Conservation of Complex Power

1. Lecture 8
Conservation of Complex
Power Law

2. After completing this unit you will be able to:
1. Apply complex power conservation law to compute active and
reactive power consumption in a circuit
2. Apply complex power conservation law to determine the value of
the capacitance needed to correct the power factor of a circuit.
Learning Outcomes

3. Consider the parallel connected circuit shown below. The complex
power supplied to the circuit is given by the expression
Conservation of Complex Power in Parallel Circuit
*
2
1
I
V
S 

*
2
*
*
1 I
I
I 

where I* is the conjugate of the input
current I. For the circuit given, the
input current I is related to the branch
currents I1 and I2 via the equatian
2
1 I
I
I 

Hence
S

4. This result shows that for a parallel-connected circuit the complex
power supplied by a voltage source is equal to the sum of the
complex power consumed by the load impedances.
Thus we can write
Conservation of Complex Power in Parallel Circuit
*
2
1
I
V
S 
  
*
2
*
1
2
1
I
I
V 


*
*
2
1
2
1
2
1
I
V
I
V 



2
1 S
S 

S

5. Alternatively, we can rewrite the previous expression as follows:
0


 2
1 S
S
S
In words, this result tells us the sum of complex power absorbed by
all the elements of the circuit (i.e., voltage source and impedances) is
zero. This statement is the complex power conservation law.
Conservation of Complex Power in Parallel Circuit

6. Consider next the series-connected circuit shown below. Complex
power supplied by voltage source V to the circuit is given by the
expression
*
2
1
I
V
S 

2
1 V
V
V 

Conservation of Complex Power in Series Circuit
Now, for the series-connected
impedances, their voltage drops V1
and V2 are related to the source
voltage V via the expression
Hence,
*
2
*
*
1 V
V
V 

S

7. Therefore, we can write
*
2
1
I
V
S 
   *
*
2
*
1
2
1
I
V
V 


*
2
*
1
2
1
2
1
I
V
I
V 



2
1 S
S 

S
This can be written as 0


 2
1 S
S
S
Conservation of Complex Power in Series Circuit
Thus, like the parallel circuit discussed earlier, this results tells us
that the sum of complex power absorbed by all the elements of the
series circuit (voltage source V and impedances Z1 and Z2) is zero.

8. Consider a closed circuit consisting of n elements. If the current
flowing through the kth circuit element is Ik and Vk is the voltage drop
across it, then, we can write
Now, the complex power is conserved implies that both average
power and reactive power are conserved. That is,
0 and 0
k k
all all
elements elements
P Q
 
 
Conservation of Complex Power in General
0
2
I
V
S
elements
all
*
k
*
k
elements
all
k 
 

k
I
k
V
+
-

9. Worked Example
Find the total complex power supplied by the source to the three
loads.
Load 2
700 VAR (L), 200 W
Load 1
Load 3
1500 VAR (C), 300 W
100 W, 0 VAR
V
0
100 


E
+
_
I

10. Solution
Complex power consumed by load 1,
 VA
0
100
1 j
jQ
P
S 




Complex power consumed by load 2,
 VA
700
200
2 j
jQ
P
S 




Complex power consumed by load 3,
 VA
1500
300
3 j
jQ
P
S 




Load 2
700 VAR (L), 200 W
Load 1
Load 3
1500 VAR (C), 300 W
100 W, 0 VAR
V
0
100 


E
+
_
I

11. 1 2 1 2 1 2
( ) ( )
S P jQ S S P P j Q Q
       
Total complex power consumed by the loads,
VA
)
1500
700
(
)
300
200
100
( 



 j
VA
o
j 13
.
53
1000
800
600 




Solution

12. The 60  resistor absorbs 240 Watt of average power. Calculate V
and the complex power of each branch. What is the total complex
power?
Worked Example

13. Solution
Phasor domain circuit:
2
2
60
240 I


A
2
60
240
2 

I
Solving for I2, we obtain
Let be the current through the 60-Ω resistor. Now ,
therefore
2
I R
I
P 2
2


14. (rms)
A
o
j
I 0
2
0
2
2 



Let be the reference phasor. Therefore, we can write
2
I
    (rms)
V
40
120
0
2
20
60 j
j
V o
o 





Application of Ohm’s law to the right branch impedance gives us the
voltage drop
Application of Ohm’s law to the left branch impedance gives us the
branch current
(rms)
A




 4
.
2
2
.
3
10
30
1 j
j
V
I o
Next, KCL gives us the current equation
A
.
j
.
j
I
I
I 4
2
2
5
2
4
.
2
2
.
3
2
1 






Solution

15.     V
40
120
48
104
20 j
j
V
I
V o 






V
o
j 45
.
21
67
.
240
88
224 



For the 20-Ω resistor,
V
8
.
4
5
.
114
48
204
20
3
o
j
I
V 





   A
4
.
2
2
.
5
4
2
2
5
*
*
j
.
j
.
I 



 
4
.
2
2
.
5
8
.
4
5
.
114
*
3
3 j
I
V
S o






A
654 o
S 97
.
19
3 


Therefore,
Solution

16. For the (30 - j10) Ω impedance,
V
o
o j
V 8
.
24
5
.
126
40
120 




A
o
j
I 87
.
36
4
4
.
2
2
.
3
1 



VA
o
o
o I
V
S 87
.
36
4
43
.
18
5
.
126
*
1
1 






  VA
-j
S o
160
480
44
.
18
506
1 



Therefore,
Solution

17. VA
43
.
18
253
0
2
43
.
18
5
.
126
*
2
2 






 o
o
o I
V
S
VA
.
.
.
.
S o
o
T 8
24
727
5
45
21
67
240 




Therefore,
For the (60 + j20) Ω impedance, A
o
I 0
2
2 

  A
V
j
S 80
240
2 

The overall complex power supplied by the source is
  VA
j
ST 80
1736 

giving
Solution

18. Exercise
Two loads are connected in parallel. Load 1 has 2 kW, p.f. = 0.75
leading, and Load 2 has 4 kW, p.f. = 0.95 lagging. Calculate the p.f.
of the two loads and the complex power supplied by the source.
Load 2
Load 1

19. Solution
Power triangle for Load 1:
jQ1
S1
P1 = 2 kW
1
  

 41
.
41
75
.
0
acos
1

1
1
1
tan
P
Q


From power triangle, we get
Therefore,
 
VAR
1764
41
.
41
tan
2000
tan 1
1
1




 
P
Q
Complex power consumed by
load 1 is
 VA
1764
2000
1
1
1
j
jQ
P
S





20. Solution
Power triangle for Load 2:
-jQ2
S2
2
  

 195
.
18
95
.
0
acos
2

2
2
2
tan
P
Q


From power triangle, we get
Therefore,
 
VAR
1315
195
.
18
tan
4000
tan 2
2
2




 
P
Q
P2 = 4 kW
Complex power consumed by
load 2 is
 VA
1315
4000
2
2
2
j
jQ
P
S





21. Solution
Load 2
S2
Load 1
S1
S
Sum of complex power consumed by the two loads is
   
 
VA
4.28
6017
449
6000
1315
4000
1764
2000
2
1











j
j
j
S
S
S L
jQ
SL

P
Power triangle for the complex power consumed by the two loads:
997
.
0
6017
6000
cos



L
S
P

Power factor of the two loads,

22. Solution
loads
two
by the
consumed
power
complex
S 
L
By the complex power conservation law,
where
S
L S
S 
 VA
449
6000
SS j


source
by the
suuplied
power
complex
S 
S
Therefore, complex power supplied by the source is

23. Power Factor Correction
• The objective to adjust the power factor by adding a compensating
impedance (usually a capacitive reactance) to the load.
• The goal of power factor correction is to deliver maximum power
to the load using the lowest source current.
Load with low
power factor
compensating
capacitor
C
Source
current

24. Worked Example
For the circuit shown, calculate
(i) the supply current,
(ii) the reactive power that needs to be
supplied by a capacitor bank to
increase the power factor to unity.
(iii) the value of the supply current at
unity power factor.
P = 120 kW
Q = 160 kVAr
600 V Load
IS
In the following worked example we will first determine the current
that the generator needs to supply to load when its power factor is
not corrected to unity. Then, we will demonstrate the advantage of
correcting the power factor of the load on the magnitude of the
current that needs to be supplied by the generator to the load.
Power Factor Correction

25. Solution
600 V Load
P = 120 kW
Q = 160 kVAr
IS
L
S
Let be the reference phasor.
Complex power supplied by the generator to the load is
000
,
160
000
,
120
0
600 *
*
j
I
I
V
S
S
o
S
S
L






S
V
Therefore,
(rms)
A
3
.
51
33
.
333
0
600
000
,
160
000
,
120
*
o
o
S
j
I





and
(rms)
A
3
.
51
33
.
333 o
S
I 


This result shows that the supply current lags behind the supply voltage
by 51.3. The power in this case is cos(51.3) = 0.625
VS = 6000 V
IS = 333.3-51.3 A
51.3

26. To achieve unity power factor, we need to connect a compensating
reactive load in parallel with the original load that cancels out the
reactive power.
600 V Load
P = 120 kW
Q = 160 kVAr
IS
Reactive
load
To obtain unity power factor, we need ensure that has no
imaginary part. Now, since
000
,
160
000
,
120 j
SL 

L
C
G S
S
S 

Let be the complex power consumed
by the compensating load. Then,
complex power supplied by the generator
to the two loads is
C
S
G
S
Solution

27. Therefore, we require
000
,
160
j
SC 

so that
kW
120
000
,
160
000
,
160
000
,
120






j
j
S
S
S L
C
G
Therefore, after the deployment of the compensating capacitor, the
supply current
S
o
o
S
G
S I
V
S
I 




 A
0
200
0
600
kW
120
*
Solution

28. Thus, by correcting the load power factor, we have managed to deliver
the active power required by the load and at the same time significantly
reduced the supply current. This reduces the size of the cable used to
supply the load current and also the required VA rating of the generator.
The VA rating of the generator for the power factor corrected load is
S = VSIS = 600 x 200 = 120 kVA
Solution
whereas the VA rating of the generator for the uncorrected load
power factor is
   
kVA
200
10
160
10
120
2
3
2
3





S
(A reduction of about 40% in the capacity of the generator required!)

29. What is the value of C that which will
correct the power factor angle to zero,
that is, reduce Is to a minimum value in
phase with Vo? Determine the reduction
of current which resulted from
connecting the capacitor into the circuit.
Worked Example
Vo = 450∠0 V Is = 17∠−20 A
f = 50 Hz
Without the capacitor connected into the circuit shown below,

30. Solution
Consider the case where the capacitor C is not connected to the
circuit, as shown below.
ZG
Z
VS
IS
VO
Complex power supplied to the load is
 VA
j2616
7189
20
7650
20
17
0
450
)
(
*
)
(












 rms
s
rms
O
L I
V
S
Next, consider the case where the capacitor
C is now connected to the circuit, as shown
below. By definition, reactive power of the
compensated load is zero. That is, we
should have
VA
0

 C
L Q
Q
ZG
Z
VS
IS
VO
C
SL’

31. This requires us to connect a capacitance C whose capacitive
reactance – jQC will cancel out the inductive reactance of the load.
That is we require
  0

 C
L Q
Q
j
Thus, we require that
VA
2616
j
jQC 


Now,
*
C
C I
V
S 
Therefore,
A
90
81
.
5
0
450
90
2616
*










V
S
I
C
C
ZG
Z
VS
IS
VO
C
SL’
IC
and
A
90
81
.
5 


C
I
Solution

32. 


 4
.
77
81
.
5
450
C
o
C
I
V
X
Solution
Reactance of capacitor is
Since , therefore the value of capacitance required is
C
XC

1

F
7
.
40
4
.
77
50
2
1
1








C
X
C
Since , the new value of supply current is
'
S
o
L I
V
S 
A
97
.
15
450
7189
' 


o
L
S
V
S
I

33. Summary
In this study unit we have looked at
1. Complex power
2. Conservation of complex power
3. Power factor correction.