1E6 Electrical Engineering
AC Circuit Analysis and Power
Lecture 15: Power in AC Circuits II
15.1 Power Dissipation in an AC circuit
In general, an ac circuit will contain a combination of resistive and reactive
components and the reactive elements may be either inductive or capacitive as
shown in Fig.1 below. This means that at different points in the circuit the
current and voltage relationships will vary depending on the elements involved.
From the point of view of a voltage source driving such a circuit, the overall
network will have an impedance, which has a magnitude and phase and a
current will flow into the circuit which also possesses a corresponding
magnitude and phase as shown below.
i(t)
R1
C1
L2
R2
Z
v(t)
~
C2
L1
v(t)
Z
i(t)
j
v

-j
R3
i(t) 
i
i(t) 
v(t) v 0

Z
Z 
v 0
Z

Fig. 1 The Phase Relationship Associated with an AC Circuit having Reactance
1
A plot of the voltage, which is taken as the reference zero angle, and the current
with the instantaneous power is shown in Fig. 2 below. The current is seen to lag
behind the voltage by an angle  . Note that, unlike the case for resistive and
purely reactive circuits, the instantaneous power profile is not symmetrical. It
can be seen in this example that the power profile is positive for longer than it is
negative and also that it reaches a higher positive peak than negative peak. This
means that more power is delivered to the network in each cycle of the
sinusoidal source than is returned to the source. Therefore there is a net transfer
of power from the source to the circuit and this power is dissipated in the
resistive components of the network.
Vm
V(t)
Im
i(t)
+
+
_
Pi(t)
_
+
_
_
t
Fig. 2 Waveforms Showing Power Relations in an AC Circuit having Reactance
Instantaneous Power:
The instantaneous power can be found as before as the product of the voltage
and current as continuous functions of time:
If
Then
vt   Vm Sin ωt and
it   I m Sin ωt - 
Pi  Vm I mSin ωt.Sin ωt - 
2
Average Power:
1 T
1 T
PAVE   Pi dt   vt  it dt
T 0
T 0
1 T
Vm I m Sin t.Sin t - dt

0
T
PAVE 
PAVE
V I
 m m
T

T
0
Sin t.Sin t - dt
Using the trigonometric expansion
SinASinB 
1
Cos A - B - Cos A  B where A  t and B  t  
2
PAVE
V I
 m m
2T
 Cos  - Cos 2t - dt
T
0
Using the trigonometric expansion
Cos A  B  CosACos B  SinASin B where A  2t and B  
gives:
PAVE
V I
 m m
2T
 Cos- Cos 2t Cos-   Sin 2t Sin -  dt
T
0
Cos -   Cos
But
Sin -   -Sin 
and
So that:
PAVE
V I
 m m
2T
 Cos - Cos2t .Cos  Sin2 t Sin  dt
T
0
3
The factors Cos  and Sin  are constants for a given circuit where there is a
given phase shift between the supply voltage and the current drawn by the
circuit so that:
PAVE
T
T
Vm I m
Vm I m

Cos dt 
Cos Cos2t.dt
0
0
2T
2T
T
Vm I m

Sin   Sin2 t.dt
0
2T
PAVE 
Vm I m
V I
1
T
T
Cos t 0  m m Cos
Sin2 t 0
2T
2T
2

PAVE 
Vm I m
1
T
Sin 
Cos2t 0
2T
2
Vm I m
V I
Cos T - 0   m m Cos Sin4  - Sin0 
2T
4T

Vm I m
Sin  Cos4 - Cos0 
4T
The last two terms in this expression have a value of zero as before so that
finally:
PAVE 
Vm I m
V
I
Cos  m x m Cos
2
2
2
PAVE  VRMS I RMS Cos
The term Cos  is referred to as the Power Factor of the circuit. This is a
property of the ac network and is determined by the phase angle of the network
impedance.
Power Factor  Cos
4
The Power Factor varies between a value of 0 and 1.
  0o  Cos  1 PAVE  VRMS IRMS
  90o  Cos  0 PAVE  0
purely resistive circuit
purely reactive circuit
The average power calculated above is the actual power consumed from the
power delivered to the network. This is dissipated by the resistive elements of
the circuit. However, the source must be rated to generate and deliver the total
power demanded by the circuit even though not all of this is consumed. The
power dissipated is also referred to as Active Power and represents energy
consumed.
15.2 Complex Power
It has been seen from the previous waveform showing the instantaneous power
that the positive excursion is greater than the negative excursion, so that there is
a net transfer of power from the source to the load per cycle of the source
voltage. The phase of the impedance of the network results in a phase angle
between voltage and current which gives the Power Factor in the Average or
Active Power drawn by the network. However, as with purely reactive circuits,
there is also some power which is drawn from the source, stored temporarily in
the reactive elements and then returned to the source in a later part of each
cycle. This is referred to as the Reactive Power. In practice the source driving
the network must be rated to handle and deliver both the active and reactive
power, even though only the active power will be dissipated by the circuit. The
vector sum of the Active and Reactive Power is referred to as the Apparent
Power and gives the concept of Complex Power as illustrated in phasor form in
Fig. 3 below.
Apparent Power  Active Power  j Reactive Power
Apparent Power  Average Power  j Reactive Power
Fig. 3 A Phasor
Representation of
Complex Power
Apparent Power
VRMS IRMS

Active Power VRMS IRMS CosФ
5
Reactive Power
VRMS IRMS SinФ
 VRMS I RMS 
Apparent Power
Vm I m
2
Active or Average Power  VRMS I RMS Cos
 j VRMS I RMS Sin 
Reactive Power
and
2
2
2
VRMS
I 2RMS  VRMS
I 2RMS Cos 2   VRMS
I 2RMS Sin 2 
In order to avoid having to have a source which must be capable of providing
much more power than is actually going to be consumed by a network, the aim
is to minimise the amount of reactive power demanded of the source. Therefore
the aim is to make the apparent power and the active power equal. This means
making the power factor as close to unity as is possible.
Consider the network impedance shown in Fig. 4 below:
Z  R  jX
|Z|
Sin  
X
Z
Cos 
R
Z
jX

R
Power Factor  Cos 
Fig. 4 Power Factor in Complex Power
R
Z
where R is the overall equivalent resistance of the ac network as seen by the
source. This may not actually be a resistive element but can represent work done
by some piece of equipment or machine which is provided with electrical power
and consumes energy.
6
15.3 Case Study
Consider the circuit shown in Fig. 5 below.
i(t)
L
R1
50 Ω
750 mH
Z
v(t)
~V
C
220Sinωt
4.7 µF
Fig. 5 An Example Circuit for AC Power Analysis
f  50 Hz and ω  2 f  314 rad/s
then
jL  j x314 x 750 x103  j236Ω
1
1
106
-j
j
 j
  j 678Ω
ωC
314 x 4.7 x 10 6
1475.8
Then:
R2
ωC
Ζ  R1  jL 
1
R 2j
ωC
j
Ζ  50  j236 
 j678x1 50
150  j678
Ζ  50  j236  j
7
101700
150  j678
R2
150 Ω
Rationalising:
Ζ  50  j236  j
101700150  j678 
150 j678150  j678
 j15.3x10 6  69x10 6
Ζ  50  j236 
1502  6782
 j15.3x10 6  69x10 6
Ζ  50  j236 
482184
Ζ  50  j236  j32  143
Ζ 193  j236  j32
resistance
So that overall
inductive reactance
Ζ 193  j 204

capacitive reactance
Ω
The net impedance is more reactive than resistive and the reactance appears
inductive.
Ζ  1932  2042  281 
Ζ  Tan 1
204
 Tan 1 1.056  46.6
193
The current flowing into the circuit from the source can be found as:
i
V V 0
220 0


 0.78   46.6 A
Ζ Ζ Z 28146.6
8
The Power factor of the network is given as:
Power Factor  Cos  Cos46.6o  0.687
The complex power can be evaluated as:
Apparent Power  VRMS I RMS 
Active Power  VRMS I RMS 
Vm I m 220x 0.78

 85.8W
2
2
Vm I m
Cos  85.8x 0.687  58.9W
2
Reactive Power  j VRMS I RMSSin   j
9
Vm I m
Sin   j85.8x0.72 7  j62.4W
2