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Electrical Installations Unit 008
Installing Electrical Systems & Equipment
C&G 2351
Unit 008C1
Electrical Science
AC Theory: Resistance, Reactance,
Impedance and methods of measurement
Legh Richardson
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Electrical Installations Unit 008
Contents:
Title
Pages
AC Theory: The Basics ...................................................................................................3
Definitions:.........................................................................................................................3
Conversion Inductance and Capacitance into Reactance................................................5
Examples:...........................................................................................................................5
Capacitors in AC circuits ..................................................................................................6
Examples of Capacitive Reactance...................................................................................6
AC Waveforms ..................................................................................................................7
Examples of periodic time and frequency:.......................................................................7
Measurement of amplitude for AC waves .......................................................................8
Examples of Peak and r.m.s. amplitudes:.........................................................................8
AC Series Circuits .............................................................................................................9
Inductors in Series Circuits ...............................................................................................9
Circuit Diagram Phasor Diagram Waveform Diagram...................................................9
Examples of Phasor Diagrams ........................................................................................12
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AC Theory: The Basics
Definitions:
Resistance: Opposition to current flow in a DC Circuit or an AC resistive circuit where
the current flow is obstructed and effective volt drop combined with the current flow
generates a power consumption which produces heat
Power consumption is always resistive where both the current and voltage are in phase
Symbol = R; Unit =  (Ohms)
Reactance: Opposition to current flow in AC circuits only due the reactive effects of
of both Coils and Capacitors. Each have the ability to store and release energy back into
the circuit causing the current and voltage to be pulled out of phase with one another.
Reactance is then a form of AC resistance which does not generate heat and therefore
cannot consume power. However, it absorbs power within the electrical system and
causes losses
As we know a coil of wire with a current flowing through it generates a magnetic field
If the current flowing is ac then the field will expand and contract twice per cycle of ac
or 100 times per second
This we will explore later since 1 cycle = 2 ( radians )
( where 1 rad = 57.3 degs so  rads = 180 degs and 2 rads = 360 degs )
Inducing an emf in one direction and then the other ( e = B.l.v )
Where e = induced emf, B = flux density, l = length of conductor affected by the
magnetic field, v = velocity of the conductor passing through the magnetic field
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V
II
V
0
270
90
0
360
time
180
The emf builds up at the beginning of an ac cycle and the corresponding magnetic field
also builds up generating an emf causing current to attempt to flow in the opposite
direction to the applied emf
Reactance is opposition set up by electromagnetic and electrostatic fields and causes the
current and voltage to be pulled out of phase
Current leads the Voltage ( Capacitive Reactance )
lags the Voltage ( Inductive Reactance )
CIVIL
Symbol = XL; Unit = 
Impedance:
The Total opposition to current flow in an AC circuit and constitutes both resistance and
reactances.
Symbol = Z; Unit = 
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Conversion Inductance and Capacitance into Reactance
Both Inductors and Capacitors are frequency dependant
Increase the frequency and the Inductive reactance in Ohms increases
XL = 2..f.L
Where XL is the Inductive Reactance in Ohms
 = A constant of 3.14159
f = the frequency of the AC voltage and current flowing through the component
L = the Inductance of the component / inductor and is a form of measurement of force
between current carrying conductors measured in Henrys (H)
Examples:
1/ Calculate the inductive Reactance of an inductor of 10H in an AC circuit of
frequency 50Hz
2/ The frequency of the circuit in question 1 is increased to 500Hz. Calulate the new
Inductive reactance
3/ The Frequency of the circuit in Question 1 is halved to 25Hz calculate the new
inductive reactance.
4/ A variable inductor is placed in an AC circuit of 1kHz. Calculate the reactance when
the inductance is 150mH.
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Capacitors in AC circuits
Capacitors provide a pure reactance in that there is no resisitive part to it. Their purpose
is to counteract the byproduct of inductive reactance caused when coils, chokes, motor
windings, transformers are connected in AC circuits.
As the frequency increases the reactance falls off. Reducing to a closed circuit at very
high frequencies.
The Capacitive reactances Symbol = XC; Unit = 
1
XC = 2..f.C
Where XC = Capacitive Reactance
F = frequency of voltage and current
C = Capacitance
Examples of Capacitive Reactance
1/ Calculate the capacitive reactance of a 10uF Capacitor connected into a 50Hz AC
circuit.
2/ Calculate the capacitive reactance from the values in question 1 when the frequency
is increased to 1kHz.
3/ If the frequency is halved from that of question 1 what will be the new capacitive
reactance?
4/ If the current flow through a series AC circuit containing a capacitor across a 230V
nominal supply voltage is 2.5A. Find the value of the Capacitor.
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AC Waveforms
The waveform below shows one periodic cycle of an example of either voltage or
current.
Amplitude
V.p
V.rms
V.ave
Waveform ( V or I )
time
Periodic time ( 1 Cycle )
The time it takes to produce one cycle of AC is called the Periodic time
Usually measured in sub-multiples of a second. A 50Hz AC voltage/current has a
periodic time of 20mS for each cycle of sinusoidal voltage/current
The number of cycles in one second is called the frequency
This is measured in Hz
The General formula to express the relationship between Periodic time and the
frequency is:
T
1
1
or transposed to F 
F
T
Examples of periodic time and frequency:
1/ The periodic time of a waveform is 10mS. Calculate its frequency
2/ The frequency of a sinusoidal waveform is 1kHz. Calculate the periodic time.
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Measurement of amplitude for AC waves
The amplitude of any waveform can be measured in several ways
The Peak voltage or current (Vp or Ip)
The r.m.s. voltage or current (V.r.m.s. or I.r.m.s. )
The r.m.s. of any alternating waveform is used to tell us the amount of real power an
AC wave develops. R.m.s. stands for Root Mean Squared and is the method by which
the r.m.s. is calculated.
The definition of r.m.s. is: An AC waveform that produces the equivalent DC heating
effect. That is a 230V AC waveform will have a peak amplitude of approximately 325v
But it will produce the DC heating effect of 230v DC
All power voltages and currents are expressed as an r.m.s. unless otherwise stated
The formula for calculating the peak or the rms of sinusoidal waveforms is:
Vp  V .rms  2
or transposed to V .rms 
V .peak
2
1
0.7071
The Standard moving Coil analogue meter reads the peak waveform and converts the
peak waveform by the use of a calibrated scale on the meter front into the r.m.s format.
The old moving Iron Meter converts the peak value of the waveform automatically to
the r.m.s. value
Digital electronic meters ( the standard nowadays) requires special electronics to
convert the peak value to its r.m.s.
2 is a shorthand version of writing 1.414 it can also be expressed as
Examples of Peak and r.m.s. amplitudes:
1/ If a voltage has a peak value of 1V what is its r.m.s. value?
2/ An AC current has a r.m.s. value of 70.71 amps. What is its peak value?
3/ A single phase AC waveform has a peak voltage of 77.78V what is its r.m.s value?
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AC Series Circuits
Inductors in Series Circuits
If an Inductor is placed in series with a resistor in an electrical circuit then the current
flow through both the resistor and the inductor stays constant ( Kirchoff's Current Laws
). However, there will be a voltage drop across each of the components. This does not
follow Kirchoff's Voltage Law since the voltage is thrown out of phase with the current
flow by the inductive reactance of the coil.
In a pure Inductor with no resistance, the voltage is 90deg out of phase with the current
flow and is shown in the phasor diagram.
The phasor notation is such that the constant current is shown as the horizontal
component or line and the changing voltage is shown as the vertical component or line.
Circuit Diagram
VR
Waveform Diagram
V

V
Z
VXL
Phasor Diagram
I
Z
XL
0
XL
270
90
R
0
360
180
I
Vs
0
R
I
Notice that:
(a) The sum of the volt drops in a series circuit no longer adds up to the supply voltage
(b) The resistance of the 90deg out of phase component of the inductor is called the
Inductive Reactance ( XL ), and so the overall resistance of the circuit is now called the
Impedance ( Z ).
(c) The relationship between the Impedance and the resistance is now generated by the
relationship between sides of a right angled triangle (Pythagoras' Theorem), and is
separated by a 'phase' angle.  (Theta )
(d) The phase angle is directly related to the movement of the current out of phase with
the voltage in the waveform diagram, when measured in electrical degrees.
(e) The phasor diagram is indicated by the 'Omega' arrow and is shown rotating anticlockwise. This is the standard form for phasor diagrams and differs from vector
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diagrams since it shows that all the components in the phasor diagram are rotating
together but at different angles.
(f) The waveform diagram also shows each wave alternating through 360deg, holding
their respective distance from one another.
A pure Inductor in an ac circuit causes the current to lag the voltage by 90deg.
So Inductors in series with resistors in ac circuits cause the current to lag the voltage by
somewhere between 0deg and 90deg.
A typical Phasor diagram
V


Z
XL
0
R
I
The axis are shown for series circuits as the current being constant and always is placed
on the horizontal. The vertical axis always shows the voltage.
So we can say from this diagram that:
1/ the voltage and current through the resistive part of the circuit is always in phase with
the current
2/ The voltage and current through the inductive part of the circuit is always 90 apart
3/ The voltage and current through the combined resistance and inductive reactance also
known as the impedance is an angle  (theta) between 0 (unity) and 90.
4/ The Cosine of the angle  is a ratio of the Resistance over the Impedance and is also
known as the Power Factor of the circuit.
The General Formula is:
Cos 
Legh Richardson
V
R
W
or R or
Z
VS
VA
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The general rules when drawing phasor diagrams
V
I
Inductive Reactance
Capacitive Reactance
ω
ω
I
V
Series Circuits
Parallel Circuits
ω
ω
I
V
Inductive Reactance
Capacitive Reactance
I
V
In each when evaluating Z, V, W. The in phase component will lie on the horizontal
axis (X-Axis) and the out of phase component will be an angle  between the X-axis
and the Y-Axis.
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Examples of Phasor Diagrams
1/ Draw the Phasor diagram of a purely resistive circuit showing the relationship of the
voltage and current
2/ Draw the phasor diagram of a purely inductive reactive circuit showing the
relationship of the voltage and current in the circuit
3/ Draw a scaled phasor diagram of a circuit that contains a coil with an inductive
reactance of 40 and a resistance of 30. Show the relationship of the voltages in each
case. By measurement estimate the value of the impedance of the circuit.
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Capacitors in AC Series Circuits
The Capacitive reactance has no resistance so the voltage across a capacitor in a series
circuit is always shown at right angles to the voltage across a resistor in the same circuit
Again the total opposition to current flow is still called the impedance.
The relationship of Resistance and Capacitance in a Series Circuit
Vc

R
VR
I

R
X
I
XC
V
Z
V
The Current flow in a capacitor always lags the voltage. For series circuit you could say
that as the current stays constant the voltage changes. So for Capacitive circuits the
voltage lags the current flow.
1/ Draw the phasor diagram for a purely Capacitive reactive Circuit
2/ Draw the circuit Diagram with both resistance and Capacitive reactive components
showing the impedance.
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Examples of Capacitive reactance
1/ A capacitor of reactance 80 is placed in series with a 60 Resistor. Draw the
phasor diagram and calculate the impedance of the circuit.
2/ A 5F capacitor is placed in series with a 500 resistor across a 230v 50Hz supply.
Calculate: (I) the impedance, (ii) the current flow, (iii) the voltage across the resistor
and capacitor.
3/ Draw (I) the phasor diagram for question 2 showing the impedance triangle and (ii)
the phasor diagram showing the relationship between the voltages
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AC circuits including both Inductors and Capacitors
The easiest way to show this layout is to study the phasor diagram. You will notice that
because the voltage across the inductor and capacitor are each 90 out of phase with the
current in the circuit the resultant phasors can be shown together:
V
XL

R
I

Z
XC
V
The effect of placing a capacitor in an inductive circuit is to change the overall phase
angle between the resistive and reactive components and hence reduce the impedance of
the circuit
There are two methods of solution that can be used:
1/ By drawing the phasor diagram to scale and measuring the lengths of each of the
components vectors and then measuring the phase angle between the resistive
component and the overall impedance or
2/ By calculation, as seen from the above diagram vectors can be added or subracted. If
two vector quantities are pulling in opposite directions with an angle between them of
180 then the two quantities can be subtracted. Alternatively if there are in the same
direction then they can be added. If, however, they are at an angle other than 180 then
they follow the formula:
Z  R 2  ( XL  XC ) 2
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Example 1
A 12Vrms 10kHz AC circuit has a coil with a resistance of 500 and an inductance of
20mH. A 12nF capacitor is placed in series. Calculate:
(i)
The peak Voltage of the circuit
(ii) The inductive reactance
(iii) The capacitive reactance
(iv) The Impedance of the circuit
(v) The Phase angle and hence the power factor of the circuit
(vi) The current flowing in the circuit
(vii) The voltage across the inductor
(viii) The voltage across the capacitor
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Example 2a:
A 2Ω resistor, a 0.2H inductor and a 10uF capacitor are connected in series across a
220V 50Hz ac circuit.
Find:
1/ the impedance of the circuit
2 the current flow through the circuit
3/ the volt drop across each component
4/ the phase angle and pf of the circuit
5/ Draw the phasor diagram
Example 2b
A second 90uF capacitor is placed in parallel across the 10uF capacitor. Calculate:
6/ the new impedance of the circuit
7/ the new current through the circuit
8/ the volt drop across each component
9/ the new phase angle
10/ Show that VS   Vpds
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Worked Example 3:
A 24V 100Hz AC Sinusoidal circuit with a coil having a resistance of 2 and an
inductance of 40mH requires a 63.3F series capacitor to bring the overall impedance
of the circuit to a ‘reasonable’ level. Calculate the following
1/ the inductive reactance
2/ the capacitive reactance
3/ Draw the phasor diagram of the circuit
4/ the impedance of the circuit
5/ the current through the circuit
6/ the phase angle and cos of the circuit
7/ the voltage across each of the components in the circuit.
8/ State two dangers that might occur with such a circuit arrangement
9/ What other name is given to this type of circuit?
1/
XL  2..f.L  2..100.40  103  25.14
2/
XC 
4/
Z  R2  XL  XC 2  22  25.14  25.14 2  2
5/
I
6/
Cos 
1
106

 25.14
2..f.C 2.    100  63.3
V 24

 12A
Z
2
R 2
  1 0 o lag / lead
Z 2
VR  I.R  12  2  24V
7/
VXL  I.XL  12  25.14  301.7V
VXC  I.XC  12  25.14  301.7V
8/ Possible electric shock and the destruction of the component parts. (breakdown of
coil insulation and destruction of capacitor dielectric)
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