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Chapter 10 Sinusoidal Steady- State Power Calculations
In Chapter 9 , we calculated the steady state voltages and currents in electric circuits
driven by sinusoidal sources
We used phasor method to find the steady state voltages and currents
In this chapter, we consider power in such circuits
The techniques we develop are useful for analyzing many of the electric devices we
encounter daily, because sinusoidal sources are predominate means of providing electric
power in our homes, school and businesses
Examples
Electric Heater which transform electric energy to thermal energy
Electric Stove and oven
Toasters
Iron
Electric water heater
And many others
10.1 Instantaneous Power
Consider the following circuit represented by a black box
i (t )
i (t )  I m cos(t  i )

v (t )

v (t )  V m cos(t  v )
The instantaneous power assuming passive sign convention
( Current in the direction of voltage drop
p (t ) v (t )i (t )
  
)
( Watts )
If the current is in the direction of voltage rise (   ) the instantaneous power is
i (t )

v (t )

p (t )   v (t )i (t )
i (t )

v (t )

i (t )  I m cos(t  i )
v (t )  V m cos(t  v )
i (t )  I m cos(t )
v (t )  V m cos(t  v  i )
p (t ) v (t )i (t )  {V m cos(t  v  i )}{I m cos(t )}
 V m I m cos(t  v  i ) cos(t )
Since
1
1
cos cos   cos(   )  cos(   )
2
2
Therefore
p (t )  V m I m cos(v  i )  V m I m cos(2t v  i )
2
2
Since
cos(   )  cos cos   sin sin 
cos(2t v  i )  cos(v  i )cos(2t )  sin(v  i )sin(2t )
p (t )  V m I m cos(v  i )  V m I m cos(v  i )cos(2t ) V m I m sin(v  i )sin(2t )
2
2
2
i (t )

v (t )

i (t )  I m cos(t )
v (t )  V m cos(t  v  i )
p (t )  V m I m cos(v  i )  V m I m cos(v  i ) cos(2t ) V m I m sin(v  i ) sin(2t )
2
2
2
v =60o i =0o
You can see that that the frequency of the Instantaneous
power is twice the frequency of the voltage or current
10.2 Average and Reactive Power
Recall the Instantaneous power p(t)
p (t )  V m I m cos(v  i )  V m I m cos(v  i ) cos(2t ) V m I m sin(v  i ) sin(2t )
2
2
2
p (t )  P  P cos(2t ) Q sin(2t )
where
P  V m I m cos(v  i )
2
Q  V m I m sin(v  i )
2
Average Power (Real Power)
Reactive Power
Average Power P is sometimes called Real power because it describes the power in
a circuit that is transformed from electric to non electric ( Example Heat )
It is easy to see why P is called Average Power because
t0+T
1
p (t )dt
T t0

t0+T
1

{ P  P cos(2t )Q sin(2t )}dt  P
T t0

Power for purely resistive Circuits
p (t )  P  P cos(2t ) Q sin(2t )
v =i
P  V m I m cos(v  i )
2
 V m I m cos(0)
2
 V mIm
2
Q  V m I m sin(v  i )  V m I m sin(0)  0
2
2
p (t )  V m I m  V m I m cos(2t )
2
2
V mIm
2
The Instantaneous power can never be negative
power can not be extracted from a purely resistive network
V mIm
Power for purely Inductive Circuits
v =i  90o
v i = 90o
p (t )   V m I m sin(2t )
2
p (t )  P  P cos(2t ) Q sin(2t )
P  V m I m cos(v  i )  V m I m cos(90o )  0
2
2
Q  V m I m sin(v  i )  V m I m sin(90o )  V m I m
2
2
2
V mIm
2
The Instantaneous power p(t) is continuously
exchanged between the circuit and the source
driving the circuit. The average power is zero
When p(t) is positive, energy is being stored in
the magnetic field associated with the inductive
element
When p(t) is negative, energy is being extracted
from the magnetic field
The power associated with purely inductive
circuits is the reactive power Q
The dimension of reactive power Q is the same
as the average power P. To distinguish them we
use the unit VAR (Volt Ampere Reactive) for
reactive power
V mIm
2

Power for purely Capacitive Circuits
v =i  90o
v i =  90o
p (t ) V m I m sin(2t )
2
p (t )  P  P cos(2t ) Q sin(2t )
P  V m I m cos(v  i ) V m I m cos(90o )  0
2
2
Q  V m I m sin(v  i )  V m I m sin(90o )  V m I m
2
2
2
The Instantaneous power p(t) is continuously
exchanged between the circuit and the source
driving the circuit. The average power is zero
V mIm
2
When p(t) is positive, energy is being stored in
the electric field associated with the capacitive
element
When p(t) is negative, energy is being extracted
from the electric field
The power associated with purely capacitive
circuits is the reactive power Q (VAR)
V mIm
2

The power factor
Recall the Instantaneous power p(t)
p (t )  V m I m cos(v  i )  V m I m cos(v  i ) cos(2t ) V m I m sin(v  i ) sin(2t )
2
2
2
P average
P average
Q reactive
power
power
power
 P  P cos(2t ) Q sin(2t )
The angle v  i plays a role in the computation of both average and reactive power
The angle v  i is referred to as the power factor angle
We now define the following :
The power factor
The reactive factor
pf  cos(v  i )
rf  sin(v  i )
The power factor
pf  cos(v  i )
Knowing the power factor pf does not tell you the power factor angle , because
cos(v  i )  cos(i v )
To completely describe this angle, we use the descriptive phrases lagging power factor
and leading power factor
Lagging power factor implies that current lags voltage hence an inductive load
Leading power factor implies that current leads voltage hence a capacitive load
10.3 The rms Value and Power Calculations
Assume that a sinusoidal voltage is applied to the terminals of a resistor as shown
+
V m cos(t  v )
R

Suppose we want to determine the average power delivered to the resistor
P

t0+T
1
T t0


t0+T V
p (t )dt  1
T t0
 dt  1
cos(t  )
m
v
R
2
R

1

T

t0+T
t
0
V cos (t 
m
v
2
2
t0+T
1
2
2
V rms 
V cos (t  )dt
m
v
T t0

However since
2
V
P  rms
R
If the resistor carry sinusoidal current
2
P  RI rms


)dt 


Recall the Average and Reactive power
P  V m I m cos(v  i )
2
Q  V m I m sin(v  i )
2
Which can be written as
P  V m I m cos(v  i )
2 2
Q  V m I m sin(v  i )
2 2
Therefore the Average and Reactive power can be written in terms of the rms value as
P  V rmsI rms cos(v  i )
Q  V rms I rms sin(v  i )
The rms value is also referred to as the effective value eff
Therefore the Average and Reactive power can be written in terms of the eff value
as
P  V eff I eff cos(v  i )
Q  V eff I eff sin(v  i )
Example 10.3
10.4 Complex Power
Previously, we found it convenient to introduce sinusoidal voltage and current in terms
of the complex number the phasor
Definition
Let the complex power be the complex sum of real power and reactive power
S  P  jQ
were
S is the complex power
P is the average power
Q is the reactive power
Advantages of using complex power
S  P  jQ
 We can compute the average and reactive power from the complex power S
P  {S }
Q  {S }
 complex power S provide a geometric interpretation
S  P  jQ  S
e
j
S
were
S = P Q
2
2
Q
(reactive power)

Is called apparent power
P
(average power)
V I cos(   ) 
 cos(   ) 
Q 
v
v
i   tan  
i   tan  tan(   ) 
  tan   m m
 v i 




P 
V
I
sin(



)
sin(



)
v
v
m
m
i 
i 


 =tan 
v  i
power factor angle
The geometric relations for a right triangle mean the four power triangle dimensions
(|S|, P, Q,  ) can be determined if any two of the four are known
Example 10.4
10.5 Power Calculations
S  P  jQ  V m I m cos(v  i )  j V m I m sin(v  i )
2
2
j (v  i )
j (v  i )
V
I
V
I
m
m
m
m
cos(v   )  j sin(v   )  
e
 V eff I eff e

i
i 
2
2 
 V eff e
were
*
I eff
j v
I eff e
j i
*
 V eff I eff
Is the conjugate of the current phasor
I eff

V eff

Also
S
1
VI *
2
Circuit
I eff
Alternate Forms for Complex Power
I eff

V eff

The complex power was defined as
Circuit
S  P  jQ
Then complex power was calculated to be
*
OR
S  V eff I eff
S
1
VI *
2
However there several useful variations as follows:
First variation
I eff

V eff

*
*
*
S  V eff I eff
 (ZI eff )I eff
 ZI eff I eff
Z = R + jX
 (R + j X ) | I eff |2  R | I eff |2 + j X | I eff |2
P
P  R | I eff |
2

2
R I eff
 Z | I eff |2
1
 R I m2
2
Q
2 
Q  X | I eff |2  X I eff
1 2
XI
2 m
Second variation
S
I eff

V eff

*
V eff I eff
V
V eff  eff
 Z
*



|V eff |2
|V eff |2 R  j X


R  jX
R  j X R  jX
Z = R + jX

Q
R
|V eff |2
R 2X 2

R
2
V eff
R 2X 2
X
2
X
V eff
|V eff |2  2
2
2
R X 2
R X
If Z = R (pure resistive) X= 0
If Z = X (pure reactive) R= 0
P
|V eff |2

Z*
 R  j X |V eff |2
R 2 X 2
X
R
|V eff |2  j
|V eff |2
2
2
R 2 X 2
R X
P
P
*
V effV eff

Z*


Q
1 R
V m2
2 R 2 X 2
1 X
V m2
2 R 2 X 2
2
R
|V eff |2  |V eff |
R 2X 2
R
P 0
2
X
|
V
|
2
Q
|V |
eff
2
2 eff  X
R X
Q 0
Example 10.5
Line
Load
rms because
the voltage is
given in terms
of rms
Q  650 var
P  975 W
The load average power is the power absorb by the load resistor 39 W
V RI R
Recall the average Power for purely resistive Circuits P  m m V R I R
2
Another solution
eff
were
V
R
eff
P V
and I R
ef f
R
eff
IR
eff
ef f
Are the rms voltage across the resistor and the rms current through
the resistor
2
 RI eff

V
R
eff


V
Inductor
eff

P  975 W
Q  650 var
From Power for purely resistive Circuits
IL
1
P  V m I m V eff I eff
P |V R || I R | V R I R
eff
eff
eff ef f
2
39
39
VR 
V 
234.36e j 3.18  195e j 36.87
eff
39 j 26 L 39 j 26
|V effR |
P V R I R
 (195)(5)  975 W
o
eff
OR
Q V eff I eff
Inductor
eff
V
R
eff
195
eff
P V
R
eff
IR
eff
Q V
130
 R (I R )2  (39)(52 )  (39)(25)  975 W
 (RI R )I R
eff
Inductor I Inductor
eff
eff
I L 5
V
o
eff
eff
V
Inductor
eff

j 26
j 26
V 
234.36e j 3.18  130e
39 j 26 L 39 j 26
Q  (130)(5)  650 VAR
o
OR
2
Q  XI eff
 650 var
j 93o
Line
V
Line
eff

2 R
P  I eff
2
Q  I eff
X
OR using complex power
Line *
S Line  V eff
I eff
V
Line
eff

1 j 4
(250)
(1 j 4)(39 j 26)
V Line  20.6
eff
Line *
S Line  V eff
I eff
 20.6
OR
V
Line
eff
 250 V
L
39.1o V rms
39.1o
5 36.87o 103
75.97o
 25  j 100 VA
Line
Load
S
Absorb
S
Line
+S
From part (c)
Load
(25 j 100) + (975 j 650)
S
OR
S
Supply
From part (b)
Supply
  250
= S
Absorb
 (1000  j 750) VA
0o (I* )   250
L
 1000  j 750 VA
 (25  975)  j (100 + 650)1000  j 750 VA
0o
 1250
36.87o VA
Example 10.6 Calculating Power in Parallel Loads
pf  cos(v  i )
S  8000  j 6000 VA
1
S 12000  j 16000 VA
S  20000  j 10000 VA
2
The apparent power which must be supplied to these loads is
|S | | 20000  j 10000| VA  22.36 kVA