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STEADY-STATE POWER ANALYSIS
LEARNING GOALS
Instantaneous Power
For the special case of steady state sinusoidal signals
Average Power
Power absorbed or supplied during one cycle
Maximum Average Power Transfer
When the circuit is in sinusoidal steady state
Effective or RMS Values
For the case of sinusoidal signals
Power Factor
A measure of the angle between current and voltage phasors
Complex
Power
Power Factor
Correction
Measure
of
power
usingtransfer
phasorsto a load by “aligning” phasors
How to improve
power
Single Phase Three-Wire Circuits
Typical distribution method for households and small loads
INSTANTANEOUS POWER
Instantaneous
Power Supplied
to Impedance
p( t )  v ( t ) i ( t )
In steady State
Assume : v (t )  4 cos( t  60),
Z  230
Find : i (t ), p(t )
V 460
I 
 230( A)
Z 230
i (t )  2 cos( t  30)( A)
VM  4, v  60
I M  2, i  30
v (t )  VM cos( t   v )
p(t )  4 cos 30  4 cos(2 t  90)
i (t )  I M cos( t   i )
p(t )  VM I M cos( t   v ) cos( t   i )
1
cos1 cos2  cos(1  2 )  cos(1  2 )
2
V I
p(t )  M M cos( v   i )  cos(2 t   v   i )
2
constant
LEARNING EXAMPLE
Twice the
frequency
AVERAGE POWER
LEARNING EXAMPLE
For sinusoidal (and other periodic signals)
we compute averages over one period
1
P
T
t 0 T
 p(t )dt
T

VR
2
t0
Find the average
power absorbed
by impedance


VM I M
cos( v   i )  cos(2 t   v   i )
2
V I
P  M M cos( v   i ) It does not matter
who leads
2
p( t ) 
If voltage and current are in phase
1
2
 v   i  P  VM I M Purely
resistive
If voltage and current are in quadrature
 v   i  90  P  0 Purely
inductive or
capacitive
I
1060 1060

 3.5315( A)
2  j 2 2 245
VM  10, I M  3.53, v  60, i  15
P  35.3 cos(45)  12.5W
Since inductor does not absorb power
one can use voltages and currents across
the resistive part
VR 
2
1060  7.0615(V )
2  j2
1
P  7.06  3.53W
2
LEARNING EXAMPLE
Determine the average power absorbed by each resistor,
the total average power absorbed and the average power
supplied by the source
Inductors and capacitors do not absorb
power in the average
Ptotal  18  28.7W
Psupplied  Pabsorbed  Psupplied  46.7W
Verification
If voltage and current are in phase
I  I1  I 2  345  5.3671.57
I  8.1562.10( A)
2
1 2
1
1
V
M
 v   i  P  VM I M  RI1M 
2
2
2 R P  VM I M cos(   )
v
i
1245
2
I1 
 345( A)
1
4
Psupplied  12  8.15  cos(45  62.10)
1
2
P4  12  3  18W
2
1245
1245
I2 

 5.3671.57( A)
2  j1
5  26.37
1
P2   2  5.362 (W )  28.7W
2
LEARNING EXTENSION
I
I
Find average power absorbed by each resistor
I2 
1260
Zi
I 2  1.9041.6
I2
Z i  2  (4 ||  j 4)
4( j 4) 8  j8  j16 25.3  71.6


4  j4
4  j4
4 2  45
Zi  4.47  26.6
Zi  2 
1260
 2.6886.6( A)
4.47  26.6
1 2 1
P2  RI M
  2  2.682  7.20W
2
2
I
 j4
4  90
I
 2.6886.6
4  j4
4 2  45
1
P4   4  1.902 (W )
2
LEARNING EXTENSION
I1 
Vs
Find the AVERAGE power absorbed by each PASSIVE
component and the total power supplied by the source
I2
1
P4   4  4.122 (W )
2
Pj 2  0(W )

I1 
4  j2
1030
3  4  j2
4.4726.57
I1 
1030  6.1440.62( A)
7.2815.95
1 2 1
P3  RI M
  3  6.142 (W )
2
2
I 2  1030  6.1440.62
3
3030
1030 
3  4  j2
7.2815.95
 4.1214.05( A)
I2 
Power supplied by source
Method 1. Psupplied  Pabsorbed
Psupplied  P3  P4  90.50W
Method 2: P  1 V I cos(   )
M M
v
i
2
Vs  3I1  18.4240.62
1
P   18.42  10  cos(40.62  30)
2
LEARNING EXAMPLE
Determine average power absorbed or supplied by each
element
1230  60 10.39  j 6  6

 6  j 4.39
j1
j
 7.43  36.19( A)
1
P60   6  7.43 cos(0  36.19)  18W
2
I3 
Passive sign convention
I2 
1230
 630( A)
2
I1  I 2  I 3  5.20  j 3  6  j 4.39  11.2  j1.39( A)
 11.28  7.07
1 2 1
P2  RI M
  2  62  36(W )
2
2
Pj1  0
To determine power absorbed/supplied
by sources we need the currents I1, I2
Average Power
For resistors
2
1
P  VM I M cos( v   i ) P  1 RI 2  1 VM
M
2
2
2 R
1
P1230   12  11.28  cos(30  7.07)
2
 54(W )  36  18
LEARNING EXTENSION
Determine average power absorbed/supplied by each
element
V4
I2
I1
1
P120    19.92  12  cos(54.5  0)  69.4(W )
2
1
P430    4  (9.97) cos(30  204)  19.8(W )
2
Check: Power supplied =power absorbed
Loop Equations
I1  120
430   j 2 I 2  4( I 2  120)
I2 
44.58177.43
 9.97204( A)
4.47  26.57
 4( I1  I 2 )  4(12  9.97204)(V )
I2 
V4
430  480 3.46  j 2  48

4  j2
4.47  26.57
Alternative Procedure
Node Equations
V4 V4  430

0
4
 j2
430  V4
I2 
2j
 120 
 4(12  9.108  j 4.055)(V )  19.92  54.5(V )
For resistors
Average Power
1 VM2 1 19.922
P4 
 
 49.6W
2
1
2 R 2
4
1
1
V
2
M
P  VM I M cos( v   i ) P  RI M 
2
P2 j  0(W )
2
2 R
Determine average power absorbed/supplied by each
element
LEARNING EXTENSION
V
I1 
240  V 24  14.77  j1.85

 4.62  j 0.925
2
2
I1  4.71  11.32( A)
I1
I2
I2 
I2 
120  V 12  14.77  j1.85  j


j2
j2
j
 1.85  j 2.77
 0.925  j1.385( A)  1.67123.73( A)
2
1
P2   2  4.712  22.18(W ) For resistors
2
2
2
V  240 V  120 V
1
1
V
2
1
14
.
88
M

  0  j4
P  RI M 
P



27
.
67
(
W
)
4

2
j2
4
2
2 R
2
4
1
2 j (V  24)  2(V  12)  jV  0
P120    12  1.67 cos(0  123.73)  5.565(W )
2
(2  3 j )V  24  j 48
1
P240    24  4.71 cos(0  11.32)  55.42(W )
24  j 48 2  j 3 192  j 24
2
V


Check :
2  j3 2  j3
13 Average Power
Pabsorbed  22.18  27.67  5.565(W )
 14.887.125(V )
1
P  VM I M cos( v   i )
Psupplied  55.42(W )
 14.77  j1.85(V )
2
Node Equation
Sources of different frequencies
•
•
If sources have different frequencies
to find power, i.e. P=P1+P2
1  2
, must use superposition
1  2
If sources have same frequencies
, cannot use superposition
to find power, i.e. P‡P1+P2 (although you can still use superposition to
find currents and voltages).
,
Find the average power absorbed by the resistor in the circuit
shown
P2 
1 2
6 10
2
I1  460o
I 3  4  30o
I1  I 3  5.6615o
1
P13  5.66210
2
P  P13  P2  340.2W
MAXIMUM AVERAGE POWER TRANSFER
ZTH  RTH  jXTH
1 | Z L || VOC |2
PL 
2 | Z L  ZTH |2
RL
RL2  X L2
Z L  ZTH  ( RL  RTH )  j ( X L  X TH )
| Z L  ZTH |2  ( RL  RTH ) 2  ( X L  X TH ) 2
Z L  RL  jX L
1
PL  VLM I LM cos(VL   I L )
2
1
 | VL || I L | cos(VL   I L )
2
ZL
ZL
VL 
VOC | VL |
| VOC |
Z L  ZTH
Z L  ZTH
VL  I L  VL  Z L
|V |
| I L | L
Z L  V   I  Z L
| ZL |
L
L
Z L  RL  jX L  tan(Z L )  X L
RL
IL 
cos 
1
1  tan 2 
 cos(VL   I L ) 
RL
RL2  X L2
1
| VOC |2 RL
PL 
2 ( RL  RTH )2  ( X L  X TH )2
PL

 0
X L
  X L   X TH
 
PL
 0   RL  RTH

RL
*
 Z Lopt  ZTH
PLmax
1  | VOC |2 

 
2  4 RTH 
LEARNING EXAMPLE Find Z L for maximum average power transfer.
Compute the maximum average power supplied to the load
 Z Lopt

*
ZTH
PLmax
1  | VOC |2 

 
2  4 RTH 
Remove the load and determine the Thevenin equivalent of remaining circuit
8  j 4 (8  j 4)(6  j1) 52  j16



6  j1
37
37
8  j 4 8.9426.57


 1.4716.93
6  j1 6.089.64
ZTH  4 || (2  j1) 
Z L*  1.47  16.93  1.41  j 0.43
I1
VOC  4 
2
320
40 
 5.26  9.64
6  j1
6.089.64
PLmax
We are asked for the value of the
power. We need the Thevenin voltage
1 5.262
 
 2.45(W )
2 4  1.41
LEARNING EXAMPLE Find Z L for maximum average power transfer.
Compute the maximum average power supplied to the load
 Z Lopt

PLmax
*
ZTH
1  | VOC |2 

 
2  4 RTH 
Circuit with dependent sources!
ZTH 
VOC
I SC
40  Vx'  (2  j 4) I1
KVL
VX'  2 I1
40  (4  j 4) I1  (4 245) I1
I1 
KVL
40
 0.707  45( A)
4 245
VOC  2 I1  40  1  j1  4  3  j1  10  161.5
Next: the short circuit current ...
LEARNING EXAMPLE (continued)...
Original circuit
 Z Lopt

*
ZTH
LOOP EQUATIONS FOR SHORT
PLmax
1  | VOC |2 

 
2  4 RTH 
CIRCUIT CURRENT
 V x"  j 4 I  2( I  I SC )  4  0
40  2( I SC  I )  j 2 I SC  0
CONTROLLING VARIABLE
Vx"  2( I SC  I )
Substitute and rearrange
(4  j 4) I  4 I SC  4
 2 I  (2  j 2) I SC  4  I  (1  j1) I SC  2
4(1  j )(1  j ) I SC  2  4 I SC  4
I SC  1  j 2( A)  5  116.57
VOC  2 I1  40  1  j1  4  3  j1  10  161.57
ZTH  2  45  1  j1  Z Lopt  1  j1
PLmax
1 ( 10 )2
 
 1.25(W )
2
4