1E6 Electrical Engineering
AC Circuit Analysis and Power
Lecture 16: Power in AC Circuits III
16.1 Power Factor
In general, an ac circuit will contain a combination of resistive and reactive
components and the reactive elements may be inductive or capacitive as shown
in the circuit of Fig. 1. The overall impedance of such a network, as seen by a
voltage or current source driving it, has a magnitude and phase and it is this
phase angle which determines the Power Factor of the network.
i(t)
R1
L2
R2
C1
Z
V(t)
~
C2
L1
R3
Fig. 1 A General AC Circuit Containing Reactive Elements
Ζ = R ± jX
φ = Tan -1
and
±X
R
where the net reactance may be either inductive or capacitive.
Power Factor = Cosφ
In order to maximise the active power we require:
Cos φ = 1
which implies
φ=0
This means that, as seen by the source, there should be no phase shift between
the current and voltage, that is, the network should ideally look purely resistive.
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The overall impedance of an ac network can be expressed in terms of constituent
components which include an equivalent resistance, an equivalent inductance
giving a positive reactive component and an equivalent capacitance giving a
negative reactive component. Then:
Impedance = Resistance + Inductive Component + Capacitive Component
Ζ = R EQ + jX LEQ − jX CEQ
If XLEQ = XCEQ, then the net reactive component of the impedance is zero so that:
Ζ = R EQ
φ = Tan -1
±X
=0
R
Power Factor = Cosφ = 1
If this is the case then
and
Sinφ = Sin 0 = 0
Apparent Power = Active Power = VRMSI RMS =
Reactive Power = j VRMS I RMSSin φ = j
Vm I m
2
Vm I m
Sin φ = 0
2
This means that, in this case, all of the power delivered to the network by the
source is dissipated in the resistive elements. That is, all of the power delivered is
consumed and becomes useful or “active” power.
Therefore, in order to correct the power factor of any ac network to unity, the
net effective reactance must be made equal to zero. This is achieved by making
the capacitive reactive component of the impedance equal and opposite to the
inductive reactive component. If the network has only one type of reactance then
a reactance of the opposite type must be added in order to neutralise it.
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16.2 Case Study:
Consider the electrical equivalent of a motor driving a load illustrated by the
equivalent electric circuit shown in Fig. 2. Note that the resistance primarily
represents physical work done by the motor in driving a mechanical load. It also
includes the resistance of the coil windings which generate heat when a current
flows through them and consequently also dissipate power.
f = 50Hz
Z
~
L
motor
inductance
200 mH
ωL = 314 x 200 x 10 −3 = 62.8Ω
V=
310 Sin ωt
R
load
50 Ω
R = 50Ω
Fig. 2 An Equivalent Electric Circuit having Reactance
Ζ = R + jω L
Z =
R 2 + ω 2 L2
Z = 2500 + 3943 .8 = 80.3Ω
φ = Tan -1
Im =
ωL
62.8
= Tan -1
= 51.5o
R
50
Vm 310
=
= 3.86 A and Cosφ = Cos51.5 o = 0.623
Z 80.3
3
Then the power consumption is characterised as:
Apparent Power = PAPP =
Active Power = PAVE =
Vm I m 310 x 3.86
=
= 598W
2
2
Vm Im
310 x 3.86
Cosφ =
x0.623 = 373W
2
2
Reactive Power = PIMAG = j
Vm I m
310 x 3.86
x0.783 = j468W
Sinφ = j
2
2
This shows that the motor draws approximately 600W of power from the mains
supply overall. However, less than 400W of this is consumed as energy and used
by the motor to do mechanical work. A further power of more than 450W is
drawn from the supply, which can be considered as out of phase with the active
power, and is returned to the supply at a later point in each cycle.
The important point to note is that the supply must be able to generate and
provide the total apparent power. That is, the source must be rated to handle a
total power of 600W when less than 400W is used by the load. This means that
the source must have a power handling capacity which is more than 50%
greater than that required to deliver the useful power.
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16.3 Power Factor Correction
Consider including a capacitor to neutralise the effect of the motor inductance
and correct the Power Factor as shown in Fig. 3.
Z
~
L
motor
inductance
200 mH
V=
310 Sin ωt
R
load
50 Ω
Fig. 3
Correction of the Power Factor of the Circuit of Fig. 2
1
jω C
Ζ=
1
R + jωL +
jωC
(R + jωL )
Ζ = R EQ + jX LEQ − jX CEQ
Multiplying the numerator and demoninator above by jωC:
Ζ=
Ζ=
Rationalising:
C
Power
Factor
correction
capacitance
(R + jωL )
jωCR − ω2 LC + 1
(R + jωL )
(1 − ω LC)+ jωCR
2
(
R + jωL ) [(1 − ω2 LC)− jωCR ]
Ζ=
(1 − ω2LC)2 + (ωCR )2
5
Ζ=
(
)
(
)
R 1 − ω2 LC + jωL 1 − ω2 LC − jωCR 2 + ω2 LCR
(1− ω LC) + (ωCR )
2
2
Ζ=
2
R − ω2 LCR + jωL − jω3L2C − jωCR 2 + ω2 LCR
(1− ω LC) + (ωCR )
2
2
Ζ=
2
R + jωL − jω3L2C − jωCR 2
(1− ω LC) + (ωCR )
R + jωL − jωC(ω L + R )
Ζ=
(1− ω LC) + (ωCR )
2
2
2
2 2
2
2
2
2
Breaking this into its equivalent components gives:
R EQ =
R
(1− ω LC) + (ωCR )
jX LEQ =
2
2
2
jωL
(1− ω LC) + (ωCR )
- jX CEQ =
2
2
− jωC ω2 L2 + R 2
)
2
(
(1− ω LC) + (ωCR )
2
2
2
From this the equivalent resistive, inductive and capacitive components can be
identified. It is clear that cancellation of the inductive and capacitive
components requires that the values of the positive and negative imaginary
terms must be equal but opposite. This requires:
ωL
=
2
(
ωC ω2 L2 + R 2
)
(1− ω LC) + (ωCR ) (1− ω LC) + (ωCR )
2
2
6
2
2
2
Which gives:
(
ωL = ωC ω2 L2 + R 2
(
L = C ω2 L2 + R 2
C=
So that
)
)
L
ω2 L2 + R 2
For the component values given of L = 200mH, R = 50Ω at a frequency of 50Hz
200 x 10−3
200 x 10−3
0.2
0.2 x 106
C=
=
=
=
µF
2
2
62.8 + 50
3943.8 + 2500 6443.8 6443.8
C = 31µF
Then with the net reactance eliminated:
Ζ = R EQ =
R
(1− ω LC) + (ωCR )
2
2
2
For the values given of R = 50Ω, L = 200mH and C = 31µF:
R EQ =
R EQ =
In this case
50
(1 − 314
2
) (
2
x 0.2 x 31 x 10− 6 + 314 x 31 x 10− 6 x 50
)
2
50
50
50
=
=
= 128.2Ω
2
2
0.15
+
0
.24
0
.
39
(1 − 0.61) + (0.49)
Im =
Vm Vm
310
=
=
= 2.42 A
Ζ R EQ 128.2
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Then for the modified circuit with the Power Factor CosФ = 1
PAPP = PAVE =
Vm I m
2
=
310 x 2.42
= 375 W
2
and with SinФ = 0
Reactive Power = PIMAG = j
Vm I m
310 x 3.86
Sinφ = j
x 0 = 0W
2
2
Note that the average or active power has not been increased but in fact remains
the same. The apparent power, however has been reduced so that the power
which must be delivered by the source is also reduced. In fact, the inactive,
reactive or imaginary power has been eliminated from the point of view of the
source which no longer has to generate this power. This means that all of the
power drawn from the source is now only useful active power and no surplus
power is demanded which is not used.
In practice some reactive power is drawn from the source during initial
transient conditions then this power oscillates between the inductance and the
capacitance present under steady state conditions but is not dissipated.
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