Complex Power and Power Triangle
def
S = P  jQ  E I cos   j E I sin   E I p  j E I q
Iq
Ip
 E I  | S |   P 2  Q 2 
 E I  e  i  E  e I   i  EI *
• =tan-1(Q/P)
– If >0 (i <e, i.e. lagging PF), then Q>0, which
means a reactive load (inductive)
– If <0 (i >e, i.e. leading PF), then Q<0, which
means a reactive source (capacitive)
S
Q
Power triangle
• PF=cos=P/|S| (+/- sign of Q tells a lagging/leading PF)
P=|S| cos=|S| PF
• If the load impedance is Z=E/I, then
S = ZII*= Z|I|2= R|I|2+jX|I|2 = P + jQ
Q=X|I|2
P=R|I|2
 is also called the load impedance angle
Z
X
R
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Other Useful Formulas
*
E
2
2
E
S
Z *  2
|I|
S
EE
S  EI  *  *
Z
Z
 II * Z | I |2 Z
*
• If Z is purely resistive
• If Z is purely reactive
E
P
R
2
2
2
2
E
E
E

or
Q
L
1/ C
X
21
Theorem of Conservation of Complex Power
• For a power network supplied by independent sources all at the
same frequency (all voltages and currents are assumed to be
sinusoids), the sum of the complex power supplied by the
independent sources equals the sum of the complex power
received by all the other branches of the network
(SG of all actual sources)=(SL of all the others)
• In other words, the total complex power supplied (received) by
all branches is zero
(SG of all braches)=0 or (SL of all braches)=0
– For a single source with elements in series or parallel, it is proved by
Kirchhoff’s voltage or current law (KVL/KCL)
– For a general case, it is proved by Tellegen’s theorem.
– Application of the theorem: any part of the power network can be
replaced by an equivalent independent source
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• Two examples from A. Bergen and V. Vittal, Power System Analysis, Prentice Hall, 2000
23
24
Examples 7‐2 & 7‐3
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Example 2.2 on Saadat’s book
V  12000 V,
Z1  60  j 0,
Z 2  6  j12,
Z 3  30  j 30
Find the power absorbed by each load
and the total complex power
I1 
V 12000 1200


 20  j 0 A
Z1
60  j 0
60
V 12000
200
200(1  j 2)



 40  j80 A
I2 
Z2
6  j12 1  j 2
5
I3 
S2
V 12000 40
40(1  j )



 20  j 20 A
Z 3 30  j 30 1  j
2
S3
S
S1  VI1*  12000(20  j 0)  24, 000 W+j 0 var
S2  VI 2*  12000(40  j80)  48, 000 W+j 96, 000 var
S3  VI  12000(20  j 20)  24, 000 W  j 24, 000 var
*
3
S1
S  S1  S2  S3  96, 000 W+j 72, 000 var
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• Other approaches
I  I1  I 2  I 3  (20  j 0)  (40  j80)  (20  j 20)  80  j 60  100  36.87 A
S  VI *  (12000)(10036.87)  120, 00036.87 VA  96, 000 W  j 72, 000 var
or
Z / /  Z1 / / Z 2 / / Z 3 
S
V
2
Z / /*
1
 9.6  j 7.2 
1 / Z1  1 / Z 2  1 / Z 3
(1200) 2

 96, 000 W  j 72, 000 var
9.6  j 7.2
2
V
(1200) 2
 24, 000 W  j 0 var
S1  * 
Z1
60
2
V
(1200) 2
 48, 000 W  j 96, 000 var
S2  * 
Z2
6  j12
2
V
(1200) 2
S3  * 
 24, 000 W  j 24, 000 var
Z3
30  j 30
27