Download Capacitance in ac circuits

Contents
Introduction
2
Resistance in ac circuits
3
Inductance in ac circuits
7
Inductors in parallel
11
Inductors in series
13
Inductors in power circuits and systems
15
Ballasts in fluorescent light fittings
15
Fault current limiters
15
Capacitive VAR reduction
15
Inductors in electronic circuits
16
Capacitance in ac circuits
24
Capacitive reactance
26
Factors affecting capacitive reactance
27
Comparing capacitors in ac and dc circuits
31
Capacitors in series and in parallel
32
Capacitors in power circuits and systems
34
Power factor correction
34
Motors
35
Electronics
35
Summary
36
Answers
44
EGG202A: 3 Analyse single element ac circuits
 NSW DET 2017 2006/060/04/2017 LRR 3811
1
Introduction
In this section you will learn how voltage is related to current in ac circuits,
for the resistor, the capacitor and the inductor. The total opposition to
current is affected by all of the circuit components and is called the
impedance (Z).
You will see that the current through a resistor is in phase with the voltage,
but that the capacitor and inductor cause phase shifts between the applied
voltage and the resulting current.
Resistors continue to obey Ohm’s Law in an ac circuit, where voltages and
currents are expressed as rms values. Capacitors and inductors are a little
more complex, but we can still use equations very similar to Ohm’s Law for
these components. There is a quantity called the reactance of capacitors and
inductors that can be used instead of the resistance in Ohm’s Law.
At the end of this section you should be able to:
2

Determine the voltage, current and resistance from measured or
given values of any two of these quantities.

Show the relationship between voltage drops and current in a
resistive ac circuit.

List applications of resistive ac circuits.

Define 'inductive reactance' and ‘capacitive reactance’

Calculate the inductive reactance of a given inductor and show the
relationship between inductive reactance and frequency.

Determine voltage, current or inductive reactance in a purely
inductive a.c. circuit given any two of these quantities.

List applications of inductive ac circuits

Calculate the capacitive reactance of a given capacitor and show the
relationship between capacitive reactance and frequency.

Determine voltage, current or capacitive reactance in a purely
capacitive ac circuit given any two of these quantities.

List applications of capacitive a.c. circuits.
EGG202A: 3 Analyse single element ac circuits
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Resistance in ac circuits
You will remember that the two effects that are always present when an
electric current flows are the heating effect and the magnetic effect.
The heating effect is due to current flow and resistance. We can then say
that any type of load that produces heat in the circuit is a resistive load.
Some typical loads of this type are water heaters, electric ranges, toasters,
radiators and incandescent lights.
Remember that power P = VI in dc circuits and this also applies to purely
resistive ac circuits.
All materials exhibit some resistance, and we may have to consider the
resistance of the circuit wiring. Sometimes we may neglect the resistance as
other circuit properties may be so large as to make the resistance
insignificant.
In a purely resistive circuit, the impedance is equal to the circuit resistance.
As Z = R, the circuit current may be calculated:
I
V
R
Example
The hot resistance of a 60 W incandescent lamp is 960 Ω. Determine the
current if the lamp is rated at 240 V.
Solution
Given R = 960 Ω and V = 240 V, find I.
EGG202A: 3 Analyse single element ac circuits
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3
V
Z
but Z  R so
V
I
R
240

960
 0.25 A
I
Example
A water heater is rated at 3.6 kW on a 240 V 50 Hz supply. Determine the
current drawn from the supply and the resistance of the element.
Solution
Given V = 240 V and P = 3.6 kW, find I and R.
To determine the current we need to recall that the power P = VI in purely
resistive ac circuits.
(a)
P
V
3600
240
 15 A
I
(b)
V
I
240

15
 16 
R
In a purely resistive ac circuit the voltage and current will be in phase with
each other. This condition is shown in Figure 1 in both the wave diagram
and the phasor diagram.
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EGG202A: 3 Analyse single element ac circuits
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 = 0
 = 0
Figure 1: Voltage and current in phase in resistive AC circuit
The wave diagram shows instantaneous values from zero to maximum,
whereas the phasor diagram is usually drawn to represent rms values. The
rms values, as you know, are the values indicated by voltmeters and
ammeters.
EGG202A: 3 Analyse single element ac circuits
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5
Check your progress 1
Part A
In questions 1–2, place the letter matching your answer in the brackets provided.
1
In a purely resistive circuit, the current and voltage:
(a) are in phase
(b) have a phase difference of 90°
(c) are directly opposite with respect to phase angle
(d) change with respect to phase angle when resistance increases
2
(
)
(
)
Impedance of a series ac circuit is:
(a) resistance plus reactance
(b) resistance squared plus reactance squared
(c) square root of resistance squared plus reactance squared
(d) root mean squared of resistance squared and reactance squared.
Part B
1
Draw the phasor diagram for a single resistor connected to a 100 V supply and
drawing 50 A. Use a scale of 1 mm per volt and 1 mm per ampere.
Answers to Check your progress are at the end of the section.
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EGG202A: 3 Analyse single element ac circuits
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Inductance in ac circuits
The nature of an alternating current is that of a continual change in
magnitude and direction. The magnetic effect of the current then produces a
magnetic field that is also changing in the same manner as the current that
gave rise to it.
The field flux produced by an alternating current induces an emf in the
conductor that opposes the applied emf. This emf of self-induction is equal
to and directly opposite to the applied emf. In Figure 2 the two waveforms
are shown changing.
Figure 2: Voltages in an inductive ac circuit
It is this induced emf opposing the circuit current that produces inductive
reactance. Inductive reactance is the circuit property that opposes circuit
current.
The symbol for inductive reactance is XL and the unit is the ohm because XL
allows us to continue to use relations similar to Ohms-law in ac circuits
In an inductive ac circuit there is a phase shift between the applied emf and
the circuit current. The phase shift is such as to cause the circuit current to
lag the applied emf by 90°. The wave diagram and phasor diagram for this
condition are shown in Figure 3.
EGG202A: 3 Analyse single element ac circuits
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7
Figure 3: V and I relationship for a ‘pure’ inductive circuit
To explain the phase shift consider the induced emf V’. This emf is due to
the changing current in the inductor.
Recall that in a coil,
v  N

t
where:
v  instantaneous induced emf

 rate of change of flux
t
N  number of turns
Note: the negative sign indicates opposition (Lenz's law)
The rate of change of flux is directly due to the rate of change of the current
ie, the magnitude of the emf is dependent on the rate of change of the
current.
v  L
i
t
where:
v  instantaneous induced emf
i
 rate of change of current
t
L  inductance
At 0° the value of V is zero thus the rate of change of the current must be
zero so the current is at a maximum value.
Between 0° and 90° the induced emf is equal to and opposite to the applied
emf. For this to occur the change in current must be from a maximum value
negative toward zero as the induced emf opposes the current that causes it.
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EGG202A: 3 Analyse single element ac circuits
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The induced emf rises from zero to maximum negative as the current falls
from maximum negative to zero.
At 90° the induced emf is maximum and negative requiring current to be at
its maximum rate of change going positive (that is, passing through zero).
The current then follows a sine wave pattern and when plotted on the wave
diagram is seen to be lagging the applied emf by 90°.
To calculate circuit current in an inductive circuit the total impedance is
inductive reactance.
Substituting XL for Z we use:
I
V
XL
Example
What current will flow in a 240 volt, 50 hertz circuit which has an inductive
reactance of 60 ohm?
Solution
Given:
V = 240 V, 50 Hz
XL = 60 
Find I
I
V
XL
240
60
4A

Inductive reactance depends on the inductance and the frequency of the
supply voltage. The equation to calculate this value is:
X L  2 fL
where:
X L  inductive reactance in ohm
f  supply frequency in hertz
L  inductance in henry
Inductive reactance is directly proportional to both the supply frequency and
the inductance. Circuit current will be affected by a change in either.
EGG202A: 3 Analyse single element ac circuits
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9

When there is an increase in frequency, XL will increase, causing the
circuit current to decrease.

When there is a decrease in frequency, XL will decrease, causing the
circuit current to increase.

Where there is an increase in inductance, XL will increase, causing the
circuit current to decrease.

When there is a decrease in inductance, XL will decrease, causing the
circuit current to increase.
Example
1
Calculate the inductive reactance of a coil which, when connected to a
240 V, 50 Hz supply draws a current of 0.5 A .
2
What will the value of inductive reactance be if the coil is reconnected
to a 100 V 60 Hz supply?
Solution
Given:
I = 0.5A
V = 250 V 50 Hz
Find XL. First we must calculate inductive reactance of the contactor coil.
V
I
240

0.5
 480 
XL 
Now we can determine the value of inductance:
XL = 480 
f = 50 Hz
Find L
X L  2 fL
L
XL
2 f
480
314
 1.53 henry

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EGG202A: 3 Analyse single element ac circuits
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Check your progress 2
1
What is the current phase relationship to the applied voltage in a purely inductive
circuit?
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_____________________________________________________________________
_____________________________________________________________________
2
Calculate the inductive reactance of a 0.5 henry inductor on a 50 Hz supply.
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
Check your answers with those given at the end of the section.
Inductors in parallel
If inductors are connected in parallel as shown in Figure 4 the total
inductance or equivalent inductance is given as:
LT 
1
1 1 1
   .......
L1 L2 L3
which is similar to calculating equivalent resistance of parallel resistors.
From which the equivalent reactance may be calculated as follows:
X L  2 fLT
EGG202A: 3 Analyse single element ac circuits
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Figure 4: Inductors in parallel
Alternatively the reactance of each inductor can be calculated using the
general equation:
X L  2 fL
and since in a purely inductive circuit the impedance consists of reactance
only we may determine the equivalent impedance as follows:
ZT 
1
1
1
1

  ........
Z1 Z 2 Z 3
Note: this can only be used for pure inductors.
To find the total current IT use the following relationship:
IT 
V
ZT
Example
For the circuit in Figure 5 calculate:
(a)
LT
(b)
ZT
(c)
IT
Figure 5
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EGG202A: 3 Analyse single element ac circuits
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Solution
1
1 1 1
 
2 3 4
 0.92 H
LT 
(a)
(b)
(c)
ZT  XL  2    50  0.92
 289
240
289
 0.83
IT 
Inductors in series
If inductors are connected in series as shown in Figure 6 the total inductance
or equivalent inductance is given as:
LT  L1  L2  L3  ...
similar to calculating equivalent resistance of series resistors.
From which the equivalent reactance may be calculated as follows:
X L  2 fLT
Figure 6: Inductors in series
Alternatively the impedance of each inductor can be calculated using the
general equation:
X L  2 fL
and again we can calculate the equivalent impedance given that impedance
is due only to the reactance follows:
ZT  Z1  Z 2  Z3  ...
To find the total current IT use the following relationship:
EGG202A: 3 Analyse single element ac circuits
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13
IT 
V
ZT
Example
For the circuit in Figure 7 calculate:
(a)
LT
(b)
ZT
(c)
IT
Figure 7
Solution
(a)
(b)
(c)
14
LT  2  3  4
 9H
X L  2    50  9
 2827 
240
IT 
2827
 0.085 
EGG202A: 3 Analyse single element ac circuits
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Inductors in power circuits and
systems
Inductors are widely used in electrical and electronic circuits. They can be
used to:

store energy

oppose change in current flow

limit AC current with little loss of power.
Some particular applications are given below.
Ballasts in fluorescent light fittings
Fluorescent lamps, because they work on a principle of electrical current
flow through a gas do not have any built-in current limiting feature such as
incandescent lamps do. A resistor could be used for this purpose but it
would prove to be very inefficient due to heat generated in the resistor.
A better method is to use an inductor (ballast or choke) to limit the current,
and this is done with very little loss.
Fault current limiters
In high voltage systems inductors are often used in series with high voltage
lines to limit current flow in the event of a fault.
Capacitive VAR reduction
When high voltage systems are lightly loaded, the line-to-line and line-toground capacitance can cause voltages in excess of nominal system voltage
to be induced. Inductors are usually automatically brought into line to cancel
this effect.
EGG202A: 3 Analyse single element ac circuits
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Inductors in electronic circuits
In electronics, inductors are used for:
Tuning circuits—radio and television recievers employ a circuit containing
a capacitor and an inductor that allows signals of one frequency only to pass
through. By making the capacitor adjustable via a knob, the user can select
what frequency the radio receives.
Filter circuits—work in a similar way to the tuning circuits. By careful
choice of inductors and/or capacitors, a circuit can be made to block or
accept a range of frequencies. Used often to filter out unwanted noise from
electronic circuits, especially the 50 Hz mains induced noise.
Check your progress 3
For the circuit in Figure 8 calculate:
(a) LT
______________________________________________________________________
______________________________________________________________________
(b) XL
______________________________________________________________________
______________________________________________________________________
(c) IT
______________________________________________________________________
______________________________________________________________________
Figure 8
Check your answers with those given at the end of the section.
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EGG202A: 3 Analyse single element ac circuits
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Check your progress 4
Part A
In questions 1-12, place the letter matching your answer in the brackets provided.
1
In a purely resistive ac circuit, the current:
(a) lags the emf by 90°
(b) leads the emf by 90°
(c) is in phase with the emf
(d) has a phase angle of between 0° and 90°
2
(
)
(
)
The inductive reactance of a coil is directly proportional to the:
(a) circuit current
(b) applied voltage
(c) supply frequency
(d) circuit impedance
3
The angle of phase difference between the applied voltage and the current in a purely
inductive circuit is:
(a) 0°
(b) 45°
(c) 90°
(d) 180°
4
(
)
(
)
(
)
Inductive reactance is measured in ohm because it:
(a) absorbs power
(b) it is the true resistance of an ac circuit
(c) is the equivalent resistance of an ac circuit
(d) is the ratio of the emf of self induction to the current, in an ac circuit
5
The opposition to current flow in a purely inductive ac circuit is termed:
(a) inductance
(b) resistance
(c) inductive reactance
(d) impedance reactance
EGG202A: 3 Analyse single element ac circuits
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17
6
The inductance of a coil will:
(a) increase with frequency
(b) reduce with frequency
(c) reduce with applied voltage
(d) will not change with any of the above as it is a physical quantity
7
(
)
(
)
(
)
(
)
(
)
(
)
The emf of self induction is:
(a) equal to and in phase with the applied emf
(b) exactly opposite but less than the applied emf
(c) equal to and in phase opposition to the applied emf
(d) less than but directly in phase with the applied emf
8
Inductance may be defined as the circuit property which:
(a) absorbs power
(b) opposes any change in current
(c) increases with increase in frequency
(d) decreases with increase in frequency
9
As current decreases in an inductive circuit there is:
(a) an increase in magnetic flux
(b) a decrease in self induced emf
(c) an increase in self induced emf
(d) a decrease in capacitive reactance
10 A ballast is used in a fluorescent lamp fitting to:
(a) improve the power factor
(b) improve the light colour
(c) limit the current drawn by the lamp
(d) fix the fitting to a vertical surface
11 Adding an inductor in series with an existing inductor will:
(a) increase the impedance and decrease the inductance of the circuit
(b) increase the impedance and increase the inductance of the circuit
(c) decrease the impedance and decrease the inductance of the circuit
(d) decrease the impedance and increase the inductance of the circuit
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EGG202A: 3 Analyse single element ac circuits
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12 Adding an inductor in parallel with an existing inductor will:
(a) increase the impedance and decrease the inductance of the circuit
(b) increase the impedance and increase the inductance of the circuit
(c) decrease the impedance and decrease the inductance of the circuit
(d) decrease the impedance and increase the inductance of the circuit
(
)
Part B
1
A coil of negligible resistance draws a current of 4 ampere when connected to a 240
volt, 50 hertz supply. Determine the:
(a) reactance of the coil
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____________________________________________________________________
(b) inductance of the coil
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
2
A coil of negligible resistance draws a current of 10 ampere when connected to a 240
volt, 50 hertz supply. Determine the current the same coil would draw if connected to
a 120 volt, 100 hertz supply.
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
EGG202A: 3 Analyse single element ac circuits
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3
Calculate the reactance of a 3 henry coil when the supply is 240 volt, 50 hertz.
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
4
Calculate the inductance of an inductor that will limit an ac circuit to a current of 5
ampere in a 115 volt, 60 hertz supply.
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
5
For the circuit in Figure 9, determine the inductance of an ideal coil which would have
a current of 10 A through it.
Figure 9
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____________________________________________________________________
____________________________________________________________________
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6
For the circuit in Figure 10 calculate:
Figure 10
(a) LT (equivalent inductance)
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
(b) ZT (equivalent impedance)
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
(c) IT (total current)
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
7
For the circuit in Figure 11 calculate:
Figure 11
(a) LT (equivalent inductance)
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EGG202A: 3 Analyse single element ac circuits
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____________________________________________________________________
____________________________________________________________________
(b) ZT (equivalent impedance)
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
(c) IT (total current)
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
8
For the circuit in Figure 12 calculate:
Figure 12
(a) LT (equivalent inductance)
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
(b) ZT (equivalent impedance)
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
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EGG202A: 3 Analyse single element ac circuits
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(c) IT (total current)
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
Answers to check your progress are at the end of the section.
EGG202A: 3 Analyse single element ac circuits
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Capacitance in ac circuits
A capacitor has the ability to store an electric charge in the form of stress in
the dielectric materials when there is an imbalance of electrons on the
capacitor plates.
Figure 13 shows the current in a purely capacitive circuit to which a
sinusoidal voltage is applied.
Figure 13: Voltage and current in a capacitive ac circuit
You will see that I leads V by 90° when the ac circuit contains capacitance
only. This is because the current will flow from the supply to the capacitor
until the capacitor voltage equals the supply voltage and then stops. The
value of the circuit current is dependent on the difference in the voltage of
the source from the voltage across the capacitor - the greater the difference,
the greater the rate of current flow.
Current only flows in the conductors external to the capacitor while
electrons are moving from one plate of the capacitor to the other.
In Figure 13 (b) the voltage is from an ac source and is continually changing
in magnitude and direction. The rate of charge flow is related to current and
time.
I
24
Q
ampere
t
EGG202A: 3 Analyse single element ac circuits
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The circuit current will be at a maximum when the applied voltage is
changing at its maximum rate, and zero when the voltage is not changing.
At which part of the cycle do these two changes occur?
We will reconsider a single cycle of the voltage sine wave of an ac source in
Figure 14.
Figure 14: Single wave of ac voltage
The maximum rate of change in voltage occurs as the wave is passing
through zero volts. This is because the greater rise in value of voltage level
is happening in a lesser time than it does as the wave reaches a maximum.
The rate of change at various points can be easily determined as shown in
the following example.
Each period of time was over a 5° section of the wave cycle. The greater
rate is found to be at the points 1 and 3 in Figure 14.
At points 2 and 4 the rate of change is zero. The rate of change is maximum
at the instant the voltage changes direction at points 1 and 3.
We can now plot the information for current flow in a capacitive circuit as
shown in Figure 15 and you will see that the relationship is that of Figure
13.
Figure 15: Current voltage relationship in a purely capacitive circuit

At point 1 the rate of change in voltage is maximum going positive;
thus current is maximum positive.
EGG202A: 3 Analyse single element ac circuits
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25

At points 2 and 4 the rate of change of voltage is zero and current is
therefore zero.

At point 3 the rate of change of voltage is maximum going negative so
current is maximum negative.

The current wave will also be sinusoidal and, plotted to the given
points, will be seen to be leading the voltage by 90°.
In a purely capacitive circuit the current leads the voltage by 90 degrees.
Capacitive reactance
The property of a capacitor that limits current in ac circuits is known as
capacitive reactance. The symbol is XC and the unit is the ohm. If the only
component in an ac circuit is a capacitor, then the total impedance of the
circuit is equal to the capacitive reactance. To calculate circuit current in a
capacitive circuit we can use the following relationship:
I
V
ampere
XC
where V and I are rms values of voltage and current respectively.
Example
A capacitor having a capacitive reactance of 80 ohm is connected to a 240 V
50 Hz supply. Determine the circuit current.
Solution
Given:
V = 240 volt, 50 Hz
XC = 80 
Find I
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I
V
XC
240
80
3A

Example
Determine the capacitive reactance required to limit current to 0.25 A on a
240 V, 50 Hz supply.
Solution
Given
V = 240 volt 50 Hz
I = 0.25 A
Find XC.
V
I
240

0.25
 960 
XC 
As with resistance only, capacitive reactance is used with the basic Ohm’s
law equation to solve circuit problems. Values of circuit voltage and current
are rms values.
Factors affecting capacitive reactance
The factors that affect the value of the reactance of a capacitor are the
frequency of the supply and the capacitance of the capacitor.
To calculate the capacitive reactance given capacitance and supply
frequency we use:
XC 
1
2 fC
where:
X C = capacitive reactance in ohm
f = supply frequency in hertz
C = capacitance in farad
EGG202A: 3 Analyse single element ac circuits
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Capacitive reactance is inversely proportional to both frequency and
capacitance. That is, if either one is increased capacitive reactance
decreases.
Example
Consider a capacitive circuit on an ac supply which has a 5 A current. What
will be the circuit current if the frequency is doubled?
Solution
As the reactance is inversely proportional to the frequency, that is:
X C1 
1
f
the new reactance will be half, as frequency is multiplied by two.
New reactance:
1
X C 2   X C1
2
The new circuit current then must be doubled and becomes 10 amps.
Similarly, reactance is inversely proportional to capacitance and the changes
to reactance and circuit current for a change in capacitance follow those for
changes in frequency.
increase f → decrease XC → increase I
decrease f → increase XC → decrease I
increase C → decrease XC → increase I
decrease C → increase XC → decrease I
Examples

Calculate the capacitive reactance of a 60 microfarad capacitor on a
240 V, 50 Hz supply.
Solution
Given
C = 60 F (60 × 10-6 farad)
F = 50 Hz
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EGG202A: 3 Analyse single element ac circuits
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Find XC
XC 

1
2 fC
1
_
2  3.141 50  60  10 6
 53 
An alternative is to have 106 or 1 000 000 on the top line and the
capacitance in micro-farad on the bottom line.
106
ohm
2 fC
1 000 000

314  60
 53 
XC 

How much current will flow in a 240 V, 50 Hz circuit if the circuit has
an 8 F capacitor as its only load?
Solution
Given
V = 240 volt 50 Hz
C = 8 F
Find I
V
Z
Z  XC
I
I
XC 
V
XC
106
2 fC
 1 000 000
314  8
 398 
240
398
 0.603 A
I
EGG202A: 3 Analyse single element ac circuits
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Check your progress 5
1
What is the capacitive reactance of a 100 f capacitor on a 50 Hz supply?
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
2
Calculate the current if a 16 F capacitor is connected to a 240 V, 50 Hz supply.
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
Check your answers with those given at the end of the section.
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EGG202A: 3 Analyse single element ac circuits
 NSW DET 2017 2006/060/04/2017 LRR 3811
Comparing capacitors in ac and dc
circuits
When a capacitor is connected to a source of dc there will be a movement of
electrons to the uncharged plates. As the plates become charged the electron
movement will cease. When fully charged, the capacitor voltage is the same
as the supply, and there is no potential difference between the supply and
the capacitor. A charged capacitor will therefore block dc.
On an ac source, a capacitor is continually charging and discharging. No
electrons pass through the capacitor even though there is an ac current
through the circuit.
Figure 16: A capacitor passing ac current
There will be a continual flow of AC in the lamp filament causing a
continual glow. Therefore a capacitor will allow a flow of ac in the external
circuit, even though no electrons actually cross the dielectric!
EGG202A: 3 Analyse single element ac circuits
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Capacitors in series and in parallel
From the expression
CK
A
farad
L
it can be seen that if capacitors are connected in parallel the total
capacitance will increase as this is similar to increasing the plate area,
Figure 17 (a).
Total capacitance in parallel is the sum of the individual capacitances.
Figure 17: Capacitor connections
In series, capacitors will appear to have the distance between plates
increased as seen in Figure 5(b) which will decrease the total capacitance.
Total capacitance for a series connection is calculated as follows:
CT 
1
farad
1
1
1


C1 C2 C3
You will recognise the basic expressions and note that the change is
opposite to finding total resistance for series and parallel resistors.
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EGG202A: 3 Analyse single element ac circuits
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Check your progress 6
For the circuit in Figure 6 calculate:
(a) CT (equivalent capacitance)
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
(b) ZT (equivalent impedance)
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
(c) IT (total current)
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
Figure 18
Check your answers with those given at the end of the section.
EGG202A: 3 Analyse single element ac circuits
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33
Capacitors in power circuits and
systems
Capacitors are widely used in electrical and electronic circuits. The features
of capacitors are their ability to:

store energy

oppose change in voltage

pass ac while blocking dc current.
Some particular applications follow.
Power factor correction
Some installations have loads with a large inductive component. Because
the current through these loads is not in phase with the voltage, these
installations draw more current that is necessary. This condition causes what
is known as a low power factor. Power utilities place limits on the power
factor presented by each premises, which should ideally be unity (ie 1).
Power factor correction using capacitors is done at all stages of a power
distribution system.
At the transmission stage, large high voltage capacitors are placed in circuit
when the system power factor drops below a predetermined level. This is
often done automatically.
At the distribution and at high voltage customer premises, power factor
capacitors are connected to 11 kV supply lines and switched in banks to
compensate for low power factor.
At the 415/240 V level individual items of equipment sometimes
incorporate power factor correction capacitors. An example is fluorescent
light fittings which can be ordered as either HPF (high power factor) or LPF
(low power factor). The HPF fittings have a capacitor fitted.
34
EGG202A: 3 Analyse single element ac circuits
 NSW DET 2017 2006/060/04/2017 LRR 3811
Motors
Capacitors are used in single phase motors to enable the motor to start
unaided. This is done by using the capacitor to alter the phase angle between
two windings in the motor thereby producing a rotating magnetic field.
Electronics
Capacitors find use in ac circuits associated with the electronics industry.
Some examples are:
Power supplies—capacitors are used to smooth out (or filter) the output of
dc power supplies to remove ripple and make the waveform as ‘flat’ as
possible.
Oscillator circuits—used to produce an alternating output signal.
Blocking of dc—capacitors are used in circuits designed to allow ac to pass
but block dc.
EGG202A: 3 Analyse single element ac circuits
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Summary

Pythagoras’s theorem may be expressed as:
Z2 = R2 + X2
where
Z = impedance
R = resistance
X = reactance
and derived from this:
Z  R2  X 2

Ohms law still holds for an ac circuit, with V and I as rms values.

Inductance is a physical circuit property.

Inductance is that circuit property that opposes a change in current.

When the magnetic flux surrounding an inductor changes in strength, an
emf is induced in the conductors in such a direction as to oppose the
change in current.

An increase in current will induce the emf in a direction to try to stop
the current increasing.

A decrease in current induces the emf in a direction that will tend to
keep the current up.

The property of inductive reactance is that property that opposes circuit
current.

The symbol for inductive reactance is XL and is expressed in ohms.

Inductive reactance may be calculated given the inductance of the coil
and the supply frequency using the expression:
XL = 2fL ohm
36

In a purely inductive circuit the current lags the applied voltage by 90°.

For an increase in frequency or inductance or both these, XL will
increase and circuit current will decrease.
EGG202A: 3 Analyse single element ac circuits
 NSW DET 2017 2006/060/04/2017 LRR 3811

For a decrease in frequency, inductance or both of these, XL will
decrease and circuit current will increase.

Ohm’s law when applied to ac circuits uses values of impedance (Z),
rms values of voltage (V) and rms values of current (I).

Resistive loads generally are loads that produce heat.

In purely resistive ac circuits the voltage and current are in phase with
each other.

In purely capacitive ac circuit the current leads the voltage by 90
degrees.

Capacitive reactance is that property that limits circuit current in a
capacitive circuit. The symbol is XC and the unit is the ohm.

The equation to calculate XC, given C and f, is:
XC 
1
ohm
2fC
where
f is in hertz
C is in farad

XC is inversely-proportional to both f and C.

increase f → decrease XC → increase I
decrease f → increase XC → decrease I
increase C → decrease XC → increase I
decrease C → increase XC → decrease I

A capacitor allows AC to flow and will block dc.
EGG202A: 3 Analyse single element ac circuits
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Check your progress 7
Part A
In questions 1–13, place the letter matching your answer in the brackets provided.
1
When there is an imbalance of electrons on the opposite plates of a capacitor, there is:
(a) no charge on the capacitor
(b) no voltage across the capacitor
(c) a constant current flowing through the capacitor
(d) a stress in the electron orbits of the dielectric
2
(
)
The ability of a capacitor to store a charge which will oppose a change in voltage in an
ac circuit is called its:
(a) reactance
(b) resistance
(c) capacitance
(d) permissiveness
3
(
)
The relationship between voltage and current in a purely capacitive circuit, supplied by
ac, is that current:
(a) and voltage are in phase
(b) lags voltage by 90°
(c) leads voltage by 90°
(d) and voltage are in phase opposition
4
(
)
(
)
The resultant of two or more voltages differing in phase may be determined by:
(a) phasor addition
(b) adding numerically
(c) algebraic addition
(d) averaging the values
38
EGG202A: 3 Analyse single element ac circuits
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5
An equation to determine capacitive reactance is:
(a) X C 
V
Z
(b) X C 
R
Z
(c) X C  2 fC
(d) X C 
6
1
2 fC
(
)
(
)
(
)
(
)
As applied frequency on a capacitor increases, the capacitive reactance:
(a) reduces
(b) increases
(c) does not change
(d) acts to change applied frequency
7
8
In a purely capacitive circuit with ac supply, the current is:
(a) I 
V
R
(b) I 
V
C
(c) I 
V
XC
(d) I 
V
2 fC
The angle of phase difference between voltage and current in a purely capacitive
circuit is:
(a) 0°
(b) 45°
(c) 90°
(d) 180°
EGG202A: 3 Analyse single element ac circuits
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9
The equation to determine the total capacitance of three capacitors in parallel is:
(a) CT 
(b)
1
C1  C2  C3
1
1
1
1
 

CT C1 C2 C3
(c) CT 
C1  C2
C2  C3
(d) CT  C1  C2  C3
(
)
(
)
(
)
(
)
(
)
10 For a purely capacitive circuit, impedance (Z) equals:
(a) supply voltage
(b) total capacitance
(c) frequency of the supply
(d) total capacitive reactance
11 A capacitor is used in a fluorescent lamp fitting to:
(a) improve the power factor
(b) improve the light colour
(c) limit the current drawn by the lamp
(d) fix the fitting to a vertical surface
12 Adding a capacitor in series with another capacitor will:
(a) increase the impedance and decrease the capacitance of the circuit
(b) increase the impedance and increase the capacitance of the circuit
(c) decrease the impedance and decrease the capacitance of the circuit
(d) decrease the impedance and increase the capacitance of the circuit
13 Adding a capacitor in parallel with another capacitor will:
(a) increase the impedance and decrease the capacitance of the circuit
(b) increase the impedance and increase the capacitance of the circuit
(c) decrease the impedance and decrease the capacitance of the circuit
(d) decrease the impedance and increase the capacitance of the circuit.
40
EGG202A: 3 Analyse single element ac circuits
 NSW DET 2017 2006/060/04/2017 LRR 3811
Part B
1
Draw a neat wave and phasor diagram of the voltage and current for a capacitor on an
ac supply. (No scale is required.)
2
Calculate the voltage required to cause a 5 A current in a circuit having a capacitive
reactance of 40 Ω.
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
3
The current in a 240 V, 50 hertz capacitive circuit is 6 A. What current will flow if the
frequency is increased to 60 Hz ?
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
4
Calculate the capacitive reactance of a 100 µF capacitor when it is connected to a
415 V, 50 Hz supply.
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
EGG202A: 3 Analyse single element ac circuits
 NSW DET 2017 2006/060/04/2017 LRR 3811
41
5
When connected to a 240 volt, 50 hertz supply a capacitor draws a current of 10 A.
What current would be drawn if the size of the capacitor was doubled?
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
6
Two capacitors, each with a capacitance of 40 F are connected in series to a 240 volt,
50 hertz supply.
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
7
For the circuit in Figure 19 calculate:
(a) CT (equivalent capacitance)
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
(a) ZT (equivalent impedance)
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
(a) IT (total current)
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
42
EGG202A: 3 Analyse single element ac circuits
 NSW DET 2017 2006/060/04/2017 LRR 3811
Figure 19
Answers to Check your progress are at the end of the section.
EGG202A: 3 Analyse single element ac circuits
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43
Answers
Check your progress 1
Part A
1
(a)
2
(c)
Part B
1
Check your Progress 2
1
Current lags voltage by 90°.
2
157 
Check your Progress 3
44
(a)
3.7 H
(b)
1162 
(c)
0.206 A
EGG202A: 3 Analyse single element ac circuits
 NSW DET 2017 2006/060/04/2017 LRR 3811
Check your progress 4
Part A
1
(c)
2
(c)
3
(c)
4
(d)
5
(c)
6
(d)
7
(b)
8
(b)
9
(c)
10 (c)
11 (b)
12 (c)
Part B
1
(a) 60 Ω
(b) 0.19 H
2
2.5 A
3
942 Ω
4
0.061 H
5
0.032 H
6
(a) 0.217 H
(b) 81.8 Ω
(c) 1.34 A
7
(a) 6 H
(b) 1885 Ω
EGG202A: 3 Analyse single element ac circuits
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(c) 0.127 A
8
(a) 0.531 H
(b) 167 Ω
(c) 0.168 A
Check your progress 5
1
2.895 ohm
2
1.206 mA
Check your progress 6
1 (a) 9.9 F
(b) 321 
(c) 0.746 A
Check your progress 7
Part A
1
(d)
2
(c)
3
(c)
4
(a)
5
(d)
6
(a)
7
(c)
8
(c)
9
(a)
10 (d)
11 (a)
46
EGG202A: 3 Analyse single element ac circuits
 NSW DET 2017 2006/060/04/2017 LRR 3811
12 (a)
13 (d)
Part B
1
2
200 V
3
7.2 A
4
31.8 Ω
5
20 A
6
1.5 A
7
(a) 650 F
(b) 4.1 Ω
(c) 27 A
EGG202A: 3 Analyse single element ac circuits
 NSW DET 2017 2006/060/04/2017 LRR 3811
47