown in Figure 2.25.
From Figure 2.25 Th´evenin’s resistance,
Figure 2.24
(iv) The equivalent Th´evenin’s circuit is shown in Figure 2.26
From which
Figure 2.25
Figure 2.26
2.4 Norton’s Theorem
Norton’s theorem states:
‘The current that flows in any branch of a network is the same as that which would flow in the branch
if it were connected across a source of electrical energy, the short-circuit current of which is equal to
the current that would flow in a short-circuit across the branch, and the internal resistance of which is
equal to the resistance which appears across the open-circuited branch terminals.’
The procedure adopted when using Norton’s theorem is summarized below.
To determine the current flowing in a resistance R of a branch AB of an active network:
(i) short-circuit branch AB
(ii) determine the short-circuit current ISC flowing in the branch
(iii) remove all sources of e.m.f. and replace them by their internal
resistance (or, if a current source exists, replace with an open circuit),
then determine the resistance r,‘looking-in’ at a break made
between A and B
(iv) determine the current I flowing in resistance R from the Norton
equivalent network shown in Figure 13.33, i.e.
Figure 2.26
Example 2.9. Use Norton’s theorem to determine the current flowing in the 10 Ω resistance for the
circuit shown in Figure 2.27
24
Figure 2.27
Solution:
Following the procedure:
(i) The branch containing the 10 Ω resistance is short-circuited as
shown in Figure 2.28
(ii) Figure 2.29 is equivalent to Figure 2.28. Hence
Figure 2.28
(iii) If the 10 V source of e.m.f. is removed from Figure 2.29 the
resistance ‘looking-in’ at a break made between A and B is given by:
(iv) From the Norton equivalent network shown in Figure 2.30 the
current in the 10 Ω resistance, by current division, is given by:
Figure 2.29
as obtained previously in Example 2.6 using Th´evenin’s theorem.
Figure 2.30
Example 2.10. Use Norton’s theorem to determine the current flowing in the 3 resistance of the network
shown in Figure 2.31(a). The voltage source has negligible internal resistance.
Figure 2.31
25
Solution:
Following the procedure:
(i) The branch containing the 3 resistance is short-circuited as shown in Figure 2.31(b).
(ii) From the equivalent circuit shown in Figure 2.31(c),
(iii) If the 24 V source of e.m.f. is removed the resistance ‘looking-in’ at a break made between A and B
is obtained from Figure 2.31(d) and its equivalent circuit shown in Figure 2.31(e) and is given by:
(iv) From the Norton equivalent network shown in Figure 2.31(f) the current in the 3 Ω resistance is
given by:
as obtained previously in Example 2.8 using Th´evenin’s theorem.
Extra Example:
Example 2.11. Use Kirchhoff's Laws to find the current in each resistor for the circuit in Figure 2.32
Figure 2.32
Solution:
1. Currents, and their directions are shown labelled in Figure 2.33 following Kirchhoff’s current law.
Figure 2.33
Applying Kirchhoff’s current law for node A gives
I 4  I1  I 3
and for node B gives
I5  I 2  I3
2. The network is divided into three loops as shown in Figure 2.33.
Applying Kirchhoff’s voltage law for loop 1 gives:
then
15  50I1  40 I 4 ,
15  50 I1  40 ( I1  I 3 )
26
15  90 I1  40 I 3
divide by 5 : 3  18I1  8 I 3
simplify :
Applying Kirchhoff’s voltage law for loop 2 gives:
then
10  20 I 2  30 I 5 ,
10  20 I 2  30 ( I 2  I 3 )
simplify : 10  50 I 2  30 I 3
divide by 10 : 1  5I 2  3 I 3
Applying Kirchhoff’s voltage law for loop 3 gives:
0  25I 3  30 I 5  40 I 4 ,
then 0  25I 3  30( I 2  I 3 )  40 ( I1  I 3 )
simplify : 0   40 I1  30 I 2  95 I 3
divide by 5 : 0   8I1  6 I 2  19 I 3
Arrange the three equations in matrix form :
 18
 0


 8
0
5
6
 8  I1  3
 I   1 
3
  2  
19

0

 I3 
 
18 0
 0
5
8 6
8
18  8
18 0
3 5
3
 5 x 278  3x108 1066
 8 19
8 6
19
3 0
1  1
5
0 6
8
5
3 3
6
19
18
2  0
8
3
1
0
8
1
3 18
0
19
18
3  0
8
0
5
6
3
0
1 3
8
0
8
 3x77  1x 48 183
19
3
0
1
19
6
3
3 8
8
18 x19  8 x17  206
19
1
3
5
18
1
6
8
1
183

 0.172 A
 1066

206
I2  2 
 0.193 A
 1066

12
I3  3 
 0.011 A
 1066
I 4  0.172  0.011  0.161 A,
0
 3x 40  1x18 x6 12
6
I1 
I 5  0.193  0.011  0.182 A
27
UNIT 3
Single Phase AC
Introduction: In this unit inductors' and capacitor's reactances, AC circuit reactance, electrical power,
power factor, and power triangle will be explained.
3.1 Inductors
Inductance: The ability of a conductor to induce voltage in itself when the current changes is its selfinductance, or simply inductance. The symbol for inductance is L, and its unit is the henry (H). One
henry is the amount of inductance that permits one volt to be induced when the current
changes at the rate of one ampere per second. The formula for inductance is:
L
vL
i / t
(3.1)
where L = inductance, H
vL
= induced voltage across the coil, V
= rate of change of current, A/s
The self-induced voltage
vL
i / t
from Eq. (3.1)
vL  L
i
t
(3.2)
Physical Characteristics:
A coil's inductance depends on how it is wound, the core material on which it is wound, and the number
of turns of wire with which it is wound.
1. Inductance L increases as the number of turns of wire N around the core increases. Inductance
increases as the square of the turns increases. For example, if the number of turns is doubled ( 2 x ),
inductance increases
2 2 or ( 4 x ), assuming the area and length of the coil remain constant .
2. Inductance increases as the relative permeability r , of the core material increases.
3. As the area A enclosed by each turn increases, the inductance increases. Since the area is a function of
the square of the diameter of the coil, inductance increases as the square of the diameter.
4. Inductance decreases as the length of the coil increases (assuming the number of turns remains
constant) .
Inductive Reactance:
Inductive reactance XL is the opposition to ac current due to the inductance in the circuit. The unit of
inductive reactance is the ohm. The formula for inductive reactance is:
X L  2fL
(3.3)
where, XL = inductive reactance, Ω
f = frequency, Hz
L = inductance, H
Example 3.1: A choke coil of negligible resistance is to limit the current through it to 50 mA when 25 V
is applied across it at 400 kHz. Find its inductance.
Solution : Find XL by Ohm's law and then find L.
VL
25

 500
I L 50 x10  3
XL
500
L

 0.199 x10 3 H  0.2 mH
3
2f
2 xx 400 x10
XL 
28
Inductors in Series and Parallel
If inductors are spaced sufficiently far apart so that they do not interact electromagnetically with each
other, their values can be combined just like resistors when connected together. If a number of inductors
are connected in series (Fig. 3.1), the total inductance LT is the sum of the individual inductances, or
Series:
LT  L1  L2  L3  .....  Ln
(3.4)
Figure 3.1
Series:
LT  L1  L2  L3  .....  Ln
(3.4)
If a number of inductors are connected in parallel (Fig. 3.2), their total inductance LT is
Parallel:
1
1
1
1
1



 ..... 
L T L1 1 L 2 L 3
Ln
(3.5)
Figure 3.2
3.2 Capacitors
A capacitor is an electrical device which consists of two conducting plates of metal separated by an
insulating material called a dielectric (Fig. 3.3a). Schematic symbols shown (Fig. 3.3b and c) apply to all
capacitors.
Figure 3.3
Electrically, capacitance is the ability to store an electric charge. Capacitance is equal to the amount of
charge that can be stored in a capacitor divided by the voltage applied across the plates
C
Q
V
(3.6)
where C = capacitance, F
Q = amount of charge, C
29
V = voltage, V
The unit of capacitance is the farad (F). The farad is that capacitance that will store one coulomb of
charge in the dielectric when the voltage applied across the capacitor terminals is one volt.
Types of Capacitors
Commercial capacitors are named according to their dielectric. Most common are air, mica, paper, and
ceramic capacitors, plus the electrolytic type. These types are compared in Table 3-1. Most types of
capacitors can be connected to an electric circuit without regard to polarity. But electrolytic capacitors
and certain ceramic capacitors are marked to show which side must be connected to the more positive
side of a circuit.
Table 3.1
Dielectric
Construction
Capacitance Range
Air
Meshed plates
10-400 pF
Mica
Stacked sheets
10-5000 pF
Paper
Rolled foil
0.001-1 µF
Ceramic
Tubular
0.5-1600 pF
Disk
0.002-0.1 µ F
Electrolytic
Aluminum
5-1000 µ F
Tantalum
0.01-300 µ F
Capacitors in Series and Parallel
When capacitors are connected in series (Fig. 3-4), the total capacitance CT is
Series:
1
1
1
1
1



 ..... 
C T C1 C 2 C 3
Cn
(3.7)
Figure 3.4
When capacitors are connected in parallel (Fig. 3-5), the total capacitance CT is the sum of the individual
capacitances.
Parallel:
CT  C1  C2  C3  .....  Cn
(3.8)
Figure 3.5
Capacitive Reactance:
Capacitive reactance Xc is the opposition to the flow of ac current due to the capacitance in the circuit.
The unit of capacitive reactance is the ohm. Capacitive reactance can be found by using the equation:
XC 
1
2fC
(3.9)
where Xc = capacitive reactance, Ω
f = frequency, Hz
C = capacitance, F
30
Example 3.2 A 120 Hz, 25 mA ac current flows in a circuit containing a 10 µF capacitor. What is the
voltage drop across the capacitor?.
Solution : Find Xc and then Vc by Ohm's law.
XC 
1
1

 132.5 
2fC 2 xx120 x10 x10 6
VC  X C I C  132.5 x 25 x 103  3.31V
3.3 Series a.c. circuits
(a) Pure resistance
In an a.c. circuit containing resistance R only (Figure 3.6(a)), the current IR is in phase with the applied
voltage VR as shown in the phasor diagram of Figure 3.6(b). The phasor diagram may be superimposed
on the Argand diagram as shown in Figure 3.6(c). The impedance Z of the circuit is given by:
VR 0
Z
R
I R 0
(3.10)
Figure 3.6
(b) Pure inductance
In an a.c. circuit containing pure inductance L only (Figure 3.7(b)the current IL lags the applied voltage
VL by 90° as shown in the phasor diagram of Figure 3.7(b). The phasor diagram may be superimposed on
the Argand diagram as shown in Figure 3.7(c). The impedance Z of the circuit is given by :
VL 90
Z
 X L 90  jX L

I L 0
Figure 3.7
31
(3.11)
(c) Pure capacitance
In an a.c. circuit containing pure capacitance only (Figure 3.8(a)), the current IC leads the applied voltage
VC by 90° as shown in the phasor diagram of Figure 3.8(b). The phasor diagram may be superimposed on
the Argand diagram as shown in Figure 3.8(c). The impedance Z of the circuit is given by:
VC 0
Z
 X C   90   jX C

I C 90
(3.12)
Figure 3.8
(d) R–L series circuit
Figure 3.8
In an a.c. circuit containing resistance R and inductance L in series (Figure 3.9(a)), the applied voltage V
is the phasor sum of VR and VL as shown in the phasor diagram of Figure 3.9(b). The current I lags the
applied voltage V by an angle lying between 0° and 90°—the actual value depending on the values of VR
and VL, which depend on the values of R and L. The circuit phase angle, i.e., the angle between the
current and the applied voltage, is shown as angle  in the phasor diagram. In any series circuit the
current is common to all components and is thus taken as the reference phasor in Figure 3.9(b). The
phasor diagram may be superimposed on the Argand diagram as shown in Figure 3.9(c), where it may be
seen that in complex form the supply voltage V is given by:
V VR  jVL
(3.13)
Figure 3.9
Figure 3.10(a) shows the voltage triangle that is derived from the phasor diagram of Figure 3.10(b) (i.e.
triangle Oab). If each side of the voltage triangle is divided by current I then the impedance triangle of
Figure 3.10(b) is derived. The impedance triangle may be superimposed on the Argand diagram, as
shown in Figure 3.10(c), where it may be seen that in complex form the impedance Z is given by:
Z  R  jX L
(3.14)
Thus, for example, an impedance expressed as (3 + j4) Ω means that the resistance is 3 Ω and the
inductive reactance XL is 4 Ω.
32
Figure 3.10
In polar form,
Z 
Z  Z 
R 2  X L2
where, from the impedance triangle, the modulus of impedance
and the circuit phase angle
  tan 1
XL
R
lagging.
(e) R–C series circuit
In an a.c. circuit containing resistance R and capacitance C in series (Figure 3.11(a)), the applied voltage
V is the phasor sum of VR and VC as shown in the phasor diagram of Figure 3.11 (b). The current I leads
the applied voltage V by an angle lying between 0° and 90°—the actual value depending on the values of
VR and VC, which depend on the values of R and C. The circuit phase angle is shown as angle in the
phasor diagram. The phasor diagram may be superimposed on the Argand diagram as shown in Figure
3.11 (c), where it may be seen that in complex form the supply voltage V is given by:
V  VR  jVC
(3.15)
Figure 3.11
Figure 3.12a) shows the voltage triangle that is derived from the phasor diagram of Figure 3.12(b). If each
side of the voltage triangle is divided by current I, the impedance triangle is derived as shown in Figure
3.12(b). The impedance triangle may be superimposed on the Argand diagram as shown in Figure 3.12(c),
where it may be seen that in complex form the impedance Z is given by:
Z  R  jX C
(3.16)
Thus, for example, an impedance expressed as (9 - j14) Ω means that the resistance is 9 Ω and the
capacitive reactance XC is 14 Ω .In polar form,
Z 
R2  X C2
Z  Z 
and the circuit phase angle
33
  tan 1
where, from the impedance triangle,
XC
R
leading.
Figure 3.12
(f) R–L–C series circuit
In an a.c. circuit containing resistance R, inductance L and capacitance C in series (Figure 3.13(a)), the
applied voltage V is the phasor sum of VR, VL and VC as shown in the phasor diagram of Figure 3.13(b)
(where the condition VL > VC is shown). The phasor diagram may be superimposed on the Argand
diagram as shown in Figure 3.13(c), where it may be seen that in complex form the supply voltage V is
given by:
V  VR  j (VL  VC )
(3.17)
From the voltage triangle the impedance triangle is derived and superimposing this on the Argand
diagram gives, in complex form,
impedance
where,
Z  R  j ( X L  X C ) or Z  Z 
Z 
R2  ( X L  X C )2
and the circuit phase angle
(3.18)
  tan 1
X L  XC
R
When VL = VC, XL = XC and the applied voltage V and the current I are in phase. This effect is called
series resonance.
Figure 3.13
(g) General series circuit
In an a.c. circuit containing several impedances connected in series, say, Z1, Z2, Z3, . . ., Zn, then the total
equivalent impedance ZT is given by:
ZT  Z1  Z 2  Z3  .....  Z n
(3.19)
Example 3.3: A 200 V, 50 Hz supply is connected across a coil of negligible resistance and inductance
0.15 H connected in series with a 32 Ω resistor (Figure (3.14)). Determine
(a) the impedance of the circuit,
(b) the current and circuit phase angle,
(c) the p.d. across the 32 Ω resistor, and (d) the p.d. across the coil.
34
Figure 3.14
Figure 3.15
Solution :
X L  2fL  2xx50x.015  47.1
Z  R  jX L  (32  j 47.1)  or 57 55.81 
(a) Inductive reactance
Impedance
(b) Current
I
V
2000

 3.51   55.81 A
Z 5755.81
i.e., the current is 3.51 A lagging the voltage by 55.81°
VR  IR  (3.51  55.81)(32 0)
VR 112.3  55.81V
i.e.
(d) P.d. across the coil, VL  IX L  (3.51  55.81)(47.190)
VL 165.3 34.19V
i.e.
(c) P.d. across the 32Ω resistor,
The phasor sum of VR and VL is the supply voltage V as shown in the phasor diagram of Figure 3.15.
VR 112.3   55.81  (63.11  j92.89)V
VL 165.3 34.19  (136.73  j92.89)V
Hence V  VR  VL  (63.11  j92.89)  (136.73  j92.89)V
 (200  j 0)V or 2000V
Example 3.4: A The impedance of an electrical circuit is (30 - j50) ohms. Determine
(a) the resistance,
(b) the capacitance,
(c) the modulus of the impedance, and
(d) the current flowing and its phase angle, when the circuit is connected to a 240 V, 50 Hz supply.
Solution :
(a) Since impedance Z = (30 - j50), the resistance is 30 Ω and the capacitive reactance is 50 Ω
(b) Since X C 
C
1
, then
2fC
1
1

 63.66 F
2fX C 2 xx50 x50
R2  X C2  302  502  58.31
1 X C
 58.31  59.04 
(d) Impedance Z  (30  j 50)   58.31  tan
R
V
2400

 4.12   59.04 A
Hence current I 
Z 58.31  59.04
(c) The modulus of impedance
Z 
35
Example 3.5: For the circuit shown in Figure 3.16, determine the value of impedance Z2
Figure 3.16
Solution :
V
7030

 20   50   (12.86  j15.32)
I 3.5  20
Total impedance Z  Z1  Z 2
from which, impedance Z 2  Z  Z1  (12.86  j15.32)  (4.36  j 2.10)
 (8.50  j17.42)  19.3863.99
Total circuit impedance Z 
Example 3.6: A coil of resistance R ohms and inductance
L henrys is connected in series with a 50 μF capacitor (Fig 3.17).
If the supply voltage is 225 V at 50 Hz and the current flowing in the
circuit is 1.5  30° A, determine the values of R and L. Determine
also the voltage across the coil and the voltage across the capacitor.
Solution :
Circuit impedance
V
2250

150 30   (129.9  j 75.0)
I 1.5  30
1
1

 63.66 
Capacitive reactance X C 
2fC 2 xx50 x50 x10 6
Z
Circuit impedance Z = R + j(XL - XC)
i.e.
Figure 3.17
129.9 + j75.0 = R + j(XL – 63.66)
Equating the real parts gives: resistance R = 129.9 Ω
Equating the imaginary parts gives: 75.0 = XL - 63.66,
from which, XL = 75.0 + 63.66 = 138.66 Ω
Since
X L  2fL , inductance L 
XL
138.66

 0.441 H
2f 2 xx50
The circuit diagram is shown in Figure 3.17.
Voltage across coil, VCOIL = I ZCOIL
ZCOIL = R + jXL = (129.9 + j138.66) or 190  46.87° Ω
Hence VCOIL = (1.5  -30°)(190  46.87°)
= 285∟ 16.87° V or (272.74 + j 82.71)V
Voltage across capacitor, VC = I XC = (1.56  -30°)(63.66  -90°)
= 95.49  −120° V or (− 47.75 − j 82.70)V
36
Example 3.7: For the circuit shown in Figure 3.18, determine
The values of voltages V1 and V2 if the supply frequency is
4 kHz. Determine also the value of the supply voltage V and
the circuit phase angle. Draw the phasor diagram.
Solution :
For impedance Z1
XC 

1
2fC
1
15 
2 xx 4000 x 2.653 x10 6
Hence Z1 =(8 - j15) or 17  -61.93° Ω
and voltage V1 = I Z1 = (6  0°)(17  -61.93°)
= 102  −61.93° V or (48 − j 90)V
For impedance Z2,
Figure 3.18
X L  2fL  2 x  x 4000 x 0.477 x103 12
Hence Z2 = (5 + j12) Ω or 13  67.38° Ω
and voltage V2 =I Z2 = (6  0°)(13  67.38°)
= 78  67.38° V or (30 + j 72)V
Supply voltage, V = V1 + V2 = (48 - j90) + (30 + j72)
= (78 − j 18)V or 80  −13° V
Circuit phase angle,  = 13° leading.
The phasor diagram is shown in Figure 3.19.
Figure 3.19
3.4 Parallel a.c. circuits
As with series circuits, parallel networks may be analyzed by using phasor diagrams. However, with
parallel networks containing more than two branches this can become very complicated. It is with parallel
a.c. network analysis in particular that the full benefit of using complex numbers may be appreciated.
Before analyzing such networks admittance, conductance and susceptance are defined.
Admittance, conductance and susceptance
Admittance is defined as the current I flowing in an a.c. circuit divided by the supply voltage V (i.e. it is
the reciprocal of impedance Z). The symbol for admittance is Y. Thus
Y
I
1

V
Z
(3.20)
The unit of admittance is the Siemen, S.
An impedance may be resolved into a real part R and an imaginary part X, giving Z = R ± jX. Similarly,
an admittance may be resolved into two parts—the real part being called the conductance G, and the
imaginary part being called the susceptance B—and expressed in complex form.
Y  G  jB
(3.21)
(a) R–L parallel circuit
For parallel circuits with R and XL (Fig. 3.20a), the same applied voltage VT is across R and XL since
both are in parallel with VT. There is no phase difference between these voltages. Therefore, VT will be
used as the reference phasor. The resistive branch current IR = VT/R is in phase with VT The inductive
branch current IL = VT/XL lags VT by 90° (Fig. 3.20b) because the current in an inductance lags the
voltage across it by 90°. The phasor sum of IR and IL equals the total line current IT (Fig. 3.20c), or
37
IT 
IT 
I R2  I L2
(
IT  V
(3.22)
V 2
V 2
) (
)
R
XL
(
1 2
1 2
) (
)
R
XL
IT V
G 2  BL2
L
tan    L
IR
L
  tan 1 (  L )
IR
R
  tan 1 (
)
XL
(3.23)
(3.24)
Figure 3.20
(b) R–C parallel circuit
For parallel circuits with R and XC (Fig. 3.21a), the same applied voltage VT is across R and XC since
both are in parallel with VT. There is no phase difference between these voltages. Therefore, VT will be
used as the reference phasor. The resistive branch current IR = VT/R is in phase with VT The capacitive
branch current IC = VT/XC leads VT by 90° (Fig. 3.21b) because the current in an capacitance leads the
voltage across it by 90°. The phasor sum of IR and IC equals the total line current IT (Fig. 3.21c), or
IT 
IT 
I R2  I C2
(
IT  V
I T V
tan  
(3.25)
V 2
V 2
) (
)
R
XC
(
1 2
1 2
) (
)
R
XC
G 2  BC2
LC
IR
(3.26)
LC
)
IR
R
  tan 1 (
)
XC
  tan 1 (
(3.27)
38
Figure 3.21
(c) R–L–C parallel circuit
A three-branch parallel ac circuit (Fig. 3.22a) has resistance in one branch, inductance in the second
branch, and capacitance in the third branch. The voltage is the same across each parallel branch, so VT =
VR = VL = VC. The applied voltage VT is used as the reference line to measure phase angle θ. The total
current IT is the phasor sum of IR, IL, and IC. The current in the resistance IR is in phase with the applied
voltage VT (Fig. 3.22b). The current in the inductance IL lags the voltage VT by 90". The current in the
capacitor Ic leads the voltage VT by 90°. IL and IC are exactly 180° out of phase and thus acting in
opposite directions (Fig. 3.22b). When IL > IC, IT lags VT (Fig. 3.22c) so the parallel RLC circuit is
considered inductive.
Figure 3.22
If IC > IL, the current relationships and phasor triangle (Fig. 3.23) show that IT now leads VT so this type
of parallel RLC circuit is considered capacitive.
Figure 3.23
39
When IL > IC, or XL < XC , circuit is inductive and
IT 
IT 
I R2  ( I L  I C ) 2
(
(3.28)
V 2
V
V 2
) (

)
R
XL
XC
1 2
1
1 2
) (

)
R
XL
XC
I T V
(
IT V
(G) 2  ( BL  BC ) 2
LL  I C
)
IR
B B
  tan 1 ( L C )
G
  tan 1 (
(3.29)
(3.30)
When IC > IL, or XC < XL , the circuit is capacitive and
IT 
I R2  ( I C  I L ) 2
IT  (
V 2
V
V 2
) (

)
R
XC
XL
(3.31)
1 2
1
1 2
) (

)
R
XC
XL
I T V
(
IT V
(G) 2  ( BC  BL ) 2
LC  I L
)
IR
B  BL
   tan 1 ( C
)
G
   tan 1 (
(3.32)
(3.33)
Example 3.8: A 500 Ω R is in parallel with 300 Ω XL. Find IT, θ, and ZT . Assume VT = 500V
Solution:
VT
500

1 A
R
500
V
500
 T 
 1.67 A
XL
300
IR 
IL
IT 
I R2  I L2  12  1.67 2  1.95 A
  tan 1 (
ZT 
R
500
)  tan 1 
  59.1
XL
300
VT
500

 256.459.1 
IT
1.95  59.1
Example 3.9: A 15Ω resistor and a capacitor of 20Ω capacitive reactance are placed in parallel across a
120-V ac line. Calculate IR, IC, IT, θ, and Z. Draw the phasor diagram
Solution:
IR 
VT
120

8 A
R
15
40
VT
120

6 A
XC
20
IC 
IT 
I R2  I L2 
R
15
)  tan 1
 36.9
XC
20
  tan 1 (
ZT 
82  62  10 A
VT
120

 12  36.9 
IT
1036.9
Figure 3.24
Figure 3.24 shows the phasor diagram.
Example 3.10: A 400 Ω resistor, a 50 Ω inductive reactance, and a 40 Ω capacitive reactance are placed
in parallel across a 120-V ac line (Fig. 3.25a). Find the phasor branch currents, total current, phase angle,
and impedance. Draw the phasor diagram.
Solution:
(a)
(b)
Figure 3.25
Find IR, IL, IC
VT
120

 0.3 A
R
400
V
120
IL  T 
 2 .4 A
XL
50
V
120
IC  T 
3 A
XC
40
IR 
Find IT, θ. Since XL > XC (50Ω > 40Ω) or IC > lL (3.0A > 2.4A), the circuit is capacitive
IT  I R2  ( IC  I L )2 
  tan 1 (
ZT 
0.32  (3  2.4)2  0.671A
LC  I L
3  2.4
)  tan 1
 tan 1 2  63.43
IR
0.3
VT
120

 178.84  63.43 
IT
0.67163.43
Figure 3.25(b) shows the phasor diagram.
41
Example 3.11: For the circuit shown in Figure 3.26, determine
(a) The current I
(b) the voltage across the coil, and
(c) the voltage across the capacitor
Figure 3.26
Solution :
The coil is in series with the 20Ω resistor, and their impedance Z1 is:
Z1 = (20 + j40) Ω
The capacitor is in series with the 5Ω resistor, and their impedance Z2 is:
Z2 = (5 – j15) Ω
Z1 is in parallel with Z2 and their equivalent Z3 can be found by:
(20  j 40)(5  j15) 100  j 200  j 300  600 700  j100



20  j 40  5  j15
25  j 25
25  j 25
28  j 4 1  j1 28  j 4  j 28  4

x

 (12  j16)
1  j1 1  j1
11
Z3 
ZT =14 + 12 – j16 + 4 = (30 – j16) Ω = 34  -28.07° Ω
Therefore the total current I can be found by:
I
V
136
136
30  j16 4080  j 2176


x

Z 30  j16 30  j16 30  j16
900  256
 (3.53  j1.88)
It can be directly found from:
I
V
1360

 428.07 A
Z 3428.07
This current is divided into the branch of the coil and the branch of the capacitor, and the current in each
branch can be found using current division law:
I COIL  I
Z2
5  j15
 (3.53  j1.88) x
Z1  Z 2
20  j 40  5  j15

17.65  j9.40  j52.95  28.20 45.85  j 43.55 25  j 25

x
25  j 25
25  j 25
25  j 25

57.5  j 2235
 (0.046  j1.788) A
1250
I CAPACITOR  I
Z1
20  j 40
 (3.53  j1.88) x
Z1  Z 2
20  j 40  5  j15
42

70.6  j37.6  j141.2  75.2  4.6  j178.8 25  j 25

x
25  j 25
25  j 25
25  j 25

4355  j 4585
 (3.484  j 3.668) A
1250
VCOIL = ICOIL x (j40)
= (0.046 – j1.788)(j40) = (71.52 + j1.84) V or = 71.54  1.47° V
VCAPACITOR = ICAPACITOR x(–j15)
= (3.484  j 3.668) (–j15) = (55.02 – j52.26) V or = 75.88  –43.52° V
Example 3.12: For the circuit shown in Figure 3.27, determine the current I1, the current I2, the current IT,
and the equivalent impedance ZT.
Figure 3.27
Solution :
In the first branch the coil 87mH is in series with the 42Ω resistor, and their impedance is Z1
In the second branch the capacitor 10µF is in series with the 65Ω resistor and the coil 72mH, and their
impedance is Z2
Z1 is in parallel with Z2 and their equivalent is ZT
X L1  2fL1  2x50 x87 x103  27.33 
X L 2  2fL2  2x50 x72 x10 3  22.62 
XC 
1
1

 318.31
2fC 2x50 x10 x106
Z1 = (42 + j27.33) Ω = 50.11  33.05° Ω
Z2 = (65 + j22.62 – j318.31) Ω = (65 – j295.69) Ω = 302.75  –77.6° Ω
I1 
V
2200

 4.39  33.05 A
Z 1 50.1133.05
or I1  (3.68  j 2.39) A
I2 
V
2200

 0.7377.6 A
Z 2 302.75  77.6
43
or I 2  (0.157  j 0.713) A
then, IT  I1  I 2  (3.68  j 2.39)  (0.157  j0.713)  (3.837  j1.677) A
or IT  4.1923.61 A
ZT 
V
2200

 52.51  23.61 
I T 4.1923.61
Example 3.12: For the circuit shown in Figure 3.28, determine (a) the value of the impedance Z1, (b) If
the supply frequency is 5 kHz, determine the value of the components comprising impedance Z1
Figure 3.28
Solution :
(a) Total circuit admittance,
YT 
I 31.452.48

 0.62822.48 S  (0.58  j 0.24) S
V
5030
Y T = Y 1 + Y 2 + Y3
Thus 0.58  j 0.24  Y1 
1
1

8  j 6 10
0.58  j 0.24  Y1 
1
8  j6
x
 0.1
8  j6 8  j6
0.58  j 0.24  Y1 
8  j6
 0.1
82  6 2
0.58  j 0.24  Y1 
8  j6
 0.1
100
0.58  j0.24  Y1  0.08  j0.06  0.1
Y1  0.58  j0.24  0.08  j0.06  0.1 ( 0.4  j0.3)S or  0.536.87 S
Z1 
1
1

 2  36.87   (1.6  j1.2) 
Y1 0.536.87
(b) Since Z1 = (1.6 – j1.2) resistance = 1.6 Ω and capacitive reactance, XC = 1.2 Ω
1
1
1

 26.53 F
Since X C 
, then C 
2fX C 2 xx5000 x1.2
2fC
44
3.5 Power in a.c. circuits
Alternating currents and voltages change their polarity during each cycle. It is not surprising therefore to
find that power also pulsates with time. The product of voltage v and current i at any instant of time is
called instantaneous power p, and is given by:
p = vi
(3.34)
For purely resistive circuits the current and the voltage are in the same phase, and considering RMS
values of current and voltage then the average power is given by:
V2
2
P V I  I R 
watts
(3.35)
R
For purely inductive circuits the current lags the voltage by angle 90°, so the average power is zero. Also,
for purely capacitive circuits the current leads the voltage by angle 90°, so the average power is zero.
For circuits containing resistors and inductors, the current lags the voltage by angle Φ which is between
0° and 90°, and also for circuits containing resistors and capacitors, the current leads the voltage by angle
Φ which is between 0° and 90°, so the average power is given by:
P  V I cos  watts
(3.36)
A phasor diagram in which the current I lags the applied voltage V by angle  (i.e., an inductive circuit)
is shown in Figure 3.29(a). The horizontal component of V is (V cos  ), and the vertical component of V
is (V sin  ). If each of the voltage phasors of triangle Oab is multiplied by I, Figure 3.29(b) is produced
and is known as the ‘power triangle’. Each side of the triangle represents a particular type of power:
Figure 3.29
True or active power P = VI cos  watts (W)
Apparent power S = VI voltamperes (VA)
Reactive power Q = VI sin  (VAR)
The power triangle is not a phasor diagram since quantities P, Q and S are mean values and not rms
values of sinusoidally varying quantities.
Superimposing the power triangle on an Argand diagram produces a relationship between P, S and Q in
complex form, i.e.,
S = P + jQ
(3.37)
Apparent power, S, is an important quantity since a.c. apparatus, such as generators, transformers and
cables, is usually rated in voltamperes rather than in watts. The allowable output of such apparatus is
usually limited not by mechanical stress but by temperature rise, and hence by the losses in the device.
The losses are determined by the voltage and current and are almost independent of the power factor.
Thus the amount of electrical equipment installed to supply a certain load is essentially determined by the
voltamperes of the load rather than by the power alone. The rating of a machine is defined as the
maximum apparent power that it is designed to carry continuously without overheating.
The reactive power, Q, contributes nothing to the net energy transfer and yet it causes just as much
loading of the equipment as if it did so. Reactive power is a term much used in power generation,
distribution and utilization of electrical energy. Inductive reactive power, by convention, is defined as
positive reactive power; capacitive reactive power, by convention, is defined as negative reactive power.
The above relationships derived from the phasor diagram of an inductive circuit may be shown to be true
for a capacitive circuit, the power triangle being as shown in Figure 3.30.
45
Figure 3.30
Power factor is defined as:
Power Factor 
active power P
reactive power Q
(3.38)
V I cos 
 cos 
(3.39)
VI
A circuit in which current lags voltage (i.e., an inductive circuit) is said to have a lagging power factor,
and indicates a lagging reactive power Q.
A circuit in which current leads voltage (i.e., a capacitive circuit) is said to have a leading power factor,
and indicates a leading reactive power Q.
Power Factor 
Example 3.13: For the circuit shown in Figure 3.31, determine the active power developed between
points (a) A and B, (b) C and D, (c) E and F.
Figure 3.31
Solution :
Circuit impedance,
(3  j 4)(  j10)
 j 30  40
40  j 30 3  j 6
5 
5
x
3  j 4  j10
3  j6
3  j6
3  j6
120  j 90  j 240  180
300  j150
5
5 
9  36
45
 (11.67  j 3.33) or  12.1315.94
ZT  5 
I 
V
1000

 8.24  15.94 A
Z T 12.1315.94
(a) Active power developed between points A and B = I²R = (8.24)² (25) = 339.5 W
(b) Active power developed between points C and D is zero, since no power is developed in a pure
capacitor.
(c) The current I1 is calculated first to calculate the power between E and F.
46
I1  I
Z CD
 j10
10  90
 8.24  15.94
 8.24  15.94
12.28  42.51 A
Z CD  Z EF
3  j6
6.71  63.43
Hence the active power developed between points E and F  I12 R  (12.28) 2 (3)  452.4 W
Example 3.14: The circuit shown in Figure 3.32 dissipates
an active power of 400 W, and has a power factor of 0.766
lagging. Determine (a) the apparent power, (b) the reactive
power, (c) the value and phase of current I, and
(d) the value of impedance Z.
Figure 3.32
Solution :
Since power factor = 0.766 lagging, the circuit phase angle   cos 1 0.766 , i.e.,  = 40° lagging which
means that the current I lags voltage V by 40°.
(a) Since power, P = VI cos  , the magnitude of apparent power,
P
400

 522.2 VA
cos  0.766
(b) Reactive power Q = S sin  = 522.2 sin 40° = 335.66 VAR
S
Since the power factor is lagging the reactive power is inductive
i.e. positive value. Figure 3.33 shows the power triangle.
(c) Since VI = 522.2 VA,
522.2
  5.222 A
100
Figure 3.33
Since the voltage is at a phase angle of 30° (Figure 3.32) and current lags voltage by 40°, the phase
magnitude of current I 
angle of current is 30° – 40° = –10°.
Hence current I = 5.222  −10° A
(d) Total circuit impedance Z T 
or
V
10030

 19.1540 
I 5.222  10
 (14.67  j12.31) 
Hence impedance Z = ZT – 4 = (14.67 + j12.31) – 4
= (10.67 + j 12.31) Ω
or = 16.29  49.08° Ω
Example 3.15: For the circuit of example 3.12 find the active power, reactive power, and apparent power
that dissipated by the circuit, then draw the power triangle.
Solution:
In the solution of the example the total current of the circuit is found that:
IT  4.1923.61 A
Then the value of active power P can be found:
47
P = VI cos Φ = (220)(4.19) cos(23.61°) = 844.64 W
Also, the value of reactive power Q can be found:
Q = VI sin Φ = (220)(4.19) sin(23.61°) = 369.19 VAR
The value of apparent power S can be found:
S = P + jQ = 844.64 + j369.19 = 921.8  23.61° VA
Or it can be found by
S = V I = (220) (4.19) = 921.8 VA
The power triangle is illustrated in figure 3.34.
Figure 3.34
Power Factor Improvement
For a particular active power supplied, a high power factor reduces the current flowing in a supply system
and therefore reduces the cost of cables, transformers, switchgear and generators. Supply authorities use
tariffs which encourage consumers to operate at a reasonably high power factor. One method of
improving the power factor of an inductive load is to connect a bank of capacitors in parallel with the
load. Capacitors are rated in reactive voltamperes and the effect of the capacitors is to reduce the reactive
power of the system without changing the active power. Most residential and industrial loads on a power
system are inductive, i.e. they operate at a lagging power factor.
A simplified circuit diagram is shown in Figure 3.35(a) where a capacitor C is connected across an
inductive load. Before the capacitor is connected the circuit current is ILR and is shown lagging voltage V
by angle 1 in the phasor diagram of Figure 3.35 (b). When the capacitor C is connected it takes a current
IC which is shown in the phasor diagram leading voltage V by 90°. The supply current I in Figure 3.35(a)
is now the phasor sum of currents ILR and IC as shown in Figure 3.35(b). The circuit phase angle, i.e., the
angle between V and I, has been reduced from 1 to 2 and the power factor has been improved from
cos 1 to cos 2 .
Figure 3.35(a) shows the power triangle for an inductive circuit with a lagging power factor of cos 1 . In
Figure 3.35(b), the angle 1 has been reduced to 2 , i.e., the power factor has been improved from cos 1
to cos 2 by introducing leading reactive voltamperes (shown as length ab) which is achieved by
connecting capacitance in parallel with the inductive load. The power factor has been improved by
reducing the reactive voltamperes; the active power P has remained unaffected.
Power factor correction results in the apparent power S decreasing (from 0a to 0b in Figure 3.35(b)) and
thus the current decreasing, so that the power distribution system is used more efficiently.
Another method of power factor improvement, besides the use of static capacitors, is by using
synchronous motors; such machines can be made to operate at leading power factors.
Figure 3.35
48
Example 3.16: A circuit has an impedance Z = (3 + j4) Ω and a source p.d. of 50  30° V at a frequency
of 1.5 kHz. Determine
(a) the supply current,
(b) the active, apparent and reactive power,
(c) the rating of a capacitor to be connected in parallel with impedance Z to improve the power factor of
the circuit to 0.966 lagging, and
(d) the value of capacitance needed to improve the power factor to 0.966 lagging.
Solution:
V
5030
5030
(a) Supply current, I  Z  3  j 4  553.13  10  23.13 A
(b) Apparent power, S = V I = (50  30°) (10  23.13°) = 500  53.13° VA
= (300 + j400) VA = P + jQ
Hence active power, P = 300 W
apparent power, S = 500 VA and
reactive power, Q = 400 VAR lagging.
The power triangle is shown in Figure 3.36.
(c) A power factor of 0.966 means that cos  = 0.966.
Figure 3.36
Hence angle  = arccos 0.966 = 15° .
To improve the power factor from cos 53.13°, i.e. 0.60, to 0.966, the power triangle will need to change
from 0cb (Figure 3.37) to 0ab, the length (ca) representing the rating of a capacitor connected in parallel
with the circuit. From Figure 3.37, tan 15° = ab/300, from which, ab = 300 tan 15° = 80.38 VAR.
Hence the rating of the capacitor, ca = cb – ab
= 400 – 80.38
= 319.6 VAR leading.
Q 319.6
(d) Current in capacitor I C  V  50  6.39 A
V
50
Capacitive reactance, X C  I  6.39  7.82 
C
1
Thus 7.82  2fC
, from which,
1
required capacitance C  2 (1500)(7.82)  13.57 F
49
Figure 3.37
UNIT 4
Three Phase AC System
Introduction
This unit introduces the three phase AC connection, the relation between line and phase voltage or
current, and the star and delta connections and their relations.
4.1 Characteristics of Three Phase Systems
A three-phase (3-Φ) system is a combination of three single-phase (1-Φ) systems. In a 3-Φ balanced
system, the power comes from an ac generator that produces three separate but equal voltages, each of
which is out of phase with the other voltages by 120° (Figure. 4.1). Although 1-Φ circuits are widely used
in electrical systems, most generation and distribution of alternating current is 3-Φ. Three-phase circuits
require less weight of conductors than 1-Φ circuits of the same power rating; they permit flexibility in the
choice of voltages; and they can be used for single-phase loads. Also, 3-Φ equipment is smaller in size,
lighter in weight, and more efficient than 1-Φ machinery of the same rated capacity. A three-phase a.c.
supply is carried by three conductors, called ‘lines’ which are colored red, yellow and blue. The currents
in these conductors are known as line currents (IL) and the v.d.’s between them are known as line voltages
(VL). A fourth conductor, called the neutral (colored black, and connected through protective devices to
earth) is often used with a three-phase supply. The phase-sequence is given by the sequence in which the
conductors pass the point initially taken by the red conductor. The national standard phase sequence is R,
Y, B. To reduce the number of wires it is usual to interconnect the three phases. The three phases of a 3-Φ
system may be connected in two ways. If the three common ends of each phase are connected together at
a common terminal marked N for neutral, and the other three ends are connected to the 3-Φ line, the
system is wye- or Y-connected (Figure. 4.2a). If the three phases are connected in series to form a closed
loop, the system is delta- or Δ-connected (Figure. 4.2b).
(a)
(b)
Figure 4.1
(a)
(b)
Figure 4.2
50
4.2 Star Connection
(i) A star-connected load is shown in Figure 4.3 where the three line conductors are each connected to a
load and the outlets from the loads are joined together at N to form what is termed the neutral point or the
star point.
(ii) The voltages, VR , VY , and VB are called phase voltages or line to neutral voltages. Phase voltages are
generally denoted by VP
(iii) The voltages,
VRY , VYB and VBR are called line voltages and generally denoted by VL .
(iv) From Figure 4.3 it can be seen that the phase currents (generally denoted by I P ) are equal to their
respective line currents I R , I Y and I B , i.e. for a star connection:
IL  IP
(4.1)
VR = VY = VB
For a balanced system: I R = I Y = I B
VRY = VYB = VBR
ZR
IN  0
=
ZY
=
ZB
and the current in the neutral conductor,
When a star connected system is balanced, then the neutral conductor is unnecessary and is often omitted.
VRY , shown in Figure 4.4(a) is given by VRY = VR − VY ( VY is negative since
it is in the opposite direction to VRY ). In the phasor diagram of Figure 4.4(b), phasor VY is reversed
(vi) The line voltage,
(shown by the broken line) and then added phasorially to VR
trigonometry, or by measurement,
(i.e.
VRY
= VR
VRY = 3 V R , i.e. for a balanced star connection:
VL 
3 VP
(4.2)
Figure 4.3
Figure 4.4
51
+ (− VY )). By
(vii) The star connection of the three phases of a supply, together with a neutral conductor, allows the use
of two voltages—the phase voltage and the line voltage. A 4-wire system is also used when the load is not
balanced. The standard electricity supply to consumers in Palestine is 380/220 V, 50 Hz, 3-phase, 4-wire
alternating current, and a diagram of connections is shown in Figure 4.5.
Figure 4.5
Example 4.1:
A star-connected load consists of three identical coils each of resistance 30Ω and inductance 127.3 mH.
If the line current is 5.08 A, calculate the line voltage if the supply frequency is 50 Hz.
Solution:
Inductive reactance
X L  2fL  2 (50)(127.3 x 103 )  40 
Impedance of each phase
For a star connection
Hence phase voltage
Line voltage
VL 
ZP 
IL  IP
R 2  X L2 
V
 P
ZP
302  402  50 
VP  I P . Z P  (5.08)(50)  254V
3 VP 
3 ( 254)  440 V
Example 4.2:
A balanced, three-wire, star-connected, 3-phase load has a phase voltage of 240 V, a line current of 5 A
and a lagging power factor of 0.966. Draw the complete phasor diagram.
Solution:
The phasor diagram is shown in Figure 4.6
Procedure to construct the phasor diagram:
(i) Draw VR = VY = VB = 240 V and spaced 120° apart. (Note that VR is shown vertically upwards—
this however is immaterial for it may be drawn in any direction.)
(ii) Power factor = cos Φ = 0.966 lagging. Hence the load phase angle is given by arccos 0.966, i.e. 15°
lagging. Hence I R = I Y = I B = 5 A, lagging VR ,
VY
and VB respectively by 15°.
VRY = VR - VY (phasorially). Hence VY is reversed and added phasorially to VR . By measurement,
VRY = 415 V (i.e. 3 (254) and leads VR by 30°. Similarly, VYB = VY - VB and VBR = VB - VR .
(iii)
52
Figure 4.6
Example 4.3:
A 415 V, 3-phase, 4 wire, star-connected system supplies three resistive loads as shown in Figure 4.7.
Determine (a) the current in each line and (b) the current in the neutral conductor.
Solution:
For a star-connected system
VL 
3 VP
VL
415

 240V
3
3
Power P
for a resistive load
Since current I P 
VoltageV
PR 24000

 100 A
Then I R 
VR
240
Hence VP 
PY 18000

 75 A
VY
240
P
12000
Figure 4.7
 50 A
and I B  B 
VB
240
(b) The three line currents are shown in the phasor diagram of Figure 4.8. Since each load is resistive the
currents are in phase with the phase voltages and are hence mutually displaced by 120°.
The current in the neutral conductor is given by:
I N = I R + I Y + I B phasorially
IY 
Figure 4.8
53
Figure 4.9
Figure 4.9 shows the three line currents added phasorially. oa represents I R in magnitude and direction.
From the nose of oa, ab is drawn representing I Y in magnitude and direction. From the nose of ab, bc is
drawn representing I B in magnitude and direction. oc represents the resultant, I N .
By measurement, I N  43 A
Alternatively, by calculation, considering I R at 90°, I B at 210° and I Y at 330°:
Total horizontal component = 100 cos 90° + 75 cos 330° + 50 cos 210° = 21.65
Total vertical component = 100 sin 90° + 75 sin 330° + 50 sin 210°
Hence magnitude of I N 
= 37.50
21.65 2  37.50 2  43.3 A
4.3 Delta connection
(i) A delta (or mesh) connected load is shown in Figure 4.10. Where the end of one load is connected to
the start of the next load.
(ii) From Figure 4.10, it can be seen that the line voltages
voltages, i.e. for a delta connection
VRY , VYB and VBR are the respective phase
VL  VP
(4.3)
(iii) Using Kirchhoff’s current law in Figure 4.10, I R = I RY - I BR = I RY + ( I BR ) .
From the phasor diagram shown in Figure 4.11, by trigonometry or by measurement, I R = 3 I RY ,
i.e. for a delta connection:
IL 
3 IP
(4.4)
Figure 4.10
Figure 4.11
Example 4.4:
Three identical capacitors are connected in delta to a 415 V, 50 Hz, 3-phase supply. If the line current is
15 A, determine the capacitance of each of the capacitors.
Solution:
For a delta connection
Hence phase current
IL 
3 IP
I
15
IP  L 
 8.66 A
3
3
Capacitive reactance per phase, X C 
Hence X C 
VP VL

(since in delta connection VL  VP )
IP
IP
415
 47.92
8.66
from which capacitance, C 
1
1

 66.43F
2fX C 2 (50)( 47.92)
54
Example 4.5:
Three coils each having resistance 3Ω and inductive reactance 4Ω are connected (i) in star and (ii) in
delta to a 381 V, 3-phase supply. Calculate for each connection (a) the line and phase voltages and (b) the
phase and line currents.
Solution:
(i)For a star connection:
IL  IP
and VL 
3 VP
(a) A 381 V, 3-phase supply means that the line voltage, VL  381V
Phase voltage, VP 
VL
381

 220 V
3
3
(b) Impedance per phase, Z P 
Phase current, I P 
R 2  X L2 
32  42  5 
VP
220

 44 A
ZP
5
Line current, I L  I P  44 A
(ii) For a delta connection:
VL  VP
and
IL 
3 IP
(a) Line voltage, VL  381V
Phase voltage VP  VL  381V
(b) Phase current, I P 
Line current, I L 
VP
381

 76.2 A
ZP
5
3 IP 
3 (76.2)  132 A
4.4 Power in Three-Phase Systems
The power dissipated in a three-phase load is given by the sum of the power dissipated in each phase. If a
load is balanced then the total power P is given by: P = 3 x power consumed by one phase.
The power consumed in one phase =
I P2 RP  VP I Pcos  (where 
is the phase angle between
VP and I P ).
For a star connection,
IL  IP
and VP 
VL
3
V 
P  3 L  I L cos  
 3
For a delta connection,
VP  VL
and I P 
 I 
P  3VL  L  cos  
 3
hence
3 VL LL cos 
IL
3
(4.5)
hence
3 VL LL cos 
(4.6)
Hence for either a star or a delta balanced connection the total power P is given by:
P
or
Total volt-amperes,
Total reactive power,
3 VL LL cos 
P  3VP LP cos 
S
Q
3 VL LL
3 VL LL sin 
55
(4.7)
(4.8)
(4.9)
(4.10)
Example 4.6:
The input power to a 3-phase a.c. motor is measured as 5 kW. If the voltage and current to the motor are
400 V and 8.6 A respectively, determine the power factor of the system.
Solution:
Power, P = 5000 W; Line voltage
Power,
VL
= 400 V; Line current,
IL
= 8.6 A
3 VL LL cos 
P
Hence power factor = cos  
P

3 VL LL
5000
 0.839
3 (400)(8.6)
Example 4.7:
Three identical coils, each of resistance 10Ω and inductance 42 mH are connected (a) in star and (b) in
delta to a 415 V, 50 Hz, 3-phase supply. Determine the total power dissipated in each case.
Solution:
(a) Star connection
X L  2fL  2 (50)( 42 x 103 ) 13.19 
Inductive reactance
Phase impedance
Line voltage
VL
R 2  X L2  102  13.192  16.55 
= 415 V and phase voltage,
VP 
VL
415

 240V
3
3
VP
240

 14.50 A
ZP
16.55
IP 
Phase current,
Line current,
ZP 
I L  I P  14.50 A
Power factor = cos  
Power dissipated,
P
3 VL LL cos 
3 (415)(14.50)(0.6042)  6.3kW
=
(Alternatively,
RP
10

 0.6042 lagging
Z P 16.55
P  3I P2 RP  3(14.502 )(10)  6.3kW
(b) Delta connection
VL  VP  415V
Z P  16.55  , and cos   0.6042 lagging
V
415
IP  P 
 25.08 A
ZP
16.55
Phase current,
Line current,
,
IL 
Power dissipated,
3 IP 
P
=
(Alternatively,
(from above)
3 ( 25.08)  43.44 A
3 VL LL cos 
3 (415)( 43.44)(0.6042)  18.87kW
P  3I P2 RP  3(25.082 )(10)  18.87kW
Hence loads connected in delta dissipate three times the power than when connected in star, and also
take a line current three times greater.
56
Example 4.8:
A 381 V, 3-phase a.c. motor has a power output of 12.75 kW and operates at a power factor of 0.77
lagging and with an efficiency of 85%. If the motor is delta-connected, determine
(a) the power input, (b) the line current and (c) the phase current.
Solution:
Output Power
85
12750
hence

Input Power
100 Input Power
(12750)(85)
from which, power input =
= 15000 = 15 kW
100
(a) Efficiency =
(b) Power,
P  3 VL LL cos 
line current, LL 
P
15000

 29.52 A
3 VL cos 
3 (381)(0.77)
(c) For a delta connection,
IL 
hence phase current, I P 
3 IP
IL
29.52

 17.04 A
3
3
57
UNIT 5
Electrical Transformers
Introduction
This unit introduces the construction and principle of operation for single phase and three phases
transformers, star and delta connections for three phases transformers.
5.1 Transformer's principle of operation
A transformer is a device which uses the phenomenon of mutual induction to change the values of
alternating voltages and currents. In fact, one of the main advantages of a.c. transmission and distribution
is the ease with which an alternating voltage can be increased or decreased by transformers.
Losses in transformers are generally low and thus efficiency is high. Being static they have a long life and
are very stable.
Transformers range in size from the miniature units used in electronic applications to the large power
transformers used in power stations. The principle of operation is the same for each. A transformer is
represented in Figure 5.1(a) as consisting of two electrical circuits linked by a common ferromagnetic
core. Magnetic coupling is used to transfer electric energy from one coil to another. The coil which
receives energy from an ac source is called the primary. The coil which delivers energy to an ac load is
called the secondary. The core of transformers used at low frequencies is generally made of magnetic
material, usually sheet steel. Cores of transformers used at higher frequencies are made of powdered iron
and ceramics, or nonmagnetic materials. Some coils are simply wound on nonmagnetic hollow forms
such as cardboard or plastic so that the core material is actually air. A circuit diagram symbol for a
transformer is shown in Figure 5.1(b).
(a)
(b)
Figure 5.1
When the secondary is an open-circuit and an alternating voltage V1 is applied to the primary winding, a
small current—called the no-load current I o —flows, which sets up a magnetic flux in the core. This
alternating flux links with both primary and secondary coils and induces in them e.m.f.’s of E1 and E2
respectively by mutual induction. The induced e.m.f. E in a coil of N turns is given by:
E  N
where
d
volts,
dt
(5.1)
d
is the rate of change of flux. In an ideal transformer, the rate of change of flux is the same for
dt
E1
E
 2 , i.e. the induced e.m.f. per turn is constant.
N1
N2
Assuming no losses, E1  V1 and E2  V2 , Hence
V1
V
 2
(5.2)
N1
N2
V1
N
 1
or
(5.3)
V2
N2
58
both primary and secondary and thus
N1
V1
is called the voltage ratio, and
is called the turns ratio, or the ‘transformation ratio’ of the
N2
V2
transformer. If N2 is less than N1 then V2 is less than V1 and the device is termed a step-down
transformer. If N2 is greater then N1 then V2 is greater than V1 and the device is termed a step-up
transformer.
When a load is connected across the secondary winding, a current I 2 flows. In an ideal transformer losses
are neglected and a transformer is considered to be 100% efficient.
Hence input power = output power, or V1 I 1 =
secondary volt-amperes are equal.
V1
I
 2
Thus
V2
I1
Combining equations (5.3) and (5.4) gives:
V2 I 2 , i.e., in an ideal transformer, the primary and
(5.4)
V1
N
I
 1  2
(5.5)
V2
N2
I1
The rating of a transformer is stated in terms of the volt-amperes that it can transform without
overheating. With reference to Figure 5.1(a), the transformer rating is either V1 I 1 or V2 I 2 , where I 2 is
the full-load secondary current.
5.2 Transformers construction
(i) There are broadly two types of single-phase double-wound transformer constructions—the core type
and the shell type, as shown in Figure 5.2. The low and high voltage windings are wound as shown to
reduce leakage flux.
(a) Core Type
(b) Shell Type
Figure 5.2
(ii) For power transformers, rated possibly at several MVA and operating at a frequency of 50 Hz in
Palestine, the core material used is usually laminated silicon steel or stalloy, the laminations reducing
eddy currents and the silicon steel keeping hysteresis loss to a minimum.
Large power transformers are used in the main distribution system and in industrial supply circuits. Small
power transformers have many applications, examples including welding and rectifier supplies, domestic
bell circuits, imported washing machines, and so on.
(iii) For audio frequency (a.f.) transformers, rated from a few mVA to no more than 20 VA, and
operating at frequencies up to about 15 kHz, the small core is also made of laminated silicon steel. A
typical application of a.f. transformers is in an audio amplifier system.
(iv) Radio frequency (r.f.) transformers, operating in the MHz frequency region have either an air core,
a ferrite core or a dust core. Ferrite is a ceramic material having magnetic properties similar to silicon
59
steel, but having a high resistivity. Dust cores consist of fine particles of carbonyl iron or permalloy (i.e.
nickel and iron), each particle of which is insulated from its neighbors. Applications of r.f. transformers
are found in radio and television receivers.
(v) Transformer windings are usually of enamel-insulated copper or aluminum.
(vi) Cooling is achieved by air in small transformers and oil in large transformers.
5.3 E.m.f. equation of a transformer
The magnetic flux  set up in the core of a transformer when an alternating voltage is applied to its
primary winding is also alternating and is sinusoidal.
Let  m be the maximum value of the flux and f be the frequency of the supply. The time for 1 cycle of
the alternating flux is the periodic time T, where T  1 seconds.
f
The flux rises sinusoidally from zero to its maximum value in ¼ cycle, and the time for ¼ cycle is 1
4f
seconds.
Hence the average rate of change of flux =
m
 1

 4f 
= 4 fm Wb/s
and since 1 Wb/s = 1 volt, the average e.m.f. induced in each turn = 4 fm volts.
As the flux  varies sinusoidally, then a sinusoidal e.m.f. will be induced in each turn of both primary
and secondary windings.
RMS value
For a sine wave, form factor =
= 1.11
average value
Hence, RMS value = form factor x average value = 1.11 x average value
Thus RMS e.m.f. induced in each turn = 1.11 x 4 fm volts
= 4.44 fm volts
Therefore, RMS value of e.m.f. induced in primary,
E1  4.44 fm N1 volts
Therefore, RMS value of e.m.f. induced in secondary,
E2  4.44 fm N 2 volts
(5.6)
(5.7)
5.4 Equivalent circuit of a transformer
Figure 5.3 shows an equivalent circuit of a transformer. R1 and R2 represent the resistances of the
primary and secondary windings and X 1 and X 2 represent the reactances of the primary and secondary
windings, due to leakage flux.
Figure 5.3
60
The core losses due to hysteresis and eddy currents are allowed for by resistance R which takes a current
I C , the core loss component of the primary current. Reactance X takes the magnetizing component I M .
In a simplified equivalent circuit shown in Figure 5.4, R and X are omitted since the no-load current Io is
normally only about 3–5% of the full load primary current.
It is often convenient to assume that all of the resistance and reactance as being on one side of the
transformer.
Figure 5.4
Resistance R2 in Figure 5.4 can be replaced by inserting an additional resistance R2' in the primary
circuit such that the power absorbed in R2' when carrying the primary current is equal to that in R2 due
to the secondary current, i.e., I12 R2'  I 22 R2
from which
I
R  R2  2
 I1
'
2
2

V
  R2  1

 V2



2
Then the total equivalent resistance in the primary circuit Re is equal to the primary and secondary
resistances of the actual transformer.
Hence
Re  R1  R2'
V 
Re  R1  R2  1 
 V2 
2
(5.8)
By similar reasoning, the equivalent reactance in the primary circuit is given by:
X e  X 1  X 2'
V 
X e  X 1  X 2  1 
 V2 
2
(5.9)
The equivalent impedance Z e of the primary and secondary windings referred to the primary is given by:
Ze 
Re2  X e2
(5.10)
If e is the phase angle between I1 and the volt drop I1 Z e then
R
cos e  e
Ze
The simplified equivalent circuit of a transformer is shown in Figure 5.5.
(5.11)
Figure 5.5
61
5.5 Regulation of a transformer
When the secondary of a transformer is loaded, the secondary terminal voltage, V2 , falls. As the power
factor decreases, this voltage drop increases. This is called the regulation of the transformer and it is
usually expressed as a percentage of the secondary no-load voltage, E2 .
For full-load conditions:
 E  V2 
 x 100%
Re gulation   2
E
2


(5.12)
In words regulation is found from:
 No - load secondary voltage  terminal voltage on load 
 x 100%
Re gulation  
No - load secondary voltage


The fall in voltage, E2  V2 , is caused by the resistance and reactance of the windings.
Typical values of voltage regulation are about 3% in small transformers and about 1% in large
transformers.
5.6 Transformer losses and efficiency
There are broadly two sources of losses in transformers on load, these being copper losses and iron
losses.
(a) Copper losses are variable and result in a heating of the conductors, due to the fact that they possess
resistance. If R1 and R2 are the primary and secondary winding resistances then the total copper loss is
I12 R1  I 22 R2
(b) Iron losses are constant for a given value of frequency and flux density and are of two types—
hysteresis loss and eddy current loss.
(i) Hysteresis loss is the heating of the core as a result of the internal molecular structure reversals
which occur as the magnetic flux alternates. The loss is proportional to the area of the hysteresis loop
and thus low loss nickel iron alloys are used for the core since their hysteresis loops have small areas.
(ii) Eddy current loss is the heating of the core due to e.m.f.’s being induced not only in the
transformer windings but also in the core. These induced e.m.f.’s set up circulating currents, called
eddy currents. Owing to the low resistance of the core, eddy currents can be quite considerable and
can cause a large power loss and excessive heating of the core. Eddy current losses can be reduced by
increasing the resistivity of the core material or, more usually, by laminating the core (i.e., splitting it
into layers or leaves) when very thin layers of insulating material can be inserted between each pair of
laminations. This increases the resistance of the eddy current path, and reduces the value of the eddy
current.
Transformer efficiency  
  1
outputpower inputpower  losses

inputpower
inputpower
losses
inputpower
(5.13)
and is usually expressed as a percentage. It is not uncommon for power transformers to have efficiencies
of between 95% and 98%.
Output power = V2 I 2 cos2 ,
total losses = copper loss + iron losses,
and input power = output power + losses
62
5.7 Auto transformers
An auto transformer is a transformer which has part of its winding common to the primary and secondary
circuits. Figure 5.6(a) shows the circuit for a double-wound transformer and Figure 5.6(b) that for an auto
transformer. The latter shows that the secondary is actually part of the primary, the current in the
secondary being ( I 2 - I 1 ).
Since the current is less in this section,
the cross-sectional area of the winding
can be reduced, which reduces the
amount of material necessary.
Figure 5.6
Saving of copper in an auto transformer
For the same output and voltage ratio, the auto transformer requires less copper than an ordinary doublewound transformer. This is explained below.
The volume, and hence weight, of copper required in a winding is proportional to the number of turns and
to the cross-sectional area of the wire. In turn this is proportional to the current to be carried, i.e., volume
of copper is proportional to NI.
Volume of copper in an auto transformer  (( N1  N 2 ) I1  N 2 ( I 2  I1 ))
 ( N1 I1  N 2 I1  N 2 I 2  N 2 I1 )
 ( N1 I1  N 2 I 2  2N 2 I1 )
 (2N1 I1  2N 2 I1 )
(since N1 I1  N 2 I 2 )
Volume of copper in a double-wound transformer  ( N1 I1  N 2 I 2 )
 (2N1 I1 )
(Also, since N1 I1  N 2 I 2 )
Hence
volume of copper in an auto transformer
volume of copper in a double  wound transformer
2 N1 I 1  2 N 2 I 1
2 N1 I 1
2 N1 I 1 2 N 2 I 1


2 N1 I 1 2 N1 I 1
N
1 2
N1
N2
If
= x then
N1

(volume of copper in an auto transformer)
= (1 − x) (volume of copper in a double-wound transformer)
4
If, say, x = , then (volume of copper in auto transformer) =
5
4
(1- ) (volume of copper in a double-wound transformer)
5
1
= (volume in double-wound transformer) i.e., a saving of 80%
5
63
(5.14)
1
, the saving is 25%, and so on.
4
The closer N 2 is to N1 , the greater the saving in copper.
Similarly, if x =
Advantages of auto transformers
The advantages of auto transformers over double-wound transformers include:
1. a saving in cost since less copper is needed (see above)
2. less volume, hence less weight
3. a higher efficiency, resulting from lower I 2 R losses
4. a continuously variable output voltage is achievable if a sliding contact is used
5. a smaller percentage voltage regulation.
Disadvantages of auto transformers
The primary and secondary windings are not electrically separate, hence if an open-circuit occurs in the
secondary winding the full primary voltage appears across the secondary.
Uses of auto transformers
Auto transformers are used for reducing the voltage when starting induction motors and for
interconnecting systems that are operating at approximately the same voltage.
5.8 Three-phase transformers
Three-phase power may be transformed by a bank of single-phase transformers or by a single three-phase
transformer. A three-phase transformer is essentially three wound on a common core. They basically
consist of three pairs of single-phase windings mounted on one core, as shown in Figure 5.7. The primary
and secondary windings in Figure 5.7 are wound on top of each other in the form of concentric cylinders,
similar to that shown in Figure 5.2(a). The geometry of the core is such that the fluxes of the phases share
common paths. As a result, the volume of iron is less than that of three single-phase transformer of the
same total rating.
The three-phase transformer do not provide the flexibility of a set of single-phase transformer. For
example, one single-phase transformer in a bank may have a higher kVA rating than the others, to serve
an unbalanced load. In case of failure of the transformer serving one phase only that one transformer need
to replace; whereas it is most likely that three-phase transformer damaged in one phase will have to be
completely removed from service, at least temporarily.
However, three-phase transformer, in addition to being lighter than the equivalent bank of the singlephase units, take less place, are less expensive, and involve much less external wiring. Their efficiency is
slightly better. Improved construction and better means of protection from over-voltages and short
circuits have made transformer failure a very rare occurrence. As a result, three phase banks of singlephase transformers are seldom used in new installations except in distribution circuits where a
combination of single and three-phase loads to be served.
Figure 5.7
64
In three-phase transformer, the primary an secondary windings may be connected independently in either
delta or why. The possible combinations are Δ-Δ, Δ-Y, Y-Y, Y-Δ. Figure 5.8 shows diagrams of these
four connections.
The Y-Y connection is to avoided unless a very solid neutral connection is made between the primary
and the power source. If a neutral is not provided, the phase voltages tend to become severely unbalanced
when load is unbalanced. These problems do not exist when one of the sets of the windings is in delta.
When it is necessary to have a Y-Y connection with a weak primary neutral, or none, each phase
transformer is provided with a third winding in addition to the primary and secondary, called a "tertiary".
The tertiaries are connected in delta. This is a relatively expensive arrangement. The connection is called
"why-delta-why". The tertiary winding terminals are often brought out to supply auxiliary power (e.g.,
lights, fans, and pumps) for the station.
When a Y-Δ or Δ-Y connection is used, the Y is preferably on the high-voltage side, and the neutral is
grounded. The transformer insulation may be designed for 1 / 3 times the line voltage, rather than for
the full line voltage. Sometimes it is necessary to have the Y on the low-voltage side, if a neutral is
required for the low-voltage circuit.
Y-Δ and Δ-Y connections result in a 30° phase displacement between the primary and secondary
voltages. It is standard practice to connect these transformers in such a way that the secondary voltage
lags the primary voltage by 30°.
Figure 5.8
3.9 Current transformers
For measuring currents in excess of about 100 A a current transformer is normally used. With a d.c.
moving-coil ammeter the current required to give full scale deflection is very small—typically a few
milli-amperes. When larger currents are to be measured a shunt resistor is added to the circuit. However,
even with shunt resistors added it is not possible to measure very large currents. When a.c. is being
measured a shunt cannot be used since the proportion of the current which flows in the meter will depend
on its impedance, which varies with frequency.
I1
N2

I2
N1
N1
from which, secondary current I 2  I1
N2
In a double-wound transformer:
65
In current transformers the primary usually consists of one or two turns whilst the secondary can have
several hundred turns. A typical arrangement is shown in Figure 20.19. If, for example, the primary has 2
turns and the secondary 200 turns, then if the primary current is 500 A,
N
2
) 5 A
secondary current, I 2  I1 1  (500)(
N2
200
Current transformers isolate the ammeter from the main
circuit and allow the use of a standard range of ammeters
giving full-scale deflections of 1 A, 2 A or 5 A.
For very large currents the transformer core can be mounted around the conductor or bus-bar. Thus the
primary then has just one turn. It is very important
to short-circuit the secondary winding before removing
the ammeter. This is because if current is flowing
in the primary, dangerously high voltages could be induced
in the secondary should it be open circuited.
Figure 5.9
3.10 Voltage transformers
For measuring voltages in excess of about 500 V it is often safer to use a voltage transformer. These are
normal double-wound transformers with a large number of turns on the primary, which is connected to a
high voltage supply, and a small number of turns on the secondary. A typical arrangement is shown in
Figure 5.10.
Since
V1
N
 1
V2
N2
the secondary voltage, V2  V1
N2
N1
Thus if the arrangement in Figure 5.10 has 4000
primary turns and 20 secondary turns then for a
voltage of 22 kV on the primary, the voltage
on the secondary,
V2  V1
N2
20
 ( 22000)(
) 110V
N1
4000
Figure 5.10
Solved Examples:
Example 5.1 : A 100 kVA, 4000 V/200 V, 50 Hz single-phase transformer has 100 secondary turns.
Determine (a) the primary and secondary current, (b) the number of primary turns, and (c) the maximum
value of the flux.
Solution:
V1  4000V , V2  200V , f  50Hz , N 2 100 turns
(a) Transformer rating = V1 I1 V2 I 2 100000 VA
100000 100000

 25 A
V1
4000
100000 100000

 500 A
and secondary current, I 2 
V2
200
V1
4000
 (100)(
)  2000 turns
(b) primary turns, N1  N 2
V2
200
(c) As E2  4.44 fm N 2 , and assuming E2  V2
E2
200

 9.01  10 -3 Wb  9.01 mWb
from which, maximum flux m 
4.44 fN 2 4.44(50)(100)
Hence primary current, I1 
66
Example 5.2 : A single-phase, 50 Hz transformer has 25 primary turns and 300 secondary turns. The
cross-sectional area of the core is 300 cm². When the primary winding is connected to a 250 V supply,
determine (a) the maximum value of the flux density in the core, and (b) the voltage induced in the
secondary winding.
Solution:
(a) E1  4.44 fm N1 assuming E1  V1
250  4.44(50)m (25)
250
m 
 0.04505 Wb
4.44(50)( 25)
However, m  Bm . A

0.04505
Hence Bm  m 
1.5T
A 300 x10 4
N
 300 
(b) V2  V1 2  (250)
  3000 V  3kV
N1
 25 
(i.e. the induced voltage on the secondary windings = 3 kV)
Example 5.3 : A transformer has 600 primary turns and 150 secondary turns. The primary and secondary
resistances are 0.25 Ω and 0.01 Ω respectively, and the corresponding leakage reactances are 1.0 Ω and
0.04 Ω respectively. Determine (a) the equivalent resistance referred to the primary winding, (b) the
equivalent reactance referred to the primary winding, (c) the equivalent impedance referred to the primary
winding, and (d) the phase angle of the impedance..
Solution:
 V1 

(a) From equation (5.8), equivalent resistance Re  R1  R2 
V
 2
 600 
i.e. Re  0.25  0.01

 150 
= 0.41 Ω
2
as
2
V1 N1

V2 N 2
 V1 

(b) From equation (5.9), equivalent reactance, X e  X 1  X 2 
V
 2
2
2
 600 
 1.64 
i.e. X e 1.0  0.04
 150 
(c) From equation (5.10), equivalent impedance, Z e 

(d) From equation (5.11), cos e 
Re2  X e2
(0.41) 2  (1.64) 2 1.69 
Re 0.41

Z e 1.69
1  0.41 
then, e  cos 
  75.96
 1.69 
Example 5.4 : The open circuit voltage of a transformer is 240 V. A tap changing device is set to operate
when the percentage regulation drops below 2.5%. Determine the load voltage at which the mechanism
operates.
Solution:
 No - load secondary voltage  terminal voltage on load 
 x 100%
Re gulation  
No
load
secondary
voltage


67
Hence 2.5 
240  V2
x 100
240
(2.5)( 240)
 240  V2
100
from which, load voltage, V2  240  6  234V
Example 5.5 : A 400 kVA transformer has a primary winding resistance of 0.5 Ω and a secondary
winding resistance of 0.001 Ω. The iron loss is 2.5 kW and the primary and secondary voltages are
5 kV and 320 V respectively. If the power factor of the load is 0.85, determine the efficiency of the
transformer (a) on full load, and (b) on half load
Solution:
(a) Transformer rating = V1I1 V2 I 2  400000 VA
400000 400000

 80 A
V1
5000
400000 400000

 1250 A
and secondary current, I 2 
V2
320
Hence primary current, I1 
Total copper loss = I1 R1  I 2 R2
= (80)² (0.5) + (1250)² (0.001) = 4762.5 watts
On full load, total loss = copper loss + iron loss = 4762.5 + 2500 = 7262.5 watts = 7.2625 kW
Total output power on full load = V2 I 2 cos 2  (400000)(0.85)  340 kW
Input power = output power + losses = 340 kW + 7.2625 kW = 347.2625 kW
2

2
losses

 x 100%
Efficiency   1
input
power


7.2625 

 1 
 x 100% = 97.91%
347.2625 

(b) Since the copper loss varies as the square of the current, then total
2
1
copper loss on half load = ( 4762.5)   1190.625 W
2
Hence total loss on half load = 1190.625 + 2500 = 3690.625 W or 3.691 kW
1
(340)  170 kW
2
Output power on half full load = 
Input power on half full load = output power + losses
= 170 kW + 3.691 kW = 173.691 kW

losses

 x 100%
Hence efficiency at half full load,   1
input
power


3.691 

 1 
 x 100% = 97.87%
 173.691 
Example 5.6 : A 3-phase, 11kV/660V, Y-Δ transformer is connected to the far end of a distribution line
for the near-end voltage is maintained at 11kV. The effective leakage reactance and resistance per-phase
of the transformer are respectively 0.25Ω and 0.05 Ω referred to the low-voltage side. The reactance and
resistance of each line are respectively 2 Ω and 1 Ω. It is required to maintain the terminal voltage at
68
660V when a line current of 260A at 0.8 lagging power factor is drawn from the secondary winding.
What percentage tapping must be provided on the H.V. side of the transformer to permit the necessary
adjustment? The transformer magnetizing current may be neglected and approximate expression for the
regulation may be used. Neglect also the changes to the impedance due to the alteration of turns ration.
Solution:
Figure 5.11
The equivalent circuit per phase is shown on the figure (5.11). The line impedance can also be referred to
the secondary side and included in the regulation expression.
2


660
 . (1  j 2) = (0.0108 + j0.0216)Ω per phase
Referred line impedance  

 11000 3 
The total impedance referred to the secondary = (0.0108 + j0.0216)+(0.05 + 0.25) = (0.0608 + j0.2716)Ω
Re gulation  E2  V2
From figure 5.12 the value of voltage regulation
Per phase can be calculated as:
V .R. per phase  I ( R cos   X sin  )

260
(0.0608 x0.8  0.2716 x0.6)
3
= 31.76 V (as sin    0.6 )
So the open circuit of the transformer must be =
Figure 5.12
660 + 31.76 = 691.76 V
and turns ratio must be 
11000 3
11000 3
instead of
660
691.76
(i.e. 9.18 instead of 9.623)
So high voltage tapping must be at
9.18
= 95.4%
9.623
Example 5.7 : A 10 kVA,, 2400V/240V, single phase transformer is connected has the following
resistances and leakage reactances. Find the primary voltage required to produce 240V at the secondary
terminals at full load, when the power factor is : (a) 0.8 lagging , (b) 0.8 leading
R1  3.00  ,
R2  0.03 
X1 15.00  ,
X 2  0.15 
Solution:
The equivalent impedance referred to the secondary side is:
Z e 2  ( R1 a 2  R2 )  j ( X 1 a 2  X 2 )
69
V
where, a   1
 V2
 2400
 
10
 240
Z e 2  (3.00 10 2  0.03)  j (15.00 10 2  0.15)
= (0.06 + j 0.30) Ω
 0.3059 78.69 Ω
(a) 0.8 power factor lagging
I2 
S 10,000

 41.7 A
V2
240
With V2 as a references, the full load secondary current is
I 2 fL  41.7   cos 1 0.8 A
I 2 fL  41.7   36.87 A
V1
V2  I 2 Z e 2 = 2400  (41.7   36.87 )( 0.305978.69) V
a
= 2400  (41.7 x 0.3059 (78.69  36.87 ) V
= 240 + (9.506 + j8.506) V
= 249.506 + j8.506 V
= 249.651.952
V1
= 10 x 249.65 = 2496.5 V
(b) 0.8 power factor leading
I 2 fL  41.7  cos 1 0.8
I 2 fL  41.7 36.87
A
A
V1
V2  I 2 Z e 2 = 2400  (41.7 36.87 )( 0.305978.69) V
a
= 2400  (41.7 x 0.3059 (78.69  36.87 ) V
= 2400  (12.76 (115.56 ) V
= 240 + (-5.505 + j11.51) V
= 234.5 + j11.51 V
= 243.782.81
V1
= 10 x 243.78 = 2437.8 V
Example 5.8 : A, 7200V/208V, 50 kVA, 3-phase distribution transformer is connected Δ-Y . The
equivalent leakage reactance and resistance per-phase of the transformer are respectively 0.0104Ω and
0.0433 Ω referred to the low-voltage side. Find the voltage regulation at full load, 0.85 power factor,
lagging.
70
Solution:
The primary side is connected Δ, so the primary voltage and current are:
V1 VL1  7200V
S
50000

 4.009 A
3 VL1
3 7200
I L1 
I P1 
I L1 4.009

 2.315 A
3
3
The secondary side is connected Y, so the secondary voltage and current are:
V2 
VL 2 208

 120 V
3
3
I L2 
S
50000

138.8 A
3 VL1
3 208
I P 2  I L 2 138.8 A
V
 7200
P1
 
 60
Turns ratio a  
V
 P 2  120
The equivalent impedance per phase referred to the secondary side is
Z e 2  ( 0.0104  j 0.0433 )   0.0445 76.5 
At full load 0.85 power factor lagging,
I P 2  138.8  cos 1 0.85  138.8   31.8 A
Applying Kirchhoff's voltage low toe equivalent circuit:
V1
V2  I 2 Z e 2 = 1200  (138.8   31.8 )( 0.044576.5) V
a
=
1200  (6.3154 76.5  31.8 ) V
=
1200  (6.3154 44.7 ) V
=
120  (4.49  j 4.44)  (124.49  j 4.44) V
=
124.57 2.04 V
No-load secondary phase voltage = 124.57 V
No-load secondary line voltage =
3 124.57 = 215.76 V
 No - load secondary voltage  terminal voltage on load 
 x 100%
Re gulation  
No - load secondary voltage


 215.76  208 
 x 100%  3.6%
 215.76 
= 
71
UNIT 6
Electrical Motors
Introduction
This unit introduces the construction and principle of operation for DC motors, and AC single phase and
three phases motors, their equivalent circuits, and efficiency.
6.1 Basic Motor Principles
All motors can be classed into two categories, AC and DC. The basic motor principles are alike for both
the AC and DC motor.
Magnetism is the basis for all electric motor operation. It produces the forces necessary for the motor to
run. There are two basic types of magnets, the permanent magnet and the electromagnet. The
electromagnet has the advantage over the permanent magnet in that
the magnetic field can be made stronger. Also the polarity of the
electromagnet can easily be reversed. The construction of an
electromagnet is simple.
When a current is passed through a coil of wire, a magnetic field is
produced. This magnetic field can be made stronger by winding
the coil of wire on an iron core (Fig. 6.1). One end of the
electromagnet is a north pole, while the other end is a south pole.
Figure 6.1
These poles can be reversed by reversing the direction of current in
the coil of wire.
The basic principle of all motors can be easily be shown using two electromagnets and a permanent
magnet. Current is passed through coil #1 and coil #2 in such a direction that north and south poles are
generated next to the permanent magnet, as shown in Figure (6.2.a) A permanent magnet with a north and
south pole is the moving part of this simple motor. In Figure (6.2.a) the north pole of the permanent
magnet is adjacent to the north pole of the electromagnet. Similarly, the south poles are adjacent to each
other. Like magnetic poles repel each other, causing the movable permanent magnet to begin to turn.
After it turns part way around, the force of attraction between the unlike poles becomes strong enough to
keep the permanent magnet rotating. The rotating magnet continues to turn until the unlike poles are lined
up. At this point the rotor would normally stop because of the attraction between the unlike poles (Fig.
6.2 (b)). If, however, the direction of currents in the electromagnetic coils was suddenly reversed, thereby
reversing the polarity of the two coils, then the poles would again be opposites and repel each other
(Fig.6.2(c)). The movable permanent magnet would then continue to rotate. If the current direction in the
electromagnetic coils was changed every time the magnet turned 180° or halfway around, then the magnet
would continue to rotate. This device is a motor in its simplest form. An actual motor is more complex
than the simple device shown above, but the principle is the same.
(a)
(b)
Figure 6.2
72
(c)
6.2 Direct Current (DC) Motor
6.2.1 DC General Construction
A typical DC generator or motor usually consists of: An armature core, an air gap, poles, and a yoke
which form the magnetic circuit; an armature winding, a field winding, brushes and a commutator which
form the electric circuit; and a frame, end bells, bearings, brush supports and a shaft which provide the
mechanical support (Figure 6.3).
Figure 6.3
Armature Core or Stack
The armature stack is made up thin magnetic steel laminations stamped from sheet steel with a blanking
die. Slots are punched in the lamination with a slot die. Sometimes these two operations are done as one.
The laminations are welded, riveted, bolted or bonded together.
Armature Winding
The armature winding is the winding, which fits in the armature slots and is eventually connected to the
commutator. It either generates or receives the voltage depending on whether the unit is a generator or
motor. The armature winding usually consists of copper wire, either round or rectangular and is insulated
from the armature stack.
Field Poles
The pole cores can be made from solid steel castings or from laminations. At the air gap, the pole usually
fans out into what is known as a pole head or pole shoe. This is done to reduce the reluctance of the air
gap. Normally the field coils are formed and placed on the pole cores and then the whole assembly is
mounted to the yoke.
Field Coils
The field coils are those windings, which are located on the poles and set up the magnetic fields in the
machine. They also usually consist of copper wire are insulated from the poles. The field coils may be
either shunt windings (in parallel with the armature winding) or series windings (in series with the
armature winding) or a combination of both.
Yoke
The yoke is a circular steel ring, which supports the field, poles mechanically and provides the necessary
magnetic path between the pole. The yoke can be solid or laminated. In many DC machines, the yoke also
serves as the frame.
73
Commutator
The commutator is the mechanical rectifier, which changes the AC voltage of the rotating conductors to
DC voltage. It consists of a number of segments normally equal to the number of slots. The segments or
commutator bars are made of silver bearing copper and are separated from each other by mica insulation.
Brushes and Brush Holders
Brushes conduct the current from the commutator to the external circuit. There are many types of brushes.
A brush holder is usually a metal box that is rectangular in shape. The brush holder has a spring that holds
the brush in contact with the commutator. Each brush usually has a flexible copper shunt or pigtail, which
extends to the lead wires. Often, the entire brush assembly is insulated from the frame and is made
movable as a unit about the commutator to allow for adjustment.
Interpoles
Interpoles are similar to the main field poles and located on the yoke between the main field poles. They
have windings in series with the armature winding. Interpoles have the function of reducing the armature
reaction effect in the commutating zone. They eliminate the need to shift the brush assembly.
Frame, End Bells, Shaft, and Bearings
The frame and end bells are usually steel, aluminum or magnesium castings used to enclose and support
the basic machine parts. The armature is mounted on a steel shaft, which is supported between two
bearings. The bearings are either sleeve, ball or roller type. They are normally lubricated by grease or oil.
Back End, Front End
The load end of the motor is the Back End. The opposite load end, most often the commutator end, is the
Front End of the motor.
6.2.2 Shunt, series and compound windings
When the field winding of a d.c. machine is connected in parallel with the armature, as shown in Figure
6.4(a), the machine is said to be shunt wound. If the field winding is connected in series with the
armature, as shown in Figure 6.4(b), then the machine is said to be series wound. A compound wound
machine has a combination of series and shunt windings.
Depending on whether the electrical machine is series wound, shunt wound or compound wound, it
behaves differently when a load is applied. The behaviour of a d.c. machine under various conditions is
shown by means of graphs, called characteristic curves or just characteristics. The characteristics shown
in the following sections are theoretical, since they neglect the effects of armature reaction.
Armature reaction is the effect that the magnetic field produced by the armature current has on the
magnetic field produced by the field system. In a generator, armature reaction results in a reduced output
voltage, and in a motor, armature reaction results in increased speed. A way of overcoming the effect of
armature reaction is to fit compensating windings, located in slots in the pole face.
Figure 6.4
74
6.2.3 E.m.f. generated in an armature winding
Let Z = number of armature conductors,
Φ = useful flux per pole, in webers
p = number of pairs of poles
and n = armature speed in rev/s
The e.m.f. generated by the armature is equal to the e.m.f. generated by one of the parallel paths. Each
conductor passes 2p poles per revolution and thus cuts 2pΦ webers of magnetic flux per revolution.
Hence flux cut by one conductor per second = 2pΦn Wb and so the average e.m.f. E generated per
conductor is given by:
E = 2pΦn volts (since 1 volt = 1 Weber per second)
Let c = number of parallel paths through the winding between positive and negative brushes
c= 2 for a wave winding
c= 2p for a lap winding
The number of conductors in series in each path =
Z
c
The total e.m.f. between brushes = (average e.m.f./conductor)(number of conductors in series per path)
Z
= 2 pn
c
i.e. generated e.m.f., E  2 p n
Z
c
(6.1)
Since Z, p and c are constant for a given machine, then E  n .
However 2n is the angular velocity  in radians per second, hence the generated e.m.f. is
proportional to  and  , i.e., generated e.m.f., E    .
6.2.4 Motor Principle
Although the mechanical construction of dc motors and generators is very similar, their functions are
different. The function of a generator is to generate a voltage when conductors are moved through a field,
while that of a motor is to develop a turning effort, or torque, to produce mechanical rotation.
Direction of Armature Rotation
(a) single conductor
(b) two conductors of a single loop coil
Figure 6.5
The left-hand rule is used to determine the direction of rotation of the armature conductors. The lefthand rule for motors is as follows: With the forefinger, middle finger, and thumb of the left hand mutually
perpendicular, point the forefinger in the direction of the field and the middle finger in the direction of the
75
current in the conductor; the thumb will point in the direction in which the conductor tends to move (Fig.
6.5a). In a single-turn rectangular coil parallel to a magnetic field (Fig. 6.5b), the direction of current in
the left-hand conductor is out of the paper, while in the right-hand conductor it is into the paper.
Therefore, the left-hand conductor tends to move upward with a force F1, and the right-hand conductor
tends to move downward with an equal force F2. Both forces act to develop a torque which turns the coil
in a clockwise direction. A single-coil motor (Fig. 6.5b) is impractical because it has dead centers and the
torque developed is pulsating. Good results are obtained when a large number of coils is used as in a fourpole motor. As the armature rotates and the conductors move away from under a pole into the neutral
plane, the current is reversed in them by the action of the commutator. Thus, the conductors under a given
pole carry current in the same direction at all times.
6.2.5 DC Motor Equivalent Circuit
The equivalent circuit of the d.c. motor consists from
a d.c. source E and connected with it in series a
resistance Ra .
Voltage and current relationships of a dc motor
equivalent circuit (Fig. 6.6) are as follows:
VT  E  I a Ra
(6.2)
Figure 6.6
6.2.6 Torque of a d.c. Machine
From equation (6.2), for a d.c. motor, the supply voltage VT is given by:
VT  E  I a Ra
Multiplying each term by current I a gives:
VT I a  EI a  I a2 Ra
The term VT I a is the total electrical power supplied to the armature, the term I a2 Ra is the loss due to
armature resistance, and the term EI a is the mechanical power developed by the armature.
If T is the torque, in newton meters, then the mechanical power developed is given by T watts
Hence, T  2nT  EI a , from which
Torque T 
EI a
Newton.meters
2n
(6.3)
From equation (6.1), the e.m.f. E generated is given by:
E  2 p n
Z
c
Z

 2 pn  I a
c

Then, Torque T 
Newton.meters
2n
pZI a
i.e. Torque T 
Newton.meters
c
For a given machine, Z, c and p are fixed values
Hence torque, T  I a
(6.4)
(6.5)
76
6.2.7 Types of d.c. motor and their characteristics
(a) Shunt-wound motor
In the shunt wound motor the field winding is in
Parallel with the armature across the supply as
shown in Figure 6.7.
For the circuit shown in Figure 6.7,
Supply voltage, VT  E  I a Ra
or generated e.m.f., E  VT  I a Ra
Supply current, I  I a  I f
Figure 6.7
Characteristics
The two principal characteristics are the torque/armature current and speed/armature current relationships.
From these, the torque/speed relationship can be derived.
(i) The theoretical torque/armature current characteristic can be derived from the expression T  I a .
For a shunt-wound motor, the field winding is connected in parallel with the armature circuit and thus the
applied voltage gives a constant field current, i.e. a shunt-wound motor is a constant flux machine. Since
 is constant, it follows that T  I a , and the characteristic is as shown in Figure 6.8.
(ii) The armature circuit of a d.c. motor has resistance due to the armature winding and brushes, R a
ohms, and when armature current I a is flowing through it, there is a voltage drop of I aRa volts. In
Figure 6.7 the armature resistance is shown as a separate resistor in the armature circuit to help
understanding. Also, even though the machine is a motor, because conductors are rotating in a magnetic
field, a voltage, E    , is generated by the armature conductors.
From equation (6.2) VT  E  I a Ra , or E  VT  I a Ra
However, as E  n , hence, n  E /  i.e.
V I R
E
 T a a,
(6.6)


For a shunt motor, V,  and R a are constants, hence as armature current I a increases, I aRa
speed of rotation, n 
increases and VT  I a Ra decreases, and the speed is proportional to a quantity which is decreasing and is as
shown in Figure 6.9. As the load on the shaft of the motor increases, I a increases and the speed drops
slightly. In practice, the speed falls by about 10% between no-load and full-load on many d.c. shuntwound motors. Due to this relatively small drop in speed, the d.c. shunt-wound motor is taken as basically
being a constant-speed machine and may be used for driving lathes, lines of shafts, fans, conveyor belts,
pumps, compressors, drilling machines and so on.
(iii) Since torque is proportional to armature current, (see (i) above), the theoretical speed/torque
characteristic is as shown in Figure 6.10.
Figure 6.8
Figure 6.9
77
Figure 6.10
(b) Series-wound motor
In the series-wound motor the field winding is in series with the
armature across the supply as shown in Figure 6.11.
For the series motor shown in Figure 6.11,
Supply voltage
VT  E  I a ( Ra  R f )
or generated e.m.f. E  VT  I a ( Ra  R f )
Figure 6.11
Characteristics
In a series motor, the armature current flows in the field winding and is equal to the supply current, I.
(i) The torque/current characteristic
It is shown in Section (5.2.6) that torque T  ( I a) . Since the armature and field currents are the same
current, I, in a series machine, then T  (  I ) over a limited range, before magnetic saturation of the
magnetic circuit of the motor is reached, (i.e., the linear portion of the B–H curve for the yoke, poles, air
2
gap, brushes and armature in series). Thus   I and T  I . After magnetic saturation,  almost
becomes a constant and T  I . Thus the theoretical torque/current characteristic is as shown in Figure
6.12.
(ii) The speed/current characteristic
V I R
E
 T a a . In a series motor, I  I a and below the


VT  IR
magnetic saturation level,   I . Thus n 
where R is the combined resistance of the series
I
field and armature circuit. Since IR is small compared with VT , then an approximate relationship for the
It is shown in equation (6.6) that n 
VT
1
since VT is constant. Hence the theoretical speed/current characteristic is as shown

I
I
in Figure 6.13. The high speed at small values of current indicate that this type of motor must not be run
on very light loads and invariably, such motors are permanently coupled to their loads.
speed is n 
(iii) The theoretical speed/torque characteristic may be derived from (i) and (ii) above by obtaining the
torque and speed for various values of current and plotting the co-ordinates on the speed/torque
characteristics. A typical speed/torque characteristic is shown in Figure 6.14.
A d.c. series motor takes a large current on starting and the characteristic shown in Figure 6.12 shows that
the series-wound motor has a large torque when the current is large. Hence these motors are used for
traction (such as trains, milk delivery vehicles, etc.), driving fans and for cranes and hoists, where a large
initial torque is required.
Figure 6.12
Figure 6.13
78
Figure 6.14
(c) Compound-wound motor
There are two types of compound wound motor:
(i) Cumulative compound, in which the series winding is so connected that the field due to it assists that
due to the shunt winding.
(ii) Differential compound, in which the series winding is so connected that the field due to it opposes
that due to the shunt winding. Figure 6.15(a) shows a long-shunt compound motor and Figure 6.15(b) a
short-shunt compound motor.
Figure 6.15
Characteristics
A compound-wound motor has both a series and a shunt field winding, (i.e. one winding in series and one
in parallel with the armature), and is usually wound to have a characteristic similar in shape to a series
wound motor (see Figures 6.12–6.14). A limited amount of shunt winding is present to restrict the noload speed to a safe value. However, by varying the number of turns on the series and shunt windings and
the directions of the magnetic fields produced by these windings (assisting or opposing), families of
characteristics may be obtained to suit almost all applications. Generally, compound-wound motors are
used for heavy duties, particularly in applications where sudden heavy load may occur such as for driving
plunger pumps, presses, geared lifts, conveyors, hoists and so on.
Typical compound motor torque and speed characteristics are shown in Figure 6.16.
Figure 6.16
6.2.8 Changing direction of rotation of a d.c. motor
The direction of rotation of a d.c. motor can be done simply by changing the direction of one of the
magnetic fields either in stator or rotor. It is not convenient to change the direction of the magnetic field
of the stator especially in the compound wound stators. Changing direction of rotation is achieved by
changing the polarity of brushes which result in changing the direction of current in the armature making
the direction of magnetic field to be changed. There are some mechanical or electronic circuits that can
change the polarity of the brushes manually or automatically as requested. In every time of changing
direction of rotation the motor must be braked before changing circuit begins.
6.2.9 The efficiency of a d.c. motor
The efficiency of a d.c. machine is given by:
output power
Effeciency 
x 100 %
input power
Also, the total losses = I a Ra  I f VT  C (for a shunt motor) where C is the sum of the iron, friction
and windage losses.
2
79
For a motor, the input power = VT I
and the output power = VT I – losses
= VT I  I a Ra  I f VT  C
2
 VT I  I a2 Ra  I f VT  C 
 x 100 %
Hence efficiency   

V
I
T


(6.7)
The efficiency of a motor is a maximum when the load is such that: I a Ra  I f VT  C
2
Solved Examples:
Example 6.1: A six-pole lap-wound motor is connected to a 250 V d.c. supply. The armature has 500
conductors and a resistance of 1Ω. The flux per pole is 20 mWb. Calculate (a) the speed and (b) the
torque developed when the armature current is 40 A.
Solution :
VT = 250 V, Z = 500, Ra = 1 Ω,   20 x10 3 Wb, I a = 40 A, c = 2p for a lap winding
(a) Back e.m.f. E  VT  I a Ra = 250 – (40)(1) = 210 V
E.m.f E  2 p n
i.e. 210 
Z
c
2 p 20 x10 3 n 500
2p
210
 21 rev/s
20 x 10 3 x 500
or = 21 x 60 = 1260 rev/min
Hence speed n 
(b) Torque T 
EI a 210 x 40

 63.66 N.m
2n 2 x 21
Example 6.2: An 8-pole d.c. motor has a wave-wound armature with 900 conductors. The useful flux per
pole is 25 mWb. Determine the torque exerted when a current of 30 A flows in each armature conductor.
Solution :
p = 4, Z = 900,   25 x10 3 Wb, I a = 30 A, c = 2 for a wave winding
From equation (6.4), torque T 
pZI a
c
4 x 25 x 10 3 x 900 x 30

 429.7 N.m
 x2
Example 6.3: A 220 V, d.c. shunt-wound motor runs at 800 rev/min and the armature current is 30 A.
The armature circuit resistance is 0.4 Ω. Determine (a) the maximum value of armature current if the flux
is suddenly reduced by 10% and (b) the steady state value of the armature current at the new value of
flux, assuming the shaft torque of the motor remains constant.
Solution :
(a) For a d.c. shunt-wound motor, E  VT  I a Ra .
Hence initial generated e.m.f., E1  220  30 x 0.4  208 V.
The generated e.m.f. is also such that E   n , so at the instant the flux is reduced, the speed has not
had time to change, and E = 208 x 90/100 = 187.2 V.
Hence, the voltage drop due to the armature resistance is (220 – 187.2), i.e., 32.8 V. The instantaneous
value of the current is 32.8/0.4, i.e., 82 A. This increase in current is about three times the initial value
80
and causes an increase in torque, ( T  I a ). The motor accelerates because of the larger torque value
until steady state conditions are reached.
(b) T  (I a) , and since the torque is constant, then
1 I a1   2 I a 2 , The flux  is reduced by 10%, hence
 2  0.9 1
Thus, 1 x 30  0.91 I a 2
i.e. the steady state value of armature current, I a 2 
30
 33.33 A
0.9
Example 6.4: A series motor has an armature resistance of 0.2 Ω and a series field resistance of 0.3 Ω. It
is connected to a 240 V supply and at a particular load runs at 24 rev/s when drawing 15 A from the
supply.
(a) Determine the generated e.m.f. at this load.
(b) Calculate the speed of the motor when the load is changed such that the current is increased to 30 A.
Assume that this causes a doubling of the flux.
Solution :
(a) With reference to Figure 6.11, generated e.m.f., E, at initial load, is given by:
E  VT  I a ( Ra  R f )
= 240 – 15 (0.2 + 0.3) = 240 – 7.5 = 232.5 V
(b) When the current is increased to 30 A, the generated e.m.f. is given by:
E  VT  I a ( Ra  R f )
= 240 – 30 (0.2 + 0.3) = 240 – 15 = 225 V
Now e.m.f. E   n
thus
E1  1n1

E 2  2 n2
i.e.,
232.5  1 ( 24)
as  2  21

225
2 1 n 2
(2250(24)
11.6 rev/s
2(232.5)
As the current has been increased from 15 A to 30 A, the speed has decreased from 24 rev/s to 11.6 rev/s.
Its speed/current characteristic is similar to Figure 6.13.
Hence speed of motor, n2 
Example 6.5: A 320 V shunt motor takes a total current of 80 A and runs at 1000 rev/min. If the iron,
friction and windage losses amount to 1.5 kW, the shunt field resistance is 40 Ω and the armature
resistance is 0.2 Ω, determine the overall efficiency of the motor.
Solution :
Field current, I f 
VT 320

8 A
Rf
40
Armature current I a  I  I f = 80 – 8 = 72 A
C = iron, friction and windage losses = 1500 W
 VT I  I a2 Ra  I f VT  C 
 x 100 %
Efficiency,   

V
I
T



(320)(80)  (72) 2 (0.2)  (8)(320)  1500
x100  80.1%
(320)(80)
81
Example 6.6: A d.c. series motor drives a load at 30 rev/s and takes a current of 10 A when the supply
voltage is 400 V. If the total resistance of the motor is 2 Ω and the iron, friction and windage losses
amount to 300 W, determine the efficiency of the motor.
Solution :
 VT I  I 2 R  C 
 x 100 %
V
I
T


Efficiency,   

(400)(10)  (10) 2 (2)  300

x100  87.5%
(400)(10)
6.3 Alternating Current (AC) Motor
The AC motors are mainly two types, synchronous motors and induction motors or asynchronous
motors. Also, induction motors may be three phases or single phase. Only induction motors, three phases
and single phase will be explained.
In d.c. motors, introduced in section 5.2, conductors on a rotating armature pass through a stationary
magnetic field. In a three-phase induction motor, the magnetic field rotates and this has the advantage
that no external electrical connections to the rotor need be made. Its name is derived from the fact that the
current in the rotor is induced by the magnetic field instead of being supplied through electrical
connections to the supply. The result is a motor which: (i) is cheap and robust, (ii) is explosion proof, due
to the absence of a commutator or slip-rings and brushes with their associated sparking, (iii) requires little
or no skilled maintenance, and (iv) has self-starting properties when switched to a supply with no
additional expenditure on auxiliary equipment. The principal disadvantage of a three-phase induction
motor is that its speed cannot be readily adjusted.
6.3.1 Basic AC Motor Operation
An AC motor has two basic electrical parts: a "stator" and a "rotor" as shown in Figure 6.17. The stator is
in the stationary electrical component. It consists of a group of individual electro-magnets arranged in
such a way that they form a hollow cylinder, with one pole of each magnet facing toward the center of the
group. The term, "stator" is derived from the word stationary. The stator then is the stationary part of the
motor. The rotor is the rotating electrical component. It also consists of a group of electro-magnets
arranged around a cylinder, with the poles facing toward the stator poles. The rotor, obviously, is located
inside the stator and is mounted on the motor's shaft. The term "rotor" is derived from the word rotating.
The rotor then is the rotating part of the motor. The objective of these motor components is to make the
rotor rotate which in turn will rotate the motor shaft. This rotation will occur because of the previously
discussed magnetic phenomenon that unlike magnetic poles attract each other and like poles repel. If we
progressively change the polarity of the stator poles in such a way that their combined magnetic field
rotates, then the rotor will follow and rotate with the magnetic field of the stator.
Figure
Figure 6.17
6.8
82
This "rotating magnetic fields of the stator can be better understood by examining Figure 6.18. As shown,
the stator has six magnetic poles and the rotor has two poles. At time 1, stator poles A-1 and C-2 are north
poles and the opposite poles, A-2 and C-1, are south poles. The S-pole of the rotor is attracted by the two
N-poles of the stator and the N-pole of the rotor is attracted by the two south poles of the stator. At time
2, the polarity of the stator poles is changed so that now C-2 and B-1 and N-poles and C-1 and B-2 are Spoles. The rotor then is forced to rotate 60 degrees to line up with the stator poles as shown. At time 3, B1 and A-2 are N. At time 4, A-2 and C-1 are N. As each change is made, the poles of the rotor are
attracted by the opposite poles on the stator. Thus, as the magnetic field of the stator rotates, the rotor is
forced to rotate with it.
Figure 6.18
6.3.2 Production of a rotating magnetic field
One way to produce a rotating magnetic field in the stator of an AC motor is to use a three-phase power
supply for the stator coils. What, you may ask, is three-phase power? The answer to that question can be
better understood if we first examine single-phase power. Figure 7 is the visualization of single-phase
power. The associated AC generator is producing just one flow of electrical current whose direction and
intensity varies as indicated by the single solid line on the graph. From time 0 to time 3, current is flowing
in the conductor in the positive direction. From time 3 to time 6, current is flowing in the negative. At any
one time, the current is only flowing in one direction. But some generators produce three separate current
flows (phases) all superimposed on the same circuit. This is referred to as three-phase power. At any one
instant, however, the direction and intensity of each separate current flow are not the same as the other
phases. This is illustrated in Figure 6.19. The three separate phases (current flows) are labeled A, B and
C. At time 1, phase A is at zero amps, phase B is near its maximum amperage and flowing in the positive
direction, and phase C is near to its maximum amperage but flowing in the negative direction. At time 2,
the amperage of phase A is increasing and flow is positive, the amperage of phase B is decreasing and its
flow is still negative, and phase C has dropped to zero amps. A complete cycle (from zero to maximum in
one direction, to zero and to maximum in the other direction, and back to zero) takes one complete
revolution of the generator. Therefore, a complete cycle, is said to have 360 electrical degrees. In
examining Figure 10, we see that each phase is displaced 120 degrees from the other two phases.
Therefore, we say they are 120 degrees out of phase.
83
Figure 6.19
To produce a rotating magnetic field in the stator of a
three-phase AC motor, all that needs to be done is wind
the stator coils properly and connect the power supply
leads correctly. The connection for a 6 pole stator is
shown in Figure 6.20. Each phase of the three-phase
power supply is connected to opposite poles and the
associated coils are wound in the same direction. As
you will recall from Figure 4, the polarity of the poles
of an electro-magnet are determined by the direction of
the current flow through the coil. Therefore, if two
opposite stator electro-magnets are wound in the same
direction, the polarity of the facing poles must be
Figure 6.20
opposite. Therefore, when pole A1 is N, pole A2 is S.
When pole B1 is N, B2 is S and so forth.
Figure 6.21 shows how the rotating magnetic field is produced. At time1, the current flow in the phase
"A" poles is positive and pole A-1 is N. The current flow in the phase "C" poles is negative, making C-2 a
N-pole and C-1 is S. There is no current flow in phase "B", so these poles are not magnetized. At time 2,
the phases have shifted 60 degrees, making poles C-2 and B-1 both N and C-1 and B-2 both S. Thus, as
the phases shift their current flow, the resultant N and S poles move clockwise around the stator,
producing a rotating magnetic field. The rotor acts like a bar magnet, being pulled along by the rotating
magnetic field.
Figure 6.21
84
Up to this point not much has been said about the rotor. In the previous examples, it has been assumed the
rotor poles were wound with coils, just as the stator poles, and supplied with DC to create fixed polarity
poles. This, by the way, is exactly how a synchronous AC motor works. However, most AC motors being
used today are not synchronous motors. Instead, so-called "induction" motors are the workhorses of
industry.
6.3.2 Synchronous speed
The rotating magnetic field produced by three phase windings could have been produced by rotating a
permanent magnet’s north and south pole at synchronous speed, (shown as N and S at the ends of the flux
phasors in Figures 6.18). For this reason, it is called a 2-pole system and an induction motor using three
phase windings only is called a 2-pole induction motor.
If six windings displaced from one another by 60° are used, as shown in Figure 6.22 (a), by drawing the
current and resultant magnetic field diagrams at various time values, it may be shown that one cycle of
the supply current to the stator windings causes the magnetic field to move through half a revolution. The
current distribution in the stator windings are shown in Figure 6.22(a), for the time t shown in Figure
6.22(b).
It can be seen that for six windings on the stator, the magnetic flux produced is the same as that produced
by rotating two permanent magnet north poles and two permanent magnet south poles at synchronous
speed. This is called a 4-pole system and an induction motor using six phase windings is called a 4-pole
induction motor. By increasing the number of phase windings the number of poles can be increased to
any even number. In general, if f is the frequency of the currents in the stator windings and the stator is
wound to be equivalent to p pairs of poles, the speed of revolution of the rotating magnetic field, i.e., the
synchronous speed,
nS
is given by:
nS 
f
p
rev/s
(6.8)
Figure 6.22
6.4 Three Phase Induction Motors
6.4.1 Basic Construction and Operating Principles
Like most motors, an AC induction motor has a fixed outer portion, called the stator and a rotor that spins
inside with a carefully engineered air gap between the two. Virtually all electrical motors use magnetic
field rotation to spin their rotors. A three-phase AC induction motor is the only type where the rotating
magnetic field is created naturally in the stator because of the nature of the supply. DC motors depend
either on mechanical or electronic commutation to create rotating magnetic fields. A single-phase AC
induction motor depends on extra electrical components to produce this rotating magnetic field.
85
Stator
The stator is made up of several thin laminations of aluminum or cast iron. They are punched and
clamped together to form a hollow cylinder (stator core) with slots as shown in Figure 6.23. Coils of
insulated wires are inserted into these slots. Each grouping of coils, together with the core it surrounds,
forms an electromagnet (a pair of poles) on the application of AC supply. The number of poles of an AC
induction motor depends on the internal connection of the stator windings. The stator windings are
connected directly to the power source. Internally they are connected in such a way, that on applying AC
supply, a rotating magnetic field is created.
The number of slots must be even number and so that if it is divided on three (no. of phases) then the
result is divided on the number of poles must produce a result without fraction. The result of division is
the number of slots per phase per pole (so it must be integer). For example if the number of slots is 36,
and the motor has 4 poles, then the number of slots per phase per pole is 3.
Figure 6.23
Squirrel-Cage Rotor
The rotor is made up of several thin steel laminations with evenly spaced bars, which are made up of
aluminum or copper, along the periphery. In the most popular type of rotor (squirrel cage rotor), these
bars are connected at ends mechanically and electrically by the use of rings. Almost 90% of induction
motors have squirrel cage rotors. This is because the squirrel cage rotor has a simple and rugged
construction. The rotor consists of a cylindrical laminated core with axially placed parallel slots for
carrying the conductors. Each slot carries a copper, aluminum, or alloy bar. These rotor bars are
permanently short-circuited at both ends by means of the end rings, as shown in Figure 6.24. This total
assembly resembles the look of a squirrel cage, which gives the rotor its name. The rotor slots are not
exactly parallel to the shaft. Instead, they are given a skew for two main reasons. The first reason is to
make the motor run quietly by reducing magnetic hum and to decrease slot harmonics. The second reason
is to help reduce the locking tendency of the rotor. The rotor teeth tend to remain locked under the stator
teeth due to direct magnetic attraction between the two. This happens when the number of stator teeth are
equal to the number of rotor teeth. The rotor is mounted on the shaft using bearings on each end; one end
of the shaft is normally kept longer than the other for driving the load. Some motors may have an
accessory shaft on the non-driving end for mounting speed or position sensing devices. Between the stator
and the rotor, there exists an air gap, through which due to induction, the energy is transferred from the
stator to the rotor. The generated torque forces the rotor and then the load to rotate. Regardless of the type
of rotor used, the principle employed for rotation remains the same.
86
Figure 6.24
Wound Rotor
The wound rotor or slip-ring rotor differs from the
squirrel-cage rotor in the rotor winding. The rotor
winding consists of insulated coils, grouped to form
definite polar areas of magnetic force having the same
number of poles as the stator. The ends of these coils
are brought out to slip-rings. By means of brushes, a
variable resistance is placed across the rotor winding
(Figure 6.25). By varying this resistance, the speed and
torque of the motor is varied. The wound rotor motor is
an excellent motor for use on applications that require
an adjustable-varying speed (an adjustable speed that
varies with load) and high starting torque.
Figure 6.25
6.4.2 Slip
The force exerted by the rotor bars causes the rotor to turn in the direction of the rotating magnetic field.
As the rotor speed increases, the rate at which the rotating magnetic field cuts the rotor bars is less and the
frequency of the induced e.m.f.’s in the rotor bars is less. If the rotor runs at the same speed as the rotating
magnetic field, no e.m.f.’s are induced in the rotor, hence there is no force on them and no torque on the
rotor. Thus the rotor slows down. For this reason the rotor can never run at synchronous speed.
When there is no load on the rotor, the resistive forces due to windage and bearing friction are small and
the rotor runs very nearly at synchronous speed. As the rotor is loaded, the speed falls and this causes an
increase in the frequency of the induced e.m.f.’s in the rotor bars and hence the rotor current, force and
torque increase. The difference between the rotor speed, nr , and the synchronous speed, n S , is called the
slip speed, i.e.
slip speed =
n S - nr
rev/s
(6.9)
The ratio ( n S - nr )/ n S is called the fractional slip or just the slip, s, and is usually expressed as a
percentage. Thus
 nS  nr
slip, s = 
 n
S


 x 100 %

87
(6.10)
6.4.3 Induction motor torque–speed characteristics
The torque-speed and torque-slip characteristics of an induction motor are as shown in Figure 6.26.
Starting Characteristic
Induction motors, at rest, appear just like a short circuited transformer and if connected to the full supply
voltage, draw a very high current known as the “Locked Rotor Current.” They also produce torque which
is known as the “Locked Rotor Torque”. The Locked Rotor Torque (LRT) and the Locked Rotor Current
(LRC) are a function of the terminal voltage of the motor and the motor design. As the motor accelerates,
both the torque and the current will tend to alter with rotor speed if the voltage is maintained constant.
The starting current of a motor with a fixed voltage will drop very slowly as the motor accelerates and
will only begin to fall significantly when the motor has reached at least 80% of the full speed. The actual
curves for the induction motors can vary considerably between designs but the general trend is for a high
current until the motor has almost reached full speed. The LRC of a motor can range from 500% of FullLoad Current (FLC) to as high as 1400% of FLC. Typically, good motors fall in the range of 550% to
750% of FLC.
The starting torque of an induction motor starting with a fixed voltage will drop a little to the minimum
torque, known as the pull-up torque, as the motor accelerates and then rises to a maximum torque, known
as the breakdown or pull-out torque, at almost full speed and then drop to zero at the synchronous speed.
The curve of the start torque against the rotor speed is dependant on the terminal voltage and the rotor
design.
The LRT of an induction motor can vary from as low as 60% of FLT to as high as 350% of FLT. The
pull-up torque can be as low as 40% of FLT and the breakdown torque can be as high as 350% of FLT.
Typically, LRTs for medium to large motors are in the order of 120% of FLT to 280% of FLT. The PF of
the motor at start is typically 0.1-0.25, rising to a maximum as the motor accelerates and then falling
again as the motor approaches full speed.
Figure 6.26
88
Running Characteristic
Once the motor is up to speed, it operates at a low slip, at a speed determined by the number of the stator
poles. Typically, the full-load slip for the squirrel cage induction motor is less than 5%. The actual fullload slip of a particular motor is dependant on the motor design. The typical base speed of the four pole
induction motor varies between 1420 and 1480 RPM at 50 Hz, while the synchronous speed is 1500 RPM
at 50 Hz.
The current drawn by the induction motor has two components: reactive component (magnetizing current)
and active component (working current). The magnetizing current is independent of the load but is
dependant on the design of the stator and the stator voltage. The actual magnetizing current of the
induction motor can vary, from as low as 20% of FLC for the large two pole machine, to as high as 60%
for the small eight pole machine. The working current of the motor is directly proportional to the load.
The tendency for the large machines and high-speed machines is to exhibit a low magnetizing current,
while for the low-speed machines and small machines the tendency is to exhibit a high magnetizing
current. A typical medium sized four pole machine has a magnetizing current of about 33% of FLC.
A low magnetizing current indicates a low iron loss, while a high magnetizing current indicates an
increase in iron loss and a resultant reduction in the operating efficiency.
Typically, the operating efficiency of the induction motor is highest at 3/4 load and varies from less than
60% for small low-speed motors to greater than 92% for large high-speed motors. The operating PF and
efficiencies are generally quoted on the motor data sheets.
Load Characteristic
In real applications, various kinds of loads exist with different torque-speed curves. For example,
Constant Torque, Variable Speed Load (screw compressors, conveyors, feeders), Variable Torque,
Variable Speed Load (fan, pump), Constant Power Load (traction drives), Constant Power, Constant
Torque Load (coiler drive) and High Starting/Breakaway Torque followed by Constant Torque Load
(extruders, screw pumps).
The motor load system is said to be stable when the developed motor torque is equal to the load torque
requirement. The motor will operate in a steady state at a fixed speed. The response of the motor to any
disturbance gives us an idea about the stability of the motor load system. This concept helps us in quickly
evaluating the selection of a motor for driving a particular load.
In most drives, the electrical time constant of the motor is negligible as compared to its mechanical time
constant. Therefore, during transient operation, the motor can be assumed to be in an electrical
equilibrium, implying that the steady state torque-speed curve is also applicable to the transient operation.
As an example, Figure 6.27 shows torque-speed curves of the motor with two different loads. The system
can be termed as stable, when the operation will be restored after a small departure from it, due to a
disturbance in the motor or load.
For example, disturbance causes a reduction of m in speed. In the first case, at a new speed, the motor
torque ( T ) is greater than the load torque ( T1 ). Consequently, the motor will accelerate and the operation
will be restored to X. Similarly, an increase of
m in the speed, caused by a disturbance, will make the
load torque ( T1 ) greater than the motor torque ( T ), resulting in a deceleration and restoration of the point
of operation to X. Hence, at point X, the system is stable.
In the second case, a decrease in the speed causes the load torque ( T1 ) to become greater than the motor
torque ( T ), the drive decelerates and the operating point moves away from Y. Similarly, an increase in
the speed will make the motor torque ( T ) greater than the load torque ( T1 ), which will move the
operating point further away from Y. Thus, at point Y, the system is unstable.
This shows that, while in the first case, the motor selection for driving the given load is the right one; in
the second case, the selected motor is not the right choice and requires changing for driving the given
load.
89
Figure 6.27
6.4.4 Starting methods for induction motors
Squirrel-cage rotor
(i) Direct-on-line starting
With this method, starting current is high and may
cause interference with supplies to other consumers.
(ii) Auto transformer starting
With this method, an auto transformer is used to
reduce the stator voltage, E1, and thus the starting
current. However, the starting torque is seriously
reduced, so the voltage is reduced only sufficiently
to give the required reduction of the starting current.
A typical arrangement is shown in Figure 6.28. A
double-throw switch connects the auto transformer
in circuit for starting, and when the motor is up to
Figure 6.28
speed the switch is moved to the run position which
connects the supply directly to the motor.
(iii) Star-delta starting
With this method, for starting, the connections to the stator phase winding are star-connected, so that the
voltage across each phase winding is 1 3 (i.e. 0.577) of the line voltage and the line current which
equals the phase current is one third of the line current of the delta connection . For running, the windings
are switched to delta-connection. This method of starting is less expensive than by auto transformer.
Wound rotor
When starting on load is necessary, a wound
rotor induction motor must be used. This is
because maximum torque at starting can be
obtained by adding external resistance to the
rotor circuit via slip rings. A face-plate type
starter is used, and as the resistance is
gradually
reduced,
the
machine
characteristics at each stage will be similar
to Q, S, R and P of Figure 6.29. At each
resistance step, the motor operation will
transfer from one characteristic to the next
so that the overall starting characteristic will
be as shown by the bold line in Figure 6.29.
For very large induction motors, very
gradual and smooth starting is achieved by a
liquid type resistance.
Figure 6.29
90
6.4.5 Changing direction of rotation of a three phase induction motor
Changing the direction of rotation in three phase induction motors is simply done by changing the
direction of the rotating field. It simply achieved by changing two feeding phases by each other keeping
the third feeding phase in its place. As an example changing phase R with phase Y keeping phase B in its
place. There are some mechanical or electronic circuits that can change two phases manually or
automatically as requested. In every time of changing direction of rotation the motor must be braked
before changing circuit begins.
6.4.6 Induction motor losses and efficiency
Figure 6.30 summarizes losses in induction motors.
Figure 6.30
P2  P1  Stator losses
(6.11)
Pm  P2  Rotor copper losses
(6.12)
PO  Pm  ( friction and windage loss  core loss  stray load loss )
where, P1 is the input power to the motor,
P2 is the input power to the rotor, or the air-gap power
Pm is the developed mechanical power of the rotor ,
and PO is the output power of the rotor
Rotor copper losses = s (Rotor input power ( P2 ))
Pm = (1 - s) Rotor input power
P1  3 VL I L cos  for 3-phase induction motor
PO  T  r
where , T is the developed torque in N.m
and r is rotor angular velocity
Efficiency  
(6.13)
(6.14)
(6.15)
(6.16)
Output power PO

x 100 %
Input Power P1
(6.17)
91
Solved Examples:
Example 6.7: The power supplied to a three-phase induction motor is 32 kW and the stator losses are
1200 W. If the slip is 5%, determine (a) the rotor copper loss, (b) the total mechanical power developed
by the rotor, (c) the output power of the motor if friction and windage losses are 750 W, and (d) the
efficiency of the motor, neglecting rotor iron loss.
Solution :
(a) Input power to rotor = stator input power - stator losses
= 32 kW - 1.2 kW = 30.8 kW
As, Rotor copper losses = s (Rotor input power ( P2 )) then
rotor copper loss = (0.05)(30.8) = 1.54 kW
(b) Total mechanical power developed by the rotor
= rotor input power - rotor losses
= 30.8 - 1.54 = 29.26 kW
from which,
Example 6.8: An induction motor draws 25A from a 460 V, three phase line at a power factor of 0.85,
lagging. The stator copper loss is 1000 W, and the rotor copper loss is 500 W. The rotational losses are
friction and windage = 250 W, core loss = 800 W, and stray load loss = 200 W. Calculate (a) the air-gap
power P2 , (b) the developed mechanical power Pm , (c) the output horsepower and (d) the efficiency.
Solution :
(a) the air-gap power, P2  P1  Stator losses
= 3 (460)(25)(0.85) – 1000 = 16931 – 1000 = 15931 W
(b) the developed mechanical power, Pm  P2  Rotor copper losses
= 15931 – 500 = 15431 W
P

P

Friction
and windage losses
(c) the output power, O
m
= 15431 – ( 250 + 800 + 200)
= 15431 – 1250 = 14181 W
P
the output horsepower = O = 19 hp
746
Output power PO
14181

x 100 % =
(d) Efficiency  
x 100 = 83.8 %
16931
Input Power P1
Example 6.9: If the frequency of the source in Example 6.8 is 50 Hz, and the motor has four poles, find
(a) the slip, (b) the operating speed, and (c) the output torque.
Solution:
(a) Rotor copper losses = s (Rotor input power ( P2 ))
rotor copper loss
500

Then the slip, s =
= 0.0314
rotor input power 15931
(b) the synchronous speed n S 
 nS  nr
As the slip, s = 
 n
S

f
50
=
= 25 rev/s = 1500 rev/min
p
2



Then the rotor speed, nr  ns (1  s) = 1500(1 – 0.0314) = 1452.9 rev/min
(c) the output power, PO  T  r
Then, the output torque T 
PO
r
 r   s (1  s)
92
2  (50)
2 f
=
= 157.08 rad/s
2
p
r  157.08 (1  0.0314) = 152.15 rad/s
s 
the output torque, T 
PO
r
=
14181
= 93.2 N.m
152.15
Example 6.10: In a 3-phase, 380 V, Δ-stator connection, 6-pole, 50-Hz induction motor the useful torque
on full-load is 135.66 N.m on a slip = 0.03. The total mechanical power loss is 1.2 hp and the stator
copper loss is 750 W. Calculate (a) the rotor copper loss , (b) the motor input power (c) efficiency, and (d)
the line input current if the motor operates on a 0.84 p.f .
Solution :
(a) the output power, PO  T  r
2  (50)
2 f
=
= 104.72 rad/s
s 
3
p
 r   s (1  s)
r  104.72 (1  0.03) = 101.58 rad/s
the output power, PO  T  r = 135.66 x 101.58 = 13780.3 W
the total mechanical power loss = 1.2 x 746 = 895.2 W
the developed mechanical power of the rotor Pm = PO + total mechanical power loss
= 13780.3 + 895.2 = 14675.5 W
As, Pm = (1 - s) Rotor input power , and
Rotor copper losses = s (Rotor input power ( P2 ))
rotor copper loss
s

then,
rotor output power ( Pm ) 1  s
s
Rotor copper losses =
rotor developed mechanical power Pm
1 s
0.03
=
x 14675.5 = 453.9 W
1  0.03
(b) Stator losses = 750 W
The motor input power = the developed mechanical power + rotor copper losses + stator losses
= 14675.5 + 453.9 + 750
= 15879.4 W
Output power PO
13780.3

x 100 % =
(c) The efficiency  
x 100 = 86.78 %
15879.4
Input Power P1
(d) The input power P1  3 VL I L cos 
Then, the line current I L 
P1
3 VL cos 

15879.4
3 (380)(0.84)
93
= 28.72 A
6.5 Single-Phase Induction Motor
There are probably more single-phase AC induction motors in use today than the total of all the other
types put together. It is logical that the least expensive, lowest maintenance type motor should be used
most often. The single-phase AC induction motor best fits this description.
As the name suggests, this type of motor has only one stator winding (main winding) and operates with a
single-phase power supply. In all single-phase induction motors, the rotor is the squirrel cage type.
The single-phase induction motor is not self-starting. When the motor is connected to a single-phase
power supply, the main winding carries an alternating current. This current produces a pulsating magnetic
field. Due to induction, the rotor is energized. As the main magnetic field is pulsating, the torque
necessary for the motor rotation is not generated. This will cause the rotor to vibrate, but not to rotate.
Hence, the single There are probably more single-phase AC induction motors in use today than the total
of all the other types put together. It is logical that the least expensive, lowest maintenance type motor
should be used most often. The single-phase AC induction motor best fits this description. As the name
suggests, this type of motor has only one stator winding (main winding) and operates with a single-phase
power supply. In all single-phase induction motors, the rotor is the squirrel cage type.
The single-phase induction motor is not self-starting. When the motor is connected to a single-phase
power supply, the main winding carries an alternating current. This current produces a pulsating magnetic
field. Due to induction, the rotor is energized. As the main magnetic field is pulsating, the torque
necessary for the motor rotation is not generated. This will cause the rotor to vibrate, but not to rotate.
Hence, the single phase induction motor is required to have a starting mechanism that can provide the
starting kick for the motor to rotate.
The starting mechanism of the single-phase induction motor is mainly an additional stator winding (start/
auxiliary winding) as shown in Figure 6.31. The start winding can have a series capacitor and/or a
centrifugal switch. When the supply voltage is applied, current in the main winding lags the supply
voltage due to the main winding impedance. At the same time, current in the start winding leads/lags the
supply voltage depending on the starting mechanism impedance. Interaction between magnetic fields
generated by the main winding and the starting mechanism generates a resultant magnetic field rotating in
one direction. The motor starts rotating in the direction of the resultant magnetic field.
Once the motor reaches about 75% of its rated speed, a centrifugal switch disconnects the start winding.
From this point on, the single-phase motor can maintain sufficient torque to operate on its own.
Except for special capacitor start/capacitor run types, all single-phase motors are generally used for
applications up to 3/4 hp only.
Depending on the various start techniques, single phase AC induction motors are further classified as
described in the following sections.
Single-Phase AC Induction Motor
without Start Mechanism
Single-Phase AC Induction Motor
with Start Mechanism
Figure 6.31
94
6.5.1 Split-Phase AC Induction Motor
The split-phase motor is also known as an induction start/induction run motor. It has two windings: a start
and a main winding. The start winding is made with
smaller gauge wire and fewer turns, relative to the main
winding to create more resistance, thus putting the start
winding’s field at a different angle than that of the main
winding which causes the motor to start rotating. The main
winding, which is of a heavier wire, keeps the motor
running the rest of the time. Figure 6.32 shows a typical
AC split-phase induction motor.
The starting torque is low, typically 100% to 175% of the
rated torque. The motor draws high starting current,
approximately 700% to 1,000% of the rated current. The
maximum generated torque ranges from 250% to 350% of
Figure 6.32
the rated torque (see Figure 6.37 for torque-speed curve).
Good applications for split-phase motors include small
grinders, small fans and blowers and other low starting torque applications with power needs from 1/20 to
1/3 hp. Avoid using this type of motor in any applications requiring high on/off cycle rates or high torque.
6.5.2 Capacitor Start AC Induction Motor
This is a modified split-phase motor with a capacitor in series with the start winding to provide a start
“boost.” Like the split-phase motor, the capacitor start motor also has a centrifugal switch which
disconnects the start winding and the capacitor when the motor reaches about 75% of the rated speed.
Figure 6.33 shows a typical capacitor start AC induction motor.
Since the capacitor is in series with the start circuit, it creates more starting torque, typically 200% to
400% of the rated torque. And the starting current,
usually 450% to 575% of the rated current, is much
lower than the split-phase due to the larger wire in the
start circuit. Refer to Figure 6.37 for torque-speed
curve.
A modified version of the capacitor start motor is the
resistance start motor. In this motor type, the starting
capacitor is replaced by a resistor. The resistance start
motor is used in applications where the starting torque
requirement is less than that provided by the capacitor
start motor. Apart from the cost, this motor does not
offer any major advantage over the capacitor start
Figure 6.33
motor.
They are used in a wide range of belt-drive
applications like small conveyors, large blowers and pumps, as well as many direct-drive or geared
applications.
6.5.3 Permanent Split Capacitor (Capacitor Run) AC
Induction Motor
A permanent split capacitor (PSC) motor has a run type
capacitor permanently connected in series with the start
winding. This makes the start winding an auxiliary winding
once the motor reaches the running speed. Since the run
capacitor must be designed for continuous use, it cannot
provide the starting boost of a starting capacitor. The
95
Figure 6.34
typical starting torque of the PSC motor is low, from 30% to 150% of the rated torque. PSC motors have
low starting current, usually less than 200% of the rated current, making them excellent for applications
with high on/off cycle rates. Figure 6.34 shows a typical PSC induction motor. Refer to Figure 6.37 for
torque-speed curve.
Permanent split-capacitor motors have a wide variety of applications depending on the design. These
include fans, blowers with low starting torque needs and intermittent cycling uses, such as adjusting
mechanisms, gate operators and garage door openers.
6.5.4 Capacitor Start/Capacitor Run AC
Induction Motor
This motor has a start type capacitor in series with
the auxiliary winding like the capacitor start
motor for high starting torque. Like a PSC motor,
it also has a run type capacitor that is in series
with the auxiliary winding after the start capacitor
is switched out of the circuit. This allows high
overload torque. Figure 6.35 shows a typical
capacitor start / capacitor run AC induction motor.
This type of motor can be designed for lower fullFigure 6.35
load currents and higher efficiency (see Figure
6.37 for torque-speed curve). This motor is costly
due to start and run capacitors and centrifugal switch.
It is able to handle applications too demanding for any other kind of single-phase motor. These include
woodworking machinery, air compressors, high-pressure water pumps, vacuum pumps and other high
torque applications requiring 1 to 10 hp.
6.5.5 Shaded-Pole AC Induction Motor
Shaded-pole motors have only one main winding
and no start winding. Starting is by means of a
design that rings a continuous copper loop around
a small portion of each of the motor poles. This
“shades” that portion of the pole, causing the
magnetic field in the shaded area to lag behind the
field in the un-shaded area. The reaction of the
two fields gets the shaft rotating. Figure 6.36
shows a typical shaded-pole AC induction motor.
Because the shaded-pole motor lacks a start
winding, starting switch or capacitor, it is
Figure 6.36
electrically simple and inexpensive. Also, the
speed can be controlled merely by varying
voltage, or through a multi-tap winding. Mechanically, the shaded-pole motor construction allows highvolume production. In fact, these are usually considered as “disposable” motors, meaning they are much
cheaper to replace than to repair.
The shaded-pole motor has many positive features but it also has several disadvantages. It’s low starting
torque is typically 25% to 75% of the rated torque. It is a high slip motor with a running speed 7% to 10%
below the synchronous speed. Generally, efficiency of this motor type is very low (below 20%).
The low initial cost suits the shaded-pole motors to low horsepower or light duty applications. Perhaps
their largest use is in multi-speed fans for household use. But the low torque, low efficiency and less
sturdy mechanical features make shaded-pole motors impractical for most industrial or commercial use,
where higher cycle rates or continuous duty are the norm.
Figure 6.37 shows the torque-speed curves of various kinds of single-phase AC induction motors, while
Figure 6.38 shows the phase relationship of various kinds of single-phase AC induction motors.
96
Figure 6.37
Phase Relationship
of Split Phase Motor
Phase Relationship
of Capacitor Start Motor
Phase Relationship
of PSC Motor
Figure 6.38
6.5.5 Universal Motor
A variant of the wound field DC motor is the universal motor. The name derives from the fact that it may
use AC or DC supply current, although in practice they are nearly always used with AC supplies. The
principle is that in a series wound field DC motor the current in both the field and the armature (and
hence the resultant magnetic fields) will alternate (reverse polarity) at the same time, and hence the
mechanical force generated is always in the same direction. In practice, the motor must be specially
designed to cope with the AC current (impedance must be taken into account, as must the pulsating
force), and the resultant motor is generally less efficient than an equivalent pure DC motor. Operating at
normal power line frequencies, the maximum output of universal motors is limited and motors exceeding
one kilowatt are rare. But universal motors also form the basis of the traditional railway traction motor in
electric railways. In this application, to keep their electrical efficiency high, they were operated from very
low frequency AC supplies, with 25 Hz and 16 2/3 hertz operation being common. Because they are
universal motors, locomotives using this design were also commonly capable of operating from a third
rail powered by DC.
The advantage of the universal motor is that AC supplies may be used on motors which have the typical
characteristics of DC motors, specifically high starting torque and very compact design if high running
speeds are used. The negative aspect is the maintenance and short life problems caused by the
97
commutator. As a result such motors are usually used in AC devices such as food mixers and power tools
which are used only intermittently. Continuous speed control of a universal motor running on AC is very
easily accomplished using a thyristor circuit, while stepped speed control can be accomplished using
multiple taps on the field coil. Household blenders that advertise many speeds frequently combine a field
coil with several taps and a diode that can be inserted in series with the motor (causing the motor to run
on half-wave rectified AC).
Universal motors can rotate at relatively high revolutions per minute (rpm). This makes them useful for
appliances such as blenders, vacuum cleaners, and hair dryers where high-speed operation is desired.
Many vacuum cleaner and weed trimmer motors exceed 10,000 rpm, Dremel and other similar miniature
grinders will often exceed 30,000 rpm. Motor damage may occur due to over-speed (rpm in excess of
design specifications) if the unit is operated with no significant load. On larger motors, sudden loss of
load is to be avoided, and the possibility of such an occurrence is incorporated into the motor's protection
and control schemes. Often, a small fan blade attached to the armature acts as an artificial load to limit the
motor speed to a safe value, as well as provide cooling airflow to the armature and field windings.
6.5.6 Changing direction of rotation of a single phase induction motor
Changing the direction of rotation in single phase induction motors is simply done by changing the
direction of the rotating fields. It is achieved by changing the direction of one of the magnetic fields of
two windings. It can be done by changing the direction of current in auxiliary winding by making the
current to enter from the other side of the winding. There are some mechanical or electronic circuits that
can change the current to enter from the other side of the auxiliary winding manually or automatically as
requested. In every time of changing direction of rotation the motor must be braked before changing
circuit begins.
6.5.7 Single-phase Induction motor losses and efficiency
The losses and efficiency of single phase motors is the same that of three phase motors, which can be
understood from Figure 6.30.
P2  P1  Stator losses
(6.18)
Pm  P2  Rotor copper losses
(6.19)
PO  Pm  ( friction and windage loss  core loss  stray load loss )
(6.20)
where, P1 is the input power to the motor,
P2 is the input power to the rotor, or the air-gap power
Pm is the developed mechanical power of the rotor ,
and PO is the output power of the rotor
Rotor copper losses = s (Rotor input power ( P2 ))
Pm = (1 - s) Rotor input power
(6.21)
(6.22)
P1  VT I L cos  for single-phase induction motor
PO  T  r
(6.23)
(6.24)
where , T is the developed torque in N.m
and r is rotor angular velocity
Efficiency  
Output power PO

x 100 %
Input Power P1
(6.25)
98
Solved Examples:
Example 6.11: The name plate of a single phase, 4-pole induction motor gives the following data: Output
½ hp; 230 V; 50Hz; input current 2.9 A; Power factor 0.71; speed 1410 rpm. Calculate (a) the efficiency,
(b) the slip of the motor when delivering the rated output. If a capacitor of 15.9 µF is connected across the
motor while delivering the rated output , calculate (c) the new current taken by the motor, and (d) the new
power factor.
Solution :
Rated output power PO = ½ x 746 = 373 W
Input power P1  VT I L cos  = 230 x 2.9 x 0.71 = 473.6 W
(a) the efficiency  
=
Output power PO

x 100 %
Input Power P1
373
x 100 = 78.8%
473.6
 nS  nr
(b) slip, s = 
 n
S

nS =



f
50
=
= 25 rev/s = 1500 rpm
p
2
nr = 1410 rpm
s=
1500  1410
= 0.06
1500
(c) As shown in Figure 6.39 (a), the capacitor has been connected across the motor.
(a)
(b)
Figure 6.39
XC 
IC 
1
1
=
= 200 Ω
2 f c 2  (50)(15.9 x 10 6 )
230
= 1.15 A
200
This current leads the applied voltage by 90° as shown in Figure 6.39 (b). The motor current
I m lags
the voltage by the angle m  cos (0.71) = 45°
1
The motor current
I m and capacitor current I C when combined vectorially give the line current I L
The x-component = I m cos m = 2.9 x 0.71 = 2.06 A
99
I C - I m sin m = 1.15 - 2.06 = - 0.91 A
The y-component =
I L = (20.6) 2  (0.91) 2 = 2.25 A
(d) the new power factor =
2.06
= 0.916 lagging
2.25
Example 6.12: A PSC single phase, 4-pole induction motor has the following data: Output 2.5 hp; 220 V;
50Hz; It is found that it drives a current for the main winding
I m = 10   30 A, and drives a current for
the auxiliary winding I a = 4 75 A at full load and rotor speed = 1440 rpm. Calculate (a) the efficiency,
(b) the slip of the motor when delivering the rated output.
Solution:
The x-component of
IL =
I m cos m + I a cos a
= 10 cos (-30) + 5 cos (75) = 9.954 A
The y-component of
I L = I m sin m + I a sin a
= 10 sin (-30) + 4 sin (75) = - 1.136 A
I L = (9.954) 2  (1.136) 2 = 10.02 A
  1.136 
 = - 6.5°
 9.954 
Input power P1  VT I L cos  = 220 x 10.02 x cos (6.5) = 2190.23 W
Rated output power PO = 2.5 x 746 = 1865 W
  tan 1 
(a) the efficiency  
=
Output power PO

x 100 %
Input Power P1
1865
x 100 = 85.2 %
2190.23
 nS  nr
(b) slip, s = 
 n
S

nS =



f
50
=
= 25 rev/s = 1500 rpm
p
2
nr = 1410 rpm
s=
1500  1440
= 0.04
1500
100
UNIT 7
Switching and Controlling of AC Motors
Introduction
This unit introduces the construction of relay, contactor and timer. It introduces also, switching on-off and
controlling of three phases motors, switching three phases motors star-delta, switching three phases
motors with timers and switching single phase motors
7.1 Relay
A relay is used to isolate one electrical circuit from another. It allows a low current control circuit to
make or break an electrically isolated high current circuit path.
7.1.1 Construction
The basic relay consists of a coil and a set of contacts. The most common relay coil is a length of magnet
wire wrapped around a metal core. When voltage is applied to the coil, current passes through the wire
and creates a magnetic field. This magnetic field pulls the contacts together and holds them there until the
current flow in the coil has stopped. In Figure 7.1, the relay's coil is energized by the low-voltage (12
VDC) source, while the single-pole, single-throw contact interrupts the high-voltage (220 VAC) circuit.
It is quite likely that the current required to energize the relay coil will be hundreds of times less than the
current rating of the contact. Typical relay coil currents are well below 1 amp, while typical contact
ratings for industrial relays are at least 10 amps.
Figure 7.1
One relay coil/armature assembly may be used to actuate more than one set of contacts. Those contacts
may be normally-open (NO), normally-closed (NC), or any combination of the two. As with switches, the
"normal" state of a relay's contacts is that state when the coil is de-energized, just as you would find the
relay sitting on a shelf, not connected to any circuit.
There are two specifications that you must consider when selecting a relay, the coil voltage and the
current carrying capability of contacts. The coil voltage for relays may be DC voltage with rating 6, 9, 12,
and 24 V, or AC voltages with ratings 220 V. This means that if you apply 12 volts to the coil, it will pull
in and stay there until the applied voltage is removed from the coil. The current rating on relay contacts
tells how much current can be passed through the contacts without damage to the contacts. Some relays
have different current ratings for the NC contacts (which are held together by spring tension) and the NO
contacts (which are held together by the electromagnet). If you need to pass significant current through
the NC contacts, you may want to check the manufacturers specifications for the relay.
Pull in Voltage: The pull in voltage is the minimum voltage required for the relay coil to pull the contacts
together.
Drop out Voltage: The drop out voltage is the voltage at which the energized coil will release the
movable contact.
101
Coil Resistance: In a DC relay coil, the coil resistance determines the current flow through the coil. In an
AC relay coil, the resistance does not solely determine the current flow through the coil because the coil
has inductance. The inductive reactance along with the DC resistance work together to limit the current
flow through the coil.
7.1.2 Types of relay
(a) Latching relay:
A latching relay has two relaxed states (bistable). These are also called 'keep' or 'stay' relays. When the
current is switched off, the relay remains in its last state. This is achieved with a solenoid operating a
ratchet and cam mechanism, or by having two opposing coils with an over-center spring or permanent
magnet to hold the armature and contacts in position while the coil is relaxed, or with a remnant core. In
the ratchet and cam example, the first pulse to the coil turns the relay on and the second pulse turns it off.
In the two coil example, a pulse to one coil turns the relay on and a pulse to the opposite coil turns the
relay off. This type of relay has the advantage that it consumes power only for an instant, while it is being
switched, and it retains its last setting across a power outage.
(b) Reed relay
A reed relay has a set of contacts inside a vacuum or inert gas filled glass tube, which protects the
contacts against atmospheric corrosion. The contacts are closed by a magnetic field generated when
current passes through a coil around the glass tube. Reed relays are capable of faster switching speeds
than larger types of relays, but have low switch current and voltage ratings.
(c) Mercury-wetted relay
A mercury-wetted reed relay is a form of reed relay in which the contacts are wetted with mercury. Such
relays are used to switch low-voltage signals (one volt or less) because of its low contact resistance, or for
high-speed counting and timing applications where the mercury eliminates contact bounce. Mercury
wetted relays are position-sensitive and must be mounted vertically to work properly. Because of the
toxicity and expense of liquid mercury, these relays are rarely specified for new equipment.
(d) Machine tool relay
A machine tool relay is a type standardized for industrial control of machine tools, transfer machines, and
other sequential control. They are characterized by a large number of contacts (sometimes extendable in
the field) which are easily converted from normally-open to normally-closed status, easily replaceable
coils, and a form factor that allows compactly installing many relays in a control panel. Although such
relays once were the backbone of automation in such industries as automobile assembly, the
programmable logic controller (PLC) mostly displaced the machine tool relay from sequential control
applications.
(e) Contactor relay
A contactor is a very heavy-duty relay used for switching electric motors and lighting loads. With high
current, the contacts are made with pure silver. The unavoidable arcing causes the contacts to oxidize and
silver oxide is still a good conductor. Such devices are often used for motor starters. A motor starter is a
contactor with overload protection devices attached. The overload sensing devices are a form of heat
operated relay where a coil heats a bi-metal strip, or where a solder pot melts, releasing a spring to
operate auxiliary contacts. These auxiliary contacts are in series with the coil. If the overload senses
excess current in the load, the coil is de-energized. Contactor relays can be extremely loud to operate,
making them unfit for use where noise is a chief concern.
(f) Buchholz relay
A Buchholz relay is a safety device sensing the accumulation of gas in large oil-filled transformers, which
will alarm on slow accumulation of gas or shut down the transformer if gas is produced rapidly in the
transformer oil.
102
(g) Forced-guided contacts relay
A forced-guided contacts relay has relay contacts that are mechanically linked together, so that when the
relay coil is energized or de-energized, all of the linked contacts move together. If one set of contacts in
the relay becomes immobilized, no other contact of the same relay will be able to move. The function of
forced-guided contacts is to enable the safety circuit to check the status of the relay. Forced-guided
contacts are also known as "positive-guided contacts", "captive contacts", "locked contacts", or "safety
relays".
(i) Solid-state relay
A solid state relay (SSR) is a solid state electronic component that provides a similar function to an
electromechanical relay but does not have any moving components, increasing long-term reliability. With
early SSR's, the tradeoff came from the fact that every transistor has a small voltage drop across it. This
collective voltage drop limited the amount of current a given SSR could handle. As transistors improved,
higher current SSR's, able to handle 100 to 1,200 amps, have become commercially available. Compared
to electromagnetic relays, they may be falsely triggered by transients.
(j) Overload protection relay
One type of electric motor overload protection relay is operated by a heating element in series with the
electric motor . The heat generated by the motor current operates a bi-metal strip or melts solder,
releasing a spring to operate contacts. Where the overload relay is exposed to the same environment as
the motor, a useful though crude compensation for motor ambient temperature is provided.
7.2 Contactor
A contactor is an electrically controlled switch (relay) used for switching a power circuit. A contactor is
activated by a control input which is a lower voltage / current than that which the contactor is switching.
Contactors come in many forms with varying capacities and features. Unlike a circuit breaker a contactor
is not intended to interrupt a short circuit current.( See Figure 7.2 )
Contactors range from having a breaking current of several amps and 220 volts to thousands of amps and
many kilovolts. The physical size of contactors ranges from a few inches to the size of a small car.
Contactors are used to control electric motors, lighting, heating, capacitor banks, and other electrical
loads.
Figure 7.2
7.2.1 Construction
A contactor is composed of three different systems. The contact system is the current carrying part of the
contactor. This includes Power Contacts, Auxiliary Contacts, and Contact Springs. The electromagnet
system provides the driving force to close the contacts. The enclosure system is a frame housing the
contact and the electromagnet. Enclosures are made of insulating materials like Bakelite, Nylon 6, and
thermosetting plastics to protect and insulate the contacts and to provide some measure of protection
103
against personnel touching the contacts. Open-frame contactors may have a further enclosure to protect
against dust, oil, explosion hazards and weather.
Contactors used for starting electric motors are commonly fitted with overload protection to prevent
damage to their loads. When an overload is detected the contactor is tripped, removing power
downstream from the contactor.
Some contactors are motor driven rather than relay driven and high voltage contactors (greater than 1000
volts) often have arc suppression systems fitted (such as a vacuum or an inert gas surrounding the
contacts).
Magnetic blowouts are sometimes used to increase the amount of current a contactor can successfully
break. The magnetic field produced by the blowout coils force the electric arc to lengthen and move away
from the contacts. The magnetic blowouts in the pictured Albright contactor more than double the current
it can break from 600 Amps to 1500 Amps.
Sometimes an Economizer circuit is also installed to reduce the power required to keep a contactor
closed. A somewhat greater amount of power is required to initially close a contactor than is required to
keep it closed thereafter. Such a circuit can save a substantial amount of power and allow the energized
coil to stay cooler. Economizer circuits are nearly always applied on direct-current contactor coils and on
large alternating current contactor coils.
Contactors are often used to provide central control of large lighting installations, such as an office
building or retail building. To reduce power consumption in the contactor coils, two coil latching
contactors are used. One coil, momentarily energized, closes the power circuit contacts; the second opens
the contacts.
A basic contactor will have a coil input (which may be driven by either an AC or DC supply depending
on the contactor design) and generally a minimum of two poles which are controlled.
The top three contacts switch the respective phases of the incoming 3-phase AC power, typically at least
380 Volts for motors 1 horsepower or greater. The lowest contact is an "auxiliary" contact which has a
current rating much lower than that of the large motor power contacts, but is actuated by the same
armature as the power contacts. The auxiliary contact is often used in a relay logic circuit, or for some
other part of the motor control scheme, typically switching 220 Volt AC power instead of the motor
voltage. One contactor may have several auxiliary contacts, either normally-open or normally-closed, if
required.
Three-phase, 380 volt AC power comes in to the three normally-open contacts at the top of the contactor
via screw terminals labeled "L1," "L2," and "L3". Power to the motor exits the overload heater assembly
at the bottom of this device via screw terminals labeled "T1," "T2," and "T3."
7.2.2 Operating Principle
Unlike general-purpose relays, contactors are designed to be directly connected to high-current load
devices, not other control devices. Relays tend to be of much lower capacity and are usually designed for
both Normally Closed and Normally Open applications. Devices switching more than 15 amperes or in
circuits rated more than a few kilowatts are usually called contactors. Apart from optional auxiliary low
current contacts, contactors are almost exclusively fitted with Normally Open contacts.
When current passes through the electromagnet, a magnetic field is produced which attracts ferrous
objects, in this case the moving core of the contactor is attracted to the stationary core. Since there is an
air gap initially, the electromagnet coil draws more current initially until the cores meet and reduct the
gap, increasing the inductive impedance of the circuit.
For contactors energized with alternating current, a small part of the core is surrounded with a shading
coil, which slightly delays the magnetic flux in the core. The effect is to average out the alternating pull of
the magnetic field and so prevent the core from buzzing at twice line frequency.
Most motor control contactors at low voltages (600 volts and less) are "air break" contactors, since
ordinary air surrounds the contacts and extinguishes the arc when interrupting the circuit. Modern
medium-voltage motor controllers use vacuum contactors.
Motor control contactors can be fitted with short-circuit protection (fuses or circuit breakers),
disconnecting means, overload relays and an enclosure to make a combination starter. In large industrial
plants many contactors may be assembled in motor control centers.
104
7.2.3 Ratings
Contactors are rated by designed load current per contact (pole), maximum fault withstand current, duty
cycle, voltage, and coil voltage. A general purpose motor control contactor may be suitable for heavy
starting duty on large motors; so-called "definite purpose" contactors are carefully adapted to such
applications as air-conditioning compressor motor starting. North American and European ratings for
contactors follow different philosophies, with North American contactors generally emphasizing
simplicity of application while European rating philosophy emphasizes design for the intended life cycle
of the application. A contactor basically consists of two parts; signaling and actual.
7.3 Timer
A timer is a specialized type of clock. A timer can be used to control the sequence of an event or process.
A simple digital timer. The internal components—including the circuit board with control chip and LED
display, a battery, and a buzzer—are visible.
Timers can be mechanical, electromechanical, digital, or even software, since most computers include
digital timers of one kind or another.
7.3.1 Mechanical Timers
Early mechanical timers used typical clockwork mechanisms, such as an escapement and spring to
regulate their speed. Inaccurate, cheap mechanisms use a flat beater that spins against air resistance.
Mechanical egg-timers are usually of this type.
More accurate mechanisms resemble small alarm clocks. The chief advantage is that they require little
battery/electrical power, and can be stored for long periods of time. The most widely-known application
is to control explosives.
7.3.2 Electromechanical timers
Electromechanical timers have two types. A thermal type has a metal finger made of two metals with
different rates of thermal expansion (steel and bronze are common). An electric current flows through this
finger, and heats it. One side expands less than the other, and an electrical contact on the end of the finger
moves away from an electrical switch contact, or makes a contact (both types exist). The most common
use of this type is now in the "flasher" units that flash turn signals in automobiles, or sometimes in
Christmas lights.
Another type of electromechanical timer (a cam timer) uses a small synchronous AC motor turning a cam
against a comb of switch contacts. The AC motor is turned at an accurate rate by the alternating current,
which power companies carefully regulate. Gears slow this motor down to the desired rate, and turn the
cam. The most common application of this timer now is in washers, driers and dishwashers. This type of
timer often has a friction clutch between the gear train and the cam, so that the cam can be turned to reset
the time.
Electromechanical timers survive in these applications because mechanical switch contacts are still less
expensive than the semiconductor devices needed to control powerful lights, motors and heaters.
In the past these electromechanical timers were often combined with electrical relays to create electromechanical controllers. Electromechanical timers reached a high state of development in the 1950s and
60s because of their extensive use in aerospace and weapons systems. Programmable electromechanical
timers controlled launch sequence events in early rockets and ballistic missiles.
7.3.3 Digital Timers
Digital timers can achieve higher precision than mechanical timers because they are quartz clocks with
special electronics. Integrated circuits have made digital logic so inexpensive that an electronic digital
timer is now less expensive than many mechanical and electromechanical timers. Individual timers are
implemented as a simple single-chip computer system, similar to a watch. Watch technology is used in
these devices.
However, most timers are now implemented in software. Modern controllers use a programmable logic
controller rather than a box full of electromechanical parts. The logic is usually designed as if it were
105
relays, using a special computer language called ladder logic. In PLCs, timers are usually simulated by
the software built into the controller. Each timer is just an entry in a table maintained by the software.
Digital timers can also be used in safety device such as a Gas Timer.
7.3.4 Computer timers
Most computer systems have one to sixteen electronic timers. These are usually just digital counters that
are set to a number by software, and then count down to zero. When they reach zero, they interrupt the
computer.
Another common form of timer is a number that is compared to a counter. This is somewhat harder to
program, but can be used to measure events or control motors.
Embedded systems often use a hardware timer to implement a list of software timers. Basically, the
hardware timer is set to expire at the time of the next software timer of a list of software timers. The
hardware timer's interrupt software handles the house-keeping of notifying the rest of the software,
finding the next software timer to expire, and resetting the hardware timer to the next software timer's
expiration.
7.4 Switching and Controlling of Three Phase Induction Motors
For every case there are two circuits, one is for control and the other is for power or working of motor.
Also, switching on the motor is achieved by a bush-button switch, in which its contacts are normally
opened, and switching off the motor is achieved by a bush-button switch, in which its contacts are
normally closed.
7.4.1 Switching Three Phase Induction Motors ON-OFF
Actuation of pushbutton ON energizes the coil of contactor C1. The contactor switches on the motor and
maintains itself after the button is enables via its own auxiliary contact C1/14-13 and pushbutton OFF
(three-wire control contact). Contactor C1 is de-energized, in the normal course of events, by actuation of
pushbutton OFF. In the event of an overload, it is de-energized via the normally closed contact 95-96 on
the overload relay F2. The coil current is interrupted, and contactor C1 switches the motor off. Figure 7.3
shows the control circuit while Figure 7.4 shows the power circuit.
Figure 7.3
Figure 7.4
106
7.4.2 Switching Three Phase Induction Motors ON For A Fixed Period
Actuation of pushbutton ON energizes the bridging contactor C1 which then maintains itself after the
button is enables via its own auxiliary contact C1/14-13, normally closed contacts of timing relay T1/1615 and pushbutton OFF. At the same time, voltage is applied to the timing relay T1. When the set time
has elapsed, the bridging contactor C1 is disconnected by T1/16-15 which is opened after time elapsed.
T1 is likewise disconnected and, exactly as C1, can only be energized again after the motor has been
switched off by pressing pushbutton OFF. In the event of an overload, normally closed contact 95-96 on
the overload relay O.L effects de-energizing. Figure 7.5 shows the control circuit while Figure 7.6 shows
the power circuit.
Figure 7.5
Figure 7.6
7.4.3 Switching Three Phase Induction Motors In Two Directions
Actuation of pushbutton ON Clockwise energizes
the coil of contactor C1. It switches on the motor
running clockwise and maintains itself after
button ON Clockwise is enabled via its own
auxiliary contact C1/14-13, normally closed
contacts of C2/22-21, pushbutton OFF and
normally closed contacts of O.L 96-95. The
normally closed contact C1/22-21 (which is
opened after energizing of C1) electrically inhibits
switch on of contactor C2. When pushbutton ON
Unti-Clockwise is pressed, contactor C2 closes
(the motor runs in the anticlockwise direction).
Figure 7.7
107
It maintains itself after button ON Unti-Clockwise is enabled via its own auxiliary contact C2/14-13,
normally closed contacts of C1/22-21 and pushbutton OFF. The normally closed contact C2/22-21 (which
is opened after energizing of C2) electrically inhibits switch on of contactor C1. Depending on the circuit,
direction can be changed from clockwise to anticlockwise either after pressing pushbutton OFF, then
pressing the other ON button, or by directly pressing the pushbutton for the reverse direction. In the event
of an overload, normally closed contact 95-96 on the overload relay O.L effects de-energizing. Figure 7.7
shows the control circuit while Figure 7.8 shows the power circuit. It is recommended to switch the motor
off and bringing it to stop before switching the reverse ON pushbutton.
Figure 7.8
7.4.4 Switching Three Phase Induction Motors Star Delta Manually
Star delta power circuit consists from three contactor, the main contactor C2 , star only contactor C1, and
delta only contactor C3. First, when the motor is switched on, C1 and C2 are energized and the motor
runs and accelerates in star connection. After running to about 80% from rated speed, C1 is de-energized,
and C3 is energized making the motor to run in delta connection via C2 and C3. Figure 7.9 shows the
power circuit.
Figure 7.10 shows the control circuit for running the motor in star then delta connections manually.
Actuation of pushbutton ON energizes the bridging contactor C2 which then closes its own auxiliary
contact C2/14-13. At the same time, voltage is applied to contactor C1 making it to be energized. C2
maintains itself after the button is enables via auxiliary contact C1/14-13, its own auxiliary contact C2/1413, normally closed contacts of C1/22-21 pushbutton OFF 1 and normally closed contacts of O.L 96-95,
while C1 maintains itself after the button is enables via its own auxiliary contact C1/14-13, pushbutton
OFF 1 and normally closed contacts of O.L 96-95. The motor works in star via contactors C1 and C2.
After reaching 80% from the motor speed, pushbutton OFF2 ( off star – on delta ) is actuated which deenergize C2 making its main and 14-13 contacts to released and 22-21 contacts to return to its normally
close. At the same time, voltage is applied to contactor C3 making it to be energized. The motor then
108
works delta via contactors C3 and C1. The normally closed contact C3/22-21 (which is opened after
energizing of C3) electrically inhibits switch on of contactor C2. The motor then can't work in star unless
it is switched off by OFF pushbutton and restarted by ON pushbutton again. In the event of an overload,
normally closed contact 95-96 on the overload relay O.L effects de-energizing.
Figure 7.9
Figure 7.10
109
7.4.5 Switching Three Phase Induction Motors Star Delta Automatically With Timer
The power circuit is the same in the previous section (Figure 7.9). Figure 7.11 shows the control circuit
for running the motor in star then delta connections automatically with timer. Actuation of pushbutton
ON energizes the bridging contactor C2 which then closes its own auxiliary contacts C2/14-13. At the
same time, voltage is applied to the auxiliary contactor Ch, making it to be energized which then closes
its own auxiliary contacts Ch/14-13 making contactor C1 to be energized which then closes its own
auxiliary contacts C1/14-13. Voltage is also applied to the timing relay T1 at the same time making it to
be energized. C1 maintains itself after the button is enables via auxiliary contact C2/14-13, its own
auxiliary contact C1/14-13, normally closed contacts of T1/15-16, normally closed contacts of C3/22-21,
pushbutton OFF 1 and normally closed contacts of O.L 96-95, while C2 maintains itself after the button is
enables via its own auxiliary contact C2/14-13, auxiliary contact Ch/14-13, pushbutton OFF 1 and
normally closed contacts of O.L 96-95. The motor works in star via contactors C1 and C2.
When the set changeover time has elapsed, contactor C1 is disconnected by T1/16-15 which is opened
after time elapsed making its own auxiliary normally closed contacts C1/22-21 to be closed. At the same
time contacts T1/15-18 are closed making contactor C3 to be energized. The motor works then in delta
via contactors C3 and C2. The motor cannot be started up again unless it has previously been
disconnected by pushbutton OFF., or in the event of an overload, by the normally closed contact 95–96 of
the overload relay O.L and restarted by ON pushbutton again.
Figure 7.11
110
7.5 Switching and Controlling of Single Phase Induction Motors
For motors more than 2 hp it is recommended to use contactors and pushbuttons to switch motors on or
off, but for small motors it is not necessary to use contactors for switching.
7.5.1 Switching Single Phase Induction Motors ON-OFF or For A Fixed Period
When using contactors for switching, control circuits
for switching the motor on-off or for a fixed period of
time is the same as that for three phase motors
(Figures 7.3, and 7.5), but the power circuit uses one
line as shown in Figure 7.12
Figure 7.12
7.5.2 Switching Small Motors ON-OFF
For small motors it is not necessary to use contactors to switch the motors, but relays or switches are
used. The coil of the relay is energized from 220 V AC, or from 24 or 12 or 5 V DC according to the
voltage of the control circuit. One line of the motor is connected to the N.O. point of the relay and the
other is to neutral of the mains as shown in Figure 7.13. When the switch is closed the voltage is applied
to the coil of the relay making it to be energized and changing its contacts. As a result the AC voltage is
applied to the motor making it to run. Neutral of the motor may be connected via another relay that
switched at the same time for the main relay, or by using double circuit relay to switch phase and neutral
leads.
Figure 7.13
111
7.5.3 Switching Single Phase Induction Motors in Two Directions
A double circuit relay is used in this circuit as shown in Figure 7.14. When the switch S1 is opened the
current passes from the phase lead to the auxiliary winding then to the capacitor then to the neutral lead.
After the switch S1 is closed the current passes from the phase lead to the capacitor then to the auxiliary
winding from the other side then to the neutral lead. As a result switching-on the switch S1 makes the
motor to run in the reverse direction.
Figure 7.14
112
UNIT 8
Introduction To Protection
Introduction
This unit introduces an introduction to protection of electrical transformers and motors. It introduces a
brief presentation about protection devices like fuses, circuit breakers and over-load relays. Also, ratings
of transformers and motors calculations for protection will be explained.
8.1 Protection Devices
The over-current devices are an essential part of a power distribution system to prevent fire or damage.
When too much current flows through a wire, it may overheat and be damaged, or even start a fire.
Wiring regulations give the maximum rating of a device for protection of a particular circuit. There are
three protection devices that can be used to protect transformers and motors which are fuses, circuit
breakers and over-load relays.
8.1.1 Fuses
An expendable protective device that eliminates overload on an electric circuit. The fuse is connected in
series with the circuit being protected. The components of a typical low-voltage high-power fuse are a
fuse element or wire, an insulating material support and housing, two metal end fittings, and a filler.
The fuse element is a silver strip or wire that melts when the current is higher than the rated value. The
melting of the wire generates an electric arc. The extinction of this arc interrupts the current and protects
the circuit. The fuse element is connected to the metal end fittings which serve as terminals.
The filler facilitates the arc extinction. The most commonly used filler is sand, which surrounds the fuse
element. When the fuse element melts, the heat of the arc melts the sand near the element. This removes
energy from the arc, creating a channel filled with the mixture of melted sand and metal. The metal
particles from the melting fuse wire are absorbed by the melted sand. This increases the channel
resistance, which leads to the gradual reduction of the current and the extinction of the arc. The insulating
support and the tubular housing holds the fuse elements and the filler, which also serves as insulator after
The interruption time is the sum of the melting and the arcing time. It is inversely proportional to the
current, that is, a higher current melts the wire faster. The fuse operates in a time-current band between
maximum interruption time and minimum meeting time. It protects the electric circuit if the fault current
is interrupted before the circuit elements are overheated. The arc extinction often generates over-voltages,
which produce flashovers and damage. A properly designed fuse operates without over-voltage, which is
controlled by the shape of the fuse element and by the filler.
Fuses come in a vast array of sizes & styles to cater for the immense number of applications in which
they are used. While many are manufactured in standardized package layouts to make them easily
interchangeable, a large number of new styles are released into the marketplace every year. Fuse bodies
may be made of ceramic, glass, plastic, fiberglass, Molded Mica Laminates, or molded compressed fibre
depending on application and voltage class.
Cartridge (ferrule) fuses have a cylindrical body terminated with metal end caps. Some cartridge fuses are
manufactured with end caps of different sizes to prevent accidental insertion of the wrong fuse rating in a
holder. An example of such a fuse range is the 'bottle fuse', which in appearance resembles the shape of a
bottle.
Fuses used in circuits rated 200-600 volts and between about 10 and several thousand amperes, as used
for industrial applications such as protection of electric motors, commonly have metal blades located on
each end of the fuse. Fuses may be held by a spring loaded clip or the blades may be held by screws.
Blade type fuses often require the use of a special purpose extractor tool to remove them from the fuse
holder. While glass fuses have the advantage of a fuse element visible for inspection purposes, they have
a low breaking capacity which generally restricts them to applications of 15 A or less at 250 VAC.
Ceramic fuses have the advantage of a higher breaking capacity facilitating their use in higher
voltage/ampere circuits. Filling a fuse body with sand provides additional protection against arcing in an
over-current situation.
113
Cartridge fuses are generally measured as the overall length and diameter of the fuse. Due to the large
variety of cartridge fuses available, fuse identification relies on accurate measurements as fuses can differ
by only a few millimeters between types. 'Bottle style' cartridge fuses also require the measurement of the
cap diameter as this varies between ampere ratings. Figure 8.1 shows the construction of a cartridge fuse.
Other fuse packages can require a variety of measurements such as;
 body (width x height x depth)
 blade or tag (width x height x depth)
 overall length of the fuse (when the fuse features blades
or tags)
Figure 8.1
 overall width of the fuse (when the fuse features 2
bodies)
 width of the mounting holes (when the fuse features tags)
 distance between blades (when radially configured)
 fixing centre (when the fuse features tags - see below)
Fuses fitted with tags require the fixing centre measurement. This measurement is the distance between
the tag mounting holes on either end of the fuse as measured from the centre of each mounting hole.
Most fuses are marked on the body, or end caps to markings show their ratings. Surface mount
technology "chip type" fuses feature little or no markings making identification very difficult.
When replacing a fuse, it is important to interpret these markings correctly as fuses that may look the
same, could be designed for very different applications. Fuse markings will generally convey the
following information;
 Ampere rating of the fuse
 Voltage rating of the fuse
 Time-current characteristic ie. element speed
 Approvals
 Manufacturer / Part Number / Series
 Breaking capacity
Interrupting rating
A fuse also has a rated interrupting capacity, also called breaking capacity, which is the maximum current
the fuse can safely interrupt. Generally this should be higher than the maximum prospective short circuit
current. Miniature fuses may have an interrupting rating only 10 times their rated current. Fuses for small
low-voltage wiring systems are commonly rated to interrupt 10,000 amperes. Fuses for larger power
systems must have higher interrupting ratings, with some low-voltage current-limiting "high rupturing
capacity" (HRC) fuses rated for 300,000 amperes. Fuses for high-voltage equipment, up to 115,000 volts,
are rated by the total apparent power (megavolt-amperes, MVA) of the fault level on the circuit.
Voltage rating
As well as a current rating, fuses also carry a voltage rating indicating the maximum circuit voltage in
which the fuse can be used. For example, glass tube fuses rated 32 volts should never be used in lineoperated (mains-operated) equipment even if the fuse physically can fit the fuseholder. Fuses with
ceramic cases have higher voltage ratings. Fuses carrying a 250 V rating may be safely used in a 125 V
circuit, but the reverse is not true as the fuse may not be capable of safely interrupting the arc in a circuit
of a higher voltage. Medium-voltage fuses rated for a few thousand volts are never used on low voltage
circuits, due to their expense and because they cannot properly clear the circuit when operating at very
low voltages.
Figure 8.2(a) shows a marked fuse
and (b) shows two fuses with dimensions
identifications,
while Figure 8.3 shows 4 shapes of fuses.
(a)
(b)
Figure 8.2
114
Figure 8.3
8.1.2 Circuit Breakers
A device to open or close an electric power circuit either during normal power system operation or during
abnormal conditions. A circuit breaker serves in the course of normal system operation to energize or deenergize loads. During abnormal conditions, when excessive current develops, a circuit breaker opens to
protect equipment and surroundings from possible damage due to excess current. These abnormal
currents are usually the result of short circuits created by lightning, accidents, deterioration of equipment,
or sustained overloads.
Unlike a fuse, which operates once and then has to be replaced, a circuit breaker can be reset (either
manually or automatically) to resume normal operation. Circuit breakers are made in varying sizes, from
small devices that protect an individual household appliance up to large switchgear designed to protect
high voltage circuits feeding an entire city.
Magnetic circuit breakers are implemented using a solenoid (electromagnet) whose pulling force
increases with the current. The circuit breaker's contacts are held closed by a latch and, as the current in
the solenoid increases beyond the rating of the circuit breaker, the solenoid's pull releases the latch which
then allows the contacts to open by spring action. Some types of magnetic breakers incorporate a
hydraulic time delay feature wherein the solenoid core is located in a tube containing a viscous fluid. The
core is restrained by a spring until the current exceeds the breaker rating. During an overload, the solenoid
pulls the core through the fluid to close the magnetic circuit, which then provides sufficient force to
release the latch. The delay permits brief current surges beyond normal running current for motor starting,
energizing equipment, etc. Short circuit currents provide sufficient solenoid force to release the latch
regardless of core position thus bypassing the delay feature. Ambient temperature affects the time delay
but does not affect the current rating of a magnetic breaker (see Figure 8.4 for construction).
Thermal breakers use a bimetallic strip, which heats and bends with increased current, and is similarly
arranged to release the latch. This type is commonly used with motor control circuits. Thermal breakers
often have a compensation element to reduce the effect of ambient temperature on the device rating (see
Figure 8.5 for construction).
Thermo-magnetic circuit breakers, which are the type found in most distribution boards, incorporate
both techniques with the electromagnet responding instantaneously to large surges in current (short
circuits) and the bimetallic strip responding to less extreme but longer-term over-current conditions.
Circuit breakers for larger currents are usually arranged with pilot devices to sense a fault current and to
operate the trip opening mechanism (see Figure 8.6 for construction).
115
Under short-circuit conditions, a current many times greater than normal can flow. When electrical
contacts open to interrupt a large current, there is a tendency for an arc to form between the opened
contacts, which would allow the flow of current to continue. Therefore, circuit breakers must incorporate
various features to divide and extinguish the arc. In air-insulated and miniature breakers an arc chute
structure consisting (often) of metal plates or ceramic ridges cools the arc, and blowout coils deflect the
arc into the arc chute. Larger circuit breakers such as those used in electrical power distribution may use
vacuum, an inert gas such as sulfur hexafluoride or have contacts immersed in oil to suppress the arc.
The maximum short-circuit current that a breaker can interrupt is determined by testing. Application of a
breaker in a circuit with a prospective short-circuit current higher than the breaker's interrupting capacity
rating may result in failure of the breaker to safely interrupt a fault. In a worst-case scenario the breaker
may successfully interrupt the fault, only to explode when reset, injuring the technician.
Small circuit breakers are either installed directly in equipment, or are arranged in a breaker panel. Power
circuit breakers are built into switchgear cabinets. High-voltage breakers may be free-standing outdoor
equipment or a component of a gas-insulated switchgear line-up.
Components:
1. spring
2. electrical contacts
3. tie
4. latch
5. hinge
6. iron piece
7. electromagnetic coil
8. spring
Figure 8.4
(a) Normal Condition
(b) Up-normal Condition
Figure 8.5
(a) Normal Condition
(b) Up-normal Condition
Figure 8.6
116
Domestic circuit breakers
The 10 ampere DIN rail mounted thermal-magnetic miniature circuit breaker is the most common style in
modern domestic consumer units and commercial electrical distribution boards throughout Europe. The
design includes the following components:
1. Actuator lever - used to manually trip and reset the circuit breaker. Also indicates the status of the
circuit breaker (On or Off/tripped). Most breakers are designed so they can still trip even if the
lever is held or locked in the on position. This is sometimes referred to as "free trip" or "positive
trip" operation.
2. Actuator mechanism - forces the contacts together or apart.
3. Contacts - Allow current to flow when touching and break the flow of current when moved apart.
4. Terminals
5. Bimetallic strip
6. Calibration screw - allows the manufacturer to precisely adjust the trip current of the device after
assembly.
7. Solenoid
8. Arc divider / extinguisher
Rated current
International Standard IEC 60898-1 and European Standard EN 60898-1 define the rated current In of a
circuit breaker for household applications as the current that the breaker is designed to carry continuously
(at an ambient air temperature of 30 °C). The commonly-available preferred values for the rated current
are 6 A, 10 A, 13 A, 16 A, 20 A, 25 A, 32 A, 40 A, 50 A, 63 A, 80 A and 100 A (Renard series, slightly
modified to include current limit of British BS 1363 sockets). The circuit breaker is labeled with the rated
current in ampere, but without the unit symbol "A". Instead, the ampere figure is preceded by a letter "B",
"C" or "D" that indicates the instantaneous tripping current, that is the minimum value of current that
causes the circuit-breaker to trip without intentional time delay (i.e., in less than 100 ms):
B indicates tripping above 3In up to and including 5In,
C indicates tripping above 5In up to and including 10In, and
D indicates tripping above 10In up to and including 20In
Table 8.1 contains the standard ratings of fuses, Siememens 1, 2, and 3 poles miniature circuit breakers,
Siememens one, two, and three poles circuit breakers.
8.1.3 Thermal Overload Relays
Thermal overload relays prevent an electric motor from drawing too much current and overheating.
Thermal overload conditions are the most likely faults to be encountered in industrial motor applications.
They result in a rise in the motor running current, which produces an increase in the motor's thermal
dissipation and temperature. Overload protection prevents an electric motor from drawing too much
current, overheating, and literally burning out.
Thermal overload relays can be bimetallic relays, eutectic alloy relays, temperature control or probe
relays, and solid-state relays. A bimetallic device is made up of two strips of different metals. The
dissimilar metals are permanently joined. Heating the bimetallic strip causes it to bend because the
dissimilar metals expand and contract at different rates. The bimetallic strip applies tension to a spring on
a contact. If heat begins to rise, the strip bends and the spring pulls the contacts apart, breaking the
circuit. A melting alloy (or eutectic) overload relay consists of a heater coil, a eutectic alloy, and a
mechanical mechanism to activate a tripping device when an overload occurs. The relay measures the
temperature of the motor by monitoring the amount of current being drawn. This is done indirectly
through a heater coil. Temperature control relays are used to protect the motor by directly sensing the
temperature of the windings using thermistor or RTD probes. The motor must have one or more positive
temperature coefficient (PTC) thermistor probes embedded in its windings. When the nominal operating
temperature of the probe is reached, its resistance increases rapidly. This increase is detected by a
threshold circuit, which controls a set of relay contacts. Solid-state relays have no moving or mechanical
parts. The relay calculates the average temperature within the motor by monitoring its starting and
running currents. A solid-state relay is a type of over-current relay.
117
Important performance characteristics to
consider when searching for thermal
overload relays include full current load
range, trip class, and temperature trip
range. The adjustable current value
allows the relay to be set to the full-load
current shown on the motor rating plate.
This is the current range of the thermal
component that can be adjusted to the
desired trip point. The trip class is the
maximum time in seconds at which the
overload relay will trip when the
carrying current is at 600% of its current
rating. Bimetallic overload relays can be
rated as Class 10, meaning that they can
be counted on to break the circuit no
more than 10 seconds after a locked
Figure 8.7
rotor condition begins. Melting alloy
overload relays are generally Class 20. Although thermal overload relays are designed to protect motors
against overload currents, they must be capable of handling large currents without tripping for short
periods during motor starting (run-up). They should, however, trip quickly if the starting currents last too
long. The temperature trip range is the trip point setting range for relays designed to monitor the
temperature of the motor stator windings.
Other important specifications to consider when searching for thermal overload relays include motor load
phase, motor voltage, control circuit voltage, contact or output ratings, pole specifications, features, and
environmental operating parameters. The motor load phase can be single-phase protection or three-phase
protection. The maximum motor voltage is the applicable maximum motor voltage. The control voltage,
if different from the motor voltage. This is called "separate control" and means that the control circuit
gets its power form a separate source; usually lower in voltage form the motor's power source. Contact
output ratings include the contact current rating and contact maximum rated voltage. Poles can be single
pole, double pole, triple pole, four pole, or greater than four pole. Common features for thermal overload
relays include ambient temperature compensation, automatic reset, built-in emergency override, built-in
trip indicator, hermetically sealed, programmable or adjustable trip time, ground fault detection, phase
loss detection, phase reversal detection, unbalance protection, and under-load protection. An important
environmental parameter to consider is operating temperature.
They are provided with 1NO + 1NC potentially free contacts which can be used for signaling. The relay
can be used in either Auto or Manual reset mode (see Figure 8.7).
8.2 Determining Protection Ratings
Before inducing any protective device, the current rating of the device should be determined first, which
depends mainly on the load. Current ratings of loads without starting current, like transformer can be
determined easier than rating of loads with starting current like motors.
8.2.1 Current Rating for Transformers
The name plate of the transformer contains the full load power rating, primary and secondary line
voltages, primary and secondary wires for connection, power factor, efficiency at full load, etc… The
name plate of the three phase transformers also include the type of connection for primary and secondary,
like Y-Δ .
S V1 I1 V2 I 2
(8.1)
The power transformers ratings are usually in VA or KVA. For step down transformers I1 is less than I 2 ,
so the primary winding wire is thinner than that of secondary. A short circuit in the secondary circuit
could damage primary circuit. As a result protective device for transformer should be induced before the
primary. Protection of secondary part is also preferred to be induced.
118
For three phase transformers the line current is the dominant in calculating current ratings. The power of
three phase transformer is given for star or delta connections by:
S  3 V1 I1  3 V2 I 2
(8.2)
From transformers manufactures, it is known that transformer can handle extra 20% from the full load
(i.e. it can work on 120% of full load). It means that the rated current should multiplied with 1.2 to find
the maximum rating current.
After determining primary and secondary rating currents including extra loads, the protection
device rating is chosen to be the nearest higher value from the standard ratings.
8.2.2 Current Rating for Motors
The name plate of the motor contains the full load output power rating in hp , line voltages, primary and
secondary wires for connection, power factor, efficiency at full load, type of the motor AC or DC,
induction or synchronous, single or three phase etc… The name plate of the three phase transformers also
include the type of connection for continuous , like Δ . From these numbers the full load line current can
be calculated.
For DC motors the input power is found by :
Pi  VL I L
(8.3)
For single phase induction motors the input power is found by :
Pi  VL I L cos 
(8.4)
For three phase induction motors the input power is found by :
Pi  3 VL I L cos 
(8.5)
The input power can calculated from the output power and the efficiency by:
Pi 
Po (in hp) x 746

(8.6)
After that the full load current can be determined.
Also like transformers, from motor manufactures, it is known that motors can handle extra 20% from the
full load (i.e. it can work on 120% of full load). It means that the rated current should multiplied with 1.2
to find the maximum rating current. Also, starting current should be considered for three phase induction
motors mainly. The calculated maximum rated current must be multiplied by 1.5 – 2 to consider the
starting current.
After determining rating currents including extra loads and starting, the protection device rating is
chosen to be the nearest higher value from the standard ratings.
For transformers it is recommended to choose fuses or thermal circuit breakers and for motors it
is recommended to choose fuses or thermal circuit breakers type C or D or thermal overload relays
for protection.
119
Table 8.1: Fuses and Circuit Breakers Ratings for 240V
Fuse
Ratings
(A)
2
4
6
8
10
12
16
20
25
32
40
50
80
100
125
160
200
250
315
400
500
630
800
1000
1250
---------------------
Miniature Circuit
Breaker Rating &
Types
1, 2, 3 Poles (A)
1 B, C, D
2 B, C, D
4 B, C, D
6 B, C, D
10 B, C, D
16 B, C, D
20 B, C, D
25 B, C, D
32 B, C, D
40 B, C, D
50 B, C, D
63 B, C, D
--------------------------------------------------------------------------------------
One Pole Circuit
Breaker Ratings &
Types (A)
Two Pole Circuit
Breaker Ratings &
Types (A)
Three Pole Circuit
Breaker Ratings&
Types (A)
15 B, C, D, H, Q
20 B, C, D, H, Q
30 B, C, D, H, Q
35 B, C, D, H, Q
40 B, C, D, H, Q
50 B, C, D, H, Q
60 B, C, D, Q
70 B, C,
80 B, C
90 B, C
100 B, C
-------------------------------------------------------------------------------------------
15 B, C, D, E, P, Q
20 B, C, D, E, P, Q
30 B, C, D E, P, Q
35 B, C, D, E, P, Q
40 B, C, D, E, P, Q
50 B, C, D, E, P, Q
60 B, C, D, E, P, Q
70 B, C, E, P
80 B, C, E
90 B, C, E
100 B, C, E, J, P, Q
125 E, J, Q
150 E, F, J, K, Q
175 E, F, J, K
200 E, F, J, K, Q
225 E, F, J
250 F
300 C, E
350 C, E
400 C, E, H
450 E, H
500 E, H
550 E,
600 E, H
700 E,
800 E,
----------------
15 B, C, E, P
20 B, C, E, P
30 B, C, E, P
35 B, C, E, P
40 B, C, E, P
50 B, C, E, P
60 B, C, E, P, Q
70 B, C, E, P, Q
80 B, C, E, Q
90 B, C, E, Q
100 B, C, E, J, P, Q
125 E, J, Q
150 E, F, J, K, Q
175 E, F, J, K, Q
200 E, F, J, K, L, Q
225 E, F, J, Q
250 F
275 E
300 C, E
350 C, E
400 C, E, H
450 C, E , H
500 C, E, H
550 C, E
600 C, E, H
700 C, E, H
800 C, E, H
900 H
1000 H
120