1E6 Electrical Engineering:
AC Circuit Analysis and Power
Lecture 14: Power in AC Circuits I
14.1 Resistive Power Dissipation
Consider firstly power dissipation in a resistor when supplied by a dc battery as
shown in Fig 1. In this case the voltage across the resistor is constant and
therefore also is the current flowing through it. Consequently, the power
dissipation, given as the product of the voltage and the current, is constant and
invariant with time.
V
I
+
V
R
VL = V
I
P
t
Fig. 1 Power Dissipation in a DC Driven Resistive Circuit
Power dissipated in the load is given as:
V2
P  VI 
 I2R
R
In the case of an ac voltage source as shown applied to a circuit in Fig. 2, the
situation is somewhat different. The voltage across the resistor varies with time
and as a result so does the current flowing through it. However, in a resistor the
voltage and current are in phase with each other so that the waveforms of each
are the same. That is, both have the same function of time, as can be seen in the
case of the sinusoidal source shown. It can also be seen that the power, defined
as the product of the voltage and the current, is also a function of time, varying
in a sinusoidal manner.
Vt   Vm Sin ωt
i(t)
V(t)
~
RL
I m  Vm / R
VL = V(t)
it   I m Sin ωt
Pi t   V(t)i(t)
Fig. 2 An AC Driven Resistive Circuit
1
Vm
V
I
Im
Pi
PAVE
t
1 2π
T 
f ω
T
Fig. 3 Wavefroms showing Power Dissipation in an AC driven Resistive Circuit
14.2 Instantaneous and Average Power
From the waveforms shown in Fig. 3 above for an ac voltage source driving a
resistive load, it can be seen that as the voltage and current have exactly the
same phase relationship and that the resulting power waveform is always
positive. This means that power is continuously dissipated in the load, even
though it varies as a function of time from zero to some maximum value. It can
also be seen that the power waveform varies at twice the frequency of either the
voltage or current. The manner of variation of power on a short-term cyclical
basis is rarely of significant interest and it is the longer term power delivered to
the load that is of interest, that is, the average power. As the power variation is
cyclical and therefore repetitive, it is possible to calculate the average value over
one cycle of excitation and this therefore represents the long-term value with
time. It also corresponds to the equivalent amount of constant or dc-type power,
which would be delivered by a battery driving the same load.
Instantaneous Power:
Instantaneous power is the product of the instantaneous voltage across and the
instantaneous current flowing through a load and is therefore a function of time.
Pi t   V(t)i(t)
If
Then
So that
Vt   Vm Cos ωt and
at the load
it   I m Cos ωt
Pi  Vm Im Cos2ωt
Instantaneous Power
Vm I m
Pi 
2
2
1  Cos2t 
Average Power:
Average power is the long-term or average value of the instantaneous power. For a
periodic source it is calculated over one full cycle of the source delivering it to a
load.
It is an equivalent value of constant power.
PAVE 
If
Then
1
T

0
Pidt 
1
T

PAVE
PAVE
Vt it dt
T
0
Vt   Vm Cos ωt and
it   I m Cos ωt
1 T1
  Vm I m 1  Cos 2t 
T 0 2
V I
PAVE  m m
2T
T
 1.dt 
0
Vm I m
2T

T
0
Cos 2t.dt
Vm I m T
Vm I m
T

t0 
Sin 2t 0
2T
2T2ω
Vm I m
V I
(T  0)  m m Sin 2T  Sin 0
2T
2T2ω
PAVE 
But
T
ω  2f 
so
Then
For a resistive load
2
4T
then 2T 
 4π  2x2π
T
T
Sin2 T  Sin4 π  Sin0  0
PAVE 
Vm I m
2T
V
I m  m then
R
3
xT
Vm I m
2
2
2
V I
V
I R
PAVE  m m  m  m
2
2R
2
Recall the RMS value of a
sinusoidal waveform, i. e the root of
the mean of the square.
Vm
Vrms
By definition:
T
Vave = 0
t
1 T 2
VRMS 
V t dt

0
T
VRMS 
1 T 2
2
V
Cos
t.dt 
m

0
T
VRMS 
Vm2
xT 
2T
Vm2
2T

T
0
(1  Cos2t).dt
Vm2
V
 m
2
2
It can be seen from the previous relationship for average power that if:
Vm
Im 
then
R
Vm I m Vm I m
PAVE 

 VRMS I RMS
2
2 2
This is the idea behind an rms value by definition. The rms value of an ac
sinusoidal source voltage is that value of voltage which delivers the same average
power to a load as a dc supply of the same value.
Also
PAVE  VRMS I RMS
2
VRMS

 I 2RMS R
R
Note that for the sinusoidal source the instantaneous power varies between zero,
when V(t) = 0, i(t) = 0 and a maximum which occurs when V(t) = Vm and i(t) =
Im. This means that for a sinusoidal source the average power is half of the peak
power.
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14.3 Power in a Purely Inductive Load
In an inductor the current lags the voltage by 90o as can be seen from the
waveforms shown in Fig. 4 below. If the source voltage is sinusoidal, then the
current is also sinusoidal but shifted in phase. The instantaneous power, defined
as the product of the instantaneous voltage and current, can also be seen to be
sinusoidal in time. However, in contrast to the resistive load, the instantaneous
power in the inductor goes negative for part of the cycle of the source driving it.
The average power can be determined in a similar manner to that for the
resistive load.
Vt   Vm Sin t
j
i(t)
V(t)
~
it    Im Cost
V(t)
L
i(t)
ω
XL  L
Vt   jLit 
ω
-j
Vm
V(t)
Im
i(t)
Pi(t)
_
+
+
+
_
_
_
t
Fig. 4 The Power Associated with a Purely Inductive Circuit
Note: The instantaneous power alternates positive and negative at twice the
frequency of the source supplying it.
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Instantaneous Power
Pi  Vt  it    Vm Sin ωt I m Cos ωt
 Vm I m
Sin 2t  Sin 0
Pi 
2
Pi 
Average Power
PAVE
 Vm Im
PAVE 
Sin 2t
1 T

Vm I m Sin 2t
2T 0
PAVE  
PAVE
2
Vm I m
2T

T
0
Sin 2t
Vm I m
T

Cos 2t 0
2T2 
Vm I m
Cos 4 - Cos 0
2T2 
PAVE 
Vm I m
11 0
4T
We conclude that the average power dissipated in a pure inductance is zero.
However, it can be seen that the instantaneous power is not zero, except at zero
crossings of the time axis. This means that power is drawn from the source.
What happens is that when the instantaneous power is positive, energy is drawn
from the source and stored in the inductor for a quarter of a cycle. When the
instantaneous power is negative, this stored energy is returned to the source
which reabsorbs it. However, the source must still have the capacity to provide
the power required by the inductor, even though this is not dissipated or
consumed.
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14.4 Power in a Purely Capacitive Load
In a capacitor the current leads the voltage by 90 o as can be seen from the
waveforms shown in Fig. 5 below. If the source voltage is sinusoidal, then the
current and the instantaneous power are also sinusoidal. Again, in contrast to
the resistive load, the instantaneous power in the capacitor goes negative for
part of the cycle of the source driving it, alternating between positive and
negative phases twice per cycle. The average power dissipated by the capacitor
can again be determined by integration of the instantaneous power.
Vt   Vm Sin t
j
i(t)
it   Im Cost
ω
i(t)
V(t)
~
V(t)
C
ω
XL 
Vt    j
-j
Vm
1
C
1
it 
C
V(t)
Im
i(t)
Pi(t)
+
+
_
+
+
_
_
t
Fig. 5 The Power Associated with a Purely Capacitive Circuit
Note: As in the previous case, the instantaneous power alternates between
positive and negative phases at twice the frequency of the source supplying it.
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Pi  Vt  it   Vm Sin ωt I m Cos ωt
Instantaneous Power
Vm I m
Sin 2t  Sin 0
Pi 
2
Pi 
Average Power
Vm I m
2
Vm I m

2T
PAVE

T
0
Sin 2t
Sin 2t  0
Therefore, as was the case for the inductor, the average power dissipated in a
capacitor is zero. As with the inductor, power is drawn from the supply and
stored as energy in the capacitor for a quarter of a cycle and then returned to
the source during the following quarter cycle.
14.5 Imaginary Power
The power transferred to the inductor and capacitor in reactive circuits is
energy which is temporarily stored and then returned to the source. This power
is not dissipated and can therefore be considered as imaginary power.
Consider the inductor shown in Fig. 6 below
Vt   Vm Sin t
i(t)
V(t)
~
it    Im Cost
L
ZL  j ωL
Fig. 6 A Purely Inductive Circuit
But
i(t) 
v(t) v(t)
v(t)

j
ZL
j ωL
ωL
8
Pi  
Then
1
Vm I m Sin 2t
2
2
V
Pi  j m Sin 2t
2 L
So that
which is purely imaginary
Consider the capacitor
Vt   Vm Sin t
i(t)
V(t)
~
it   Im Cost
C
ZC   j
1
C
Fig. 6 A Purely Capacitive Circuit
i(t) 
But
Then
So that
Pi 
v(t)
v(t)

 jCv(t)
ZC  j 1
C
1
Vm I m Sin 2t
2
1
Pi  j Vm2 C Sin 2t
2
which is purely imaginary
This means that both the inductance and the capacitor can be thought of as
consuming imaginary power. However, they do not dissipate the energy or
power that they draw from the supply.
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