AC Power Lecture: Average Power, Reactive Power, and Power Factor Calculations

Lecture 6
Single-Phase Power
Computations

The objective of this lecture is to introduce AC power.
Lecture Objective

1. Understand the meaning of instantaneous and average
power,
2. master AC power notation, and compute average power,
and reactive for AC circuits
3. Compute the power factor of a complex load.
LEARNING OUTCOMES
After completing this course you will be able to:

AVERAGE POWER CALCULATIONS
i. the two-port network is linear and contain no energy
sources,
ii. the two-port network is excited by a sinusoidal input,
v(t) = Vmcost
iii. The terminal voltage v(t) and terminal current i(t) have
reached their steady state values.
i(t)
v(t)
Linear
network
Consider the two-port network shown in the figure below. We
assume that :

We wish to calculate:
1. Active power consumed by the network
2. Reactive power consumed by the network
3. Power factor at the input terminals of the network.
AVERAGE POWER CALCULATIONS

 
t
V
t
v m 
sin
)
( 
   
t
I
R
t
V
R
t
v
t
i m
m


sin
sin
)
(
)
( 


Let the terminal voltage v(t) be given by
the expression
where Im =Vm/R is the peak current.
Power Absorbed by Resistive Circuit
Consider a resistive circuit with a terminal
resistance R. i(t)
v(t)
Linear
circuit
Linear resistive circuit
Ohm’s law requires that the terminal current i(t) be related to
voltage v(t) and resistance R via the equation

Then, instantaneous power supplied to the circuit is
   
 
 
 
 
 
 
t
I
V
t
I
V
t
I
V
t
t
I
V
t
i
t
v
t
p
m
m
m
m
m
m
m
m





2
2
cos
2
2
2
1
1
cos
2
1
2
1
2
cos
1
2
1
sin
sin
)
(
)
(
)
(













Further simplifications give us
 
 
t
V
p m
m 
2
cos
1
I
2
1
(t) 

Therefore,
 
 
 
 
t
I
V
t
I
V
t
I
V
t
p
m
m
m
m
m
m



2
cos
1
2
2
2
2
cos
1
1
cos
1
)
( 2














 





A plot of p(t) is shown below.

From the plot, we note the following:
1. p(t) has two components: a constant component, and a
time-varying component,
2. p(t) is always positive.
rms
rms
m
m I
V
I
V
P 

2
1
The constant component is the average power consumed by
the circuit. Thus, average power consumed by the circuit is

t
I
t
i m 
sin
)
( 
Let
Power Absorbed by a Purely Inductive Circuit
i(t)
v(t)
Linear
circuit
Consider an inductive circuit of inductance L.
Then, for an inductor,
 
t
V
t
LI
t
I
dt
d
L
dt
di
L
t
v
m
m
m




cos
cos
sin
)
(




where m
m LI
V 


Instantaneous power supplied to the circuit is
t
I
V
t
t
V
I
t
V
t
I
t
i
t
v
t
p
m
m
m
m
m
m





2
sin
2
1
cos
sin
cos
sin
)
(
)
(
)
(






1. p(t) is equally both positive and negative; i.e. power is
circulating.
2. p(t) has no constant component. Thus, the average
power consumed by the purely inductive circuit element
over one cycle of the supply voltage is zero. That is,
0

P

t
V
t
v m 
sin
)
( 
Let
Power Absorbed by a Purely Capacitive Circuit
i(t)
v(t)
Linear
circuit
Consider a purely capacitive circuit of capacitance C.
Then, for a capacitor,
 
t
I
t
CV
t
V
dt
d
C
dt
dv
C
t
i
m
m
m




cos
cos
sin
)
(






where
2
m
m CV
I 


Instantaneous power supplied to the circuit is
t
I
V
t
t
V
I
t
I
t
V
t
i
t
v
t
p
m
m
m
m
m
m





2
sin
2
1
cos
sin
sin
cos
)
(
)
(
)
(









1. p(t) is equally both positive and negative, power is
circulating.
2. p(t) has no constant component. Thus, the average
power consumed by the purely capacitive circuit element
over one cycle of the supply voltage is zero. That is,
0

P

Power absorbed by a Reactive Circuit
Consider a linear AC circuit of terminal impedance Z. Let the
the circuit be excited by a sinusoidal source, v(t) = Vmcos(ωt),
as shown in the figure below. Assume that the impedance Z
causes the terminal current i(t) to lag behind the terminal
voltage by an angle . For a linear circuit we can write
i(t)
v(t)
Linear
circuit
where Im is the amplitude of the terminal
current.
 

 
 t
I
t
i m cos
)
(

Instantaneous power dissipated by a circuit element is
given by the product of the instantaneous voltage and
current:
The above equation can be further simplified with the aid of
trigonometric identities to yield
where θ is the difference in phase between voltage and current.
   


 

 t
I
V
I
V
t
p m
m
m
m
2
cos
2
cos
2
)
(
   


 

 t
t
I
V
t
i
t
v
t
p m
m cos
cos
)
(
)
(
)
(

The equation shows that the instantaneous power dissipated
by an AC circuit element is equal to the sum of an average
component:
and a sinusoidal component
oscillating at a frequency double that of the original source
frequency.
 

cos
2
m
mI
V
 

 
t
I
V m
m
2
cos
2

A plot of the instantaneous and average power is shown below.

The average power can be obtained by integrating the
instantaneous power over one cycle of the sinusoidal signal.
   
 




cos
2
2
cos
2
1
cos
2
1
)
(
1
0
0
0
m
m
av
T
m
m
T
m
m
T
av
I
V
P
dt
t
I
V
T
dt
I
V
T
dt
t
p
T
P








since the second integral is equal
to zero and cos(θ) is a constant.
Therefore,

The same analysis can be also be carried out in the phasor
domain. In the phasor domain, the terminal current and
terminal are given by
I
V
Linear
circuit

j
m
j
m
e
I
I
e
V
V


 0
The impedance of the circuit element is defined by the phasor
voltage and current to be



j
j
m
m
j
m
j
m
Ze
e
I
V
e
I
e
V
I
V
Z 


 
0
where
m
m
I
V
Z 

The expression for the average power also be represented
using phasor notation, as follows:

 cos
2
1
cos
2
1 2
2
Z
I
Z
V
P m
m
av 

It may be expanded using trigonometric identities to obtain
the following expressions:

 cos
2
1
cos
2
1 2
2
Z
I
Z
V
P m
m
av 

In terms of rms values, we have

 cos
cos 2
2
Z
I
Z
V
P rms
rms
av 


It may be expanded using trigonometric identities to obtain
the following expressions:
Recall the expression for the instantaneous power given
below
True, Reactive, and Apparent power
   


 

 t
I
V
I
V
t
p m
m
m
m
2
cos
2
cos
2
)
(
       
 
         
 
   
     
t
Z
I
t
Z
I
t
t
Z
I
t
t
Z
V
t
p
rms
rms
rms
rms














2
sin
sin
2
cos
1
.
cos
2
sin
.
sin
2
cos
.
cos
cos
2
sin
sin
2
cos
cos
cos
)
(
2
2
2
2










Now, from the impedance triangle, we have
where R and X are the resistive and reactive components of
the load impedance, respectively. On the basis of this fact, it
becomes possible to write the instantaneous power as
R
Z 

cos
X
Z 

sin
and
 
   
   
t
X
I
t
R
I
R
I
t
X
I
t
R
I
t
p
rms
rms
rms
rms
rms




2
sin
2
cos
.
2
sin
2
cos
1
.
)
(
2
2
2
2
2






Z
X
R


The physical interpretation of this expression for the
instantaneous power is as follow:
Instantaneous power dissipated by a complex load consists
of the following three components:
1. An average component, which is constant; this is called the
average power and is denoted by the symbol Pav:
where R = Re[Z].
R
Irms
2
av
P 

2. A time-varying (sinusoidal) component with zero
average value that is contributed by the power
fluctuations in the resistive component of the load
and is denoted by PR(t):
   
t
P
t
R
I
t
p av
rms
R 
 2
cos
.
2
cos
.
)
( 2


3. A time-varying (sinusoidal) component with zero
average value, due to the power fluctuation in the
reactive component of the load and denoted by
pX(t):
   
t
Q
t
X
I
t
p rms 
 2
sin
2
sin
)
( 2


where X = Im [Z] and Q is called the reactive power.

1. Since reactive elements can only store energy and not
dissipate it, there is no net average power absorbed by X.
2. Since Pav corresponds to the power absorbed by the load
resistance, it is also called the real power, measured in
units of watts (W). On the other hand, Q takes the name
of reactive power, since it is associated with the load
reactance. The units of Q are volt-amperes reactive, or
VAR.
3. The combination of reactive power and true power is
called apparent power, and it is the product of a circuit's
voltage and current, without reference to phase angle.
Apparent power is measured in the unit of Volt-Amps (VA)
and is symbolized by the capital letter S.
Notes

There are several power equations relating the three types of
power to resistance, reactance, and impedance (all using
scalar quantities):
Summary

Worked Example
Compute the power absorbed (i) by R, (ii) by L.
Solution
  V
o
s
S t
v
P
V 15
20
)
( 








 
12
10
120
100 3
j
j
L
j

Solution
Phasor domain circuit:
R
V
L
V

12
j
V
o
S
V 15
120 


I


 12
25 j
Z
12
25
15
20
j
Z
V
I
o
S





Compute circuit current, .
I
A
64
.
40
72
.
0 o




Solution
  W
1
.
234
25
72
.
0
2
1
Re
2
2
2










 Z
I
P m
  VAr
34
.
112
12
72
.
0
2
1
Im
2
2
2












 Z
I
Q m
Therefore,
R
V
L
V

12
j
V
15
20 o
S
V 


I

4.327 A peak
120 V peak
234.1 W average
The plot shows:
1. Power flows from the supply into the load for only a part of the
cycle! For a portion of the cycle, power actually flows back to
the source from the load!
2. There is reactive power in the circuit.
A plot of the voltage and current waveforms is shown in the
figure below.

Worked Example
For the circuit below, compute :
(i) true power absorbed by the load
(ii) reactive power consumed by the load
(iii) apparent power supplied by the source

A 10  resistor and a capacitive reactance of 20  are
connected in series to a 240 V supply. Calculate the apparent
power, the true power and the reactive power supplied by the
source.
Worked Example
Solution

Solution
Impedance seen by the source





 4
.
22
20
10 2
2
2
2
C
X
R
Z
A
7
.
10
4
.
22
240



Z
V
I S
Current flowing through the circuit

Apparent Power supplied to the load
VA
2570
7
.
10
240 


VI
S leading
True power supplied to the load is the power consumed by
the resistor R.
W
1145
10
7
.
10
cos
2
1 2
2
2




 R
I
Z
I
P rms
m
av 
Reactive power supplied to the load is the power consumed
by the capacitance C.
VAR
2290
20
7
.
10
sin
2
1 2
2
2




 C
rms
m X
I
Z
I
Q 

Exercise
Calculate voltage drop across each element, the
apparent power, the true power and the reactive
power supplied by the source. Given : vS(t) =
1002cos1000t






 
20
10
20
1000 3
L
XL 





 
10
10
100
1000
1
1
6
C
XC

Impedance seen by the source
  





 14
.
14
10
10 2
2
2
2
C
L X
X
R
Z
A
07
.
7
14
.
14
100



Z
V
I S
Current flowing through the circuit
Solution

Solution
Voltage drop across each element:
V
7
.
70
10
07
.
7 


 IR
VR
V
4
.
141
20
07
.
7 


 L
L IX
V
V
7
.
70
10
07
.
7 


 C
C IX
V
Apparent power supplied to circuit elements by the voltage
source
VA
707
100
07
.
7 


 S
IV
S

True power supplied to the load is the power consumed by
the resistor R.
W
500
10
07
.
7 2
2



 R
I
P rms
av
by the inductor and the capacitance.
  VAR
500
10
07
.
7
sin
2
1 2
2
2





 C
L
rms
m X
X
I
Z
I
Q 
Solution

Calculate the total active and reactive powers supplied by the source
to the resistors.
Worked Example

The total impedance seen by the source is
Solution
    




 o
j
j
Z 62
.
20
69
.
12
6
8
//
4
10
Z
 

 47
.
4
877
.
11 j
Z
In rectangular form,

Total active supplied by the source is equal to the active
power consumed by the real part of the load.
  W
4533
69
.
12
9
.
18
Re 2
2



 Z
I
P rms
av
by the capacitance C.
  VAR
2143
6
9
.
18
sin
2
1 2
2
2




 C
rms
m X
I
Z
I
Q 
Solution
Current supplied by the source to the load is
A
9
.
18
69
.
12
240



Z
V
I S
rms

Power Factor
The ratio of the real power to the apparent power is called
the power factor (pf).
S
P


power
apparent
power
real
(p.f.)
factor
Power
 
m
m
m
m
I
V
2
1
cos
I
V
2
1





cos
.
. 
f
p
I
V -


 
where
Therefore,

Calculate the power factor seen by the source and the average power
supplied by the source.
Worked Example


cos
.
. 

S
P
f
p
Z
X
R

When we have reduced the load seen by the source into one
equivalent impedance, , then the power factor of the load is
simply equal to the cosine of the angle  in the impedance triangle.
That is,
Note
Z
Linear
network
I
V
Z
Therefore, power factor of load is
   

cos
.
.
2
2
2
2
2





Z
R
X
I
R
I
R
I
S
P
f
p
Hence, for the circuit given
  936
.
0
62
.
20
cos
.
. 
 o
f
p

To determine whether the current is leading or lagging the source
voltage, we recall that
Z
I
V 

By setting , we obtain
o
I
I 0


o
o
o
IZ
Z
I
Z
I
V 62
.
20
62
.
20
0 







This result tells us that phasor V is leading phasor I by 20.62o. More
usually, we say that phasor I is lagging phasor V by 20.62o.
  936
.
0
62
.
20
cos
.
. 
 o
f
p
Hence, for the circuit given
lagging

 
Z
I
P m
Re
2
2







Average power supplied by the source to the load is
  118
877
.
11
152
.
3
Re 2
2




 Z
I
P
In terms of rms values, we can write
Load current is
A
o
o
s
Z
V
I 62
.
20
252
.
31
62
.
20
60
.
12
0
40






Therefore,
W
118
877
.
11
152
.
3 2



P

Why is electrical apparatus is rated in VA instead of watts ?
The calculations above show both loads dissipate 120 kW, but the
current rating of generator (b) is exceeded because of the power
factor of its load.
VA Rating of Electrical Apparatus

Summary
In this study unit we have looked at
1. Active power
2. Reactive power
3. Apparent power, and
4. Power factor

AC Power Lecture: Average Power, Reactive Power, and Power Factor Calculations

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AC Power Lecture: Average Power, Reactive Power, and Power Factor Calculations