AC Power Analysis
Instructor: Chia-Ming Tsai
Electronics Engineering
National Chiao Tung University
Hsinchu, Taiwan, R.O.C.
Contents
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Introduction
Instantaneous and Average Power
Maximum Average Power Transfer
Effective or RMS Value
Apparent Power and Power Factor
Complex Power
Conservation of AC Power
Power Factor Correction
Applications
Introduction
• Every electrical device has a power rating
that indicates how much power the
equipment requires; exceeding the power
rating can cause permanent damage.
• The choice of power delivery in 50- or 60-Hz
ac form is due to the allowed high-voltage
power transformation.
Instantaneous Power
T heinstantaneous power is defined as
p(t )  v(t )i (t ) watts(W)
Assuming sinusoidal excitation,
v(t )  Vm cos(t   v )

 i (t )  I m cos(t   i )
p(t )  v(t )i (t )  Vm I m cos(t   v ) cos(t   i )
1
1
 Vm I m cos( v   i )  Vm I m cos(2t   v   i )
2
2
timeindependent
twice angular frequency
Average Power
T heaveragepower is theaverageof theinstantaneous power
over one period.
1
P
T

T
0
p (t )dt
v(t )  Vm cos(t   v )

 i (t )  I m cos(t   i )
1 T1
1 T1
P   Vm I m cos( v   i )dt   Vm I m cos(2t   v   i )dt
T 0 2
T 0 2
1
1 T
1
1 T
 Vm I m cos( v   i )  dt  Vm I m  cos(2t   v   i )dt
2
T 0
2
T 0
1
1
 Vm I m cos( v   i )  Re VI*
2
2
1 * 1
1
 VI  Vm I m  v   i   Vm I m cos v   i   j sin  v   i 
2
2
2
 
Average Power (Cont’d)
Im
V Vm
Z  
( v   i )  R  jX  cos( v   i ) 
R
I Im
Vm
 
1
1
1 2
1 2
*
 P  Vm I m cos( v   i )  Re VI  I m R  I R
2
2
2
2
Case 1 :  v   i for a purely resistivecircuit
1 2
 P  I R  0 , if R  0.
2
Case 2 :  v   i  90 for a purely reactivecircuit
1
 P  Vm I m cos90  0
2
A resistiveload absorbs power at all time,while a reactive
load ( L or C ) absorbs zero averagepower.
Example 1
Given that
v(t )  120cos(377t  45)

 i (t )  10 cos(377t  10)
find theinstantaneous power and theaveragepower.
Sol :
p  vi  1200cos(377t  45) cos(377t  10)
 600cos55  cos(754t  35)
 344.2  600cos(754t  35) W
1
P  Vm I m cos( v   i )  344.2 W
2
Example 2
Find the theaveragepower absorbed by an impedance
Z  (30  j 70)  when a voltageV  1200 is applied
acrossit..
Sol :
V 1200
1200
I 

 1.57666.8
Z 30  j 70 76.16  66.8
1
1
*
P  Re VI  Re(120)(1.576)  66.8
2
2
1
 (120)(1.576) cos(66.8)  37.24 W
2
 
Example 3
Find theaveragepower
supplied by thesource and
theaveragepower absorbed
by theresistor.
Sol :
V 530
I 
Z 4  j2
530

4.472  26.57
 1.11856.57
1
PV  (5)(1.118) cos(30  56.57)
2
 2.5 W
1
PR  (1.118) 2 (4)
2
 2.5 W
Maximum Power Transfer
Z Th  RTh  jX Th

Z L  RL  jX L
VTh
VTh
I

Z Th  Z L ( RTh  RL )  j ( X Th  X L )
2
P
VTh RL
1 2
1
I RL 
2
2 ( RTh  RL ) 2  ( X Th  X L ) 2
T o find theconditionwith maximumpower,
VTh RL ( X Th  X L )
dP

dX L
( RTh  RL ) 2  ( X Th  X L ) 2
2



2
0

VTh ( RTh  RL ) 2  ( X Th  X L ) 2  2 RL ( RTh  RL )
dP

0
2
2
2
dRL
2 ( RTh  RL )  ( X Th  X L )
2


V
 X L   X Th

 Z L  RTh  jX Th  Z*Th , Pmax  Th
8 RTh
 RL  RTh
2
Maximum Power Transfer (Cont’d)
Z Th  RTh  jX Th

Z L  RL
2
P
VTh RL
1 2
1
I RL 
2
2 ( RTh  RL ) 2  X Th 2
T o find thecondit ionwith maximumpower,
dP

dRL

VTh ( RTh  RL ) 2  X Th  2 RL ( RTh  RL )
2
2

2 ( RTh  RL )  X Th
0
 RL  RTh  X Th  Z Th
2
2
2

2 2

Effective or RMS Value
T heeffective valueof a periodiccurrent
is thedc current that delivers thesame
power toa resistoras theperiodiccurrent.
T hepower absorbed by theresistorin the
ac circuit is
1 T 2
R T 2
1 T 2
P   i Rdt   i dt 
v dt

0
0
0
T
T
RT
While thepower absorbed by theresistor
in thedc circuit is
2
V
2
P  I eff
R  eff
R
1 T 2
1
 I eff 
i dt  I rms , Veff 

0
T
T

T
0
v 2 dt  Vrms
ac circuit
dc circuit
Effective or RMS Value (Cont’d)
For any periodicfunctionx(t ), the rms valueis given by
1 T 2
X rms 
x dt

0
T
For thesinusoid i (t )  I m cost , therms valus is
1 T 2
2
I rms 
I
cos
tdt 
m

T 0
Similarly,for v(t )  Vm cost ,
I m2
T

T
0
1
1  cos 2t dt  I m
2
2
Vm
Vrms 
2
T heaveragepower can be writtenas
Vm I m
1
P  Vm I m cos( v   i ) 
cos( v   i )
2
2 2
 Vrms I rms cos( v   i )
Example 1
Determinetherms value of i (t ).
If thecurrentis passed througha
2 -  resistor,find theaverage
power absorbed by theresistor.
Sol :
T heperiodis T  4.
 5t , 0  t  2
i (t )  
  10, 2  t  4
I rms
1

T

T
0
i 2 dt
4
1 2
2
2


(
5
t
)
dt

(

10
)
dt
2

4  0
I rms
4
1  t3 2

25 |0  100t |2  8.165 A
4 3

T heaveragepower absorbed is
2
P  I rms
R  (8.165) 2 (2)  133.3 W
Example 2
Find therms value and theaverage
power absorbed by a 10 -  resistor.
Sol :
T heperiodis T  2 .
10sin t , 0  t  
v(t )  
0,   t  2

1 T 2
2
Vrms   v (t )dt
T 0
2
1  
2


(
10
sin
t
)
dt

0
dt


2  0
1   100


(
1

cos
2
t
)
dt

2  0 2

50  sin 2t  

t 
 |0  25
2 
2 
T heaveragepower absorbed
by a 10 -  resistoris
 Vrms  5 V
2
Vrms
52
P
  2.5 W
R 10
Apparent Power and Power Factor
If thecurrentand voltageare
v(t )  Vm cos(t   v )

 i (t )  I m cos(t   i )
V  Vm  v
or 
 I  I m  i
T heaveragepower is
1
P  Vm I m cos( v   i )
2
 Vrms I rms cos( v   i )
 S cos( v   i )
 S  pf
S  Vrms I rms (unit  VA) : Apparent power

where  pf  cos( v   i )
: Power factor
  
: Power factor angle
 v i
S and pf (Cont’d)
P ower FactorAngle  Angle of theLoad Impedance
pf  cos( v   i )
V Vm  v Vm
Z 

( v   i )
I I m  i I m
V

Vrms  2  Vrms v
Since 
I
 I rms 
 I rms i

2
V Vrms Vrms
Z 

( v   i )
I I rms I rms
 Leadingpf meanscurrentleads voltage a capacitiveload

 Laggingpf meanscurrentlags voltage an inductiveload
Example 1
A series - connectedload draws a currentand voltageas given below
 i(t )  4 cos(100t  10)

v(t )  120cos(100t  20)
Find theapparentpower and thepower factorof theload.
Determinetheelement values tahtform theseries - connectedload.
Sol :
T heapparentpower is
120 4
S  Vrms I rms 
 240 VA
2 2
T hepower factoris
pf  cos( v   i )
 cos(20  10)  0.886
V 120  20
Z  
I
410
 30  30
 25.95  j15
1
 X C  15  
C
1
C
 212.2 F
15
Example 2
Determinethepower factor
as seen by thesource.
Calculate theaveragepower
deliveredby thesource.
Sol :
T he totalimpedanceis
Z  6  4||(-j2)  6.8  j1.6
 7  13.24
T hepower factoris
pf  cos(13.24)  0.9734
Vrms
300

Z
7  13.24
 4.28613.24
 I rms 
 P  Vrms I rms pf  (30)(4.286)pf
 125 W
Complex Power
Considering thecurrentand voltage
are given in phasorform as
V  Vm  v

 I  I m  i
T hecomplexpower S absorbed by
theac load is given as
1 *
S  VI  VrmsI *rms
2
V

V

 Vrms v
 rms
2
where 
I
 I rms 
 I rms i

2
S  Vrms I rms( v   i )
 Vrms I rms cos( v   i )  j sin( v   i )
Z
V Vrms Vrms


( v   i )
I I rms I rms
2
V
2
 S  VrmsI *rms  I rms
Z  rms*
Z
Since Z  R  jX ,
2
 S  I rms
( R  jX )  P  jQ
2
 P  Re(S)  I rms
R : Real power

2
Q

Im(
S
)

I
rms X : Reactivepower

Complex Power (Cont’d)
• P is the average or real power.
– The power delivered to the load
– The actual power dissipated by the load
• Q is the reactive or quardrature power.
– Unit: volt-ampere reactive (VAR)
– A measure of the energy exchange between the
source and the reactive part of the load
– Q = 0 for resistive loads (unity pf)
– Q < 0 for capacitive loads (leading pf)
– Q > 0 for inductive loads (lagging pf)
Summary
1 *
VI
2
 Vrms I rms( v   i )
ComplexP ower  S  P  jQ 
ApparentP ower  S  S  Vrms I rms  P 2  Q 2
Real P ower  P  Re(S)  S cos( v   i )
ReactiveP ower  Q  Im(S)  S sin( v   i )
P ower Factor
P
 cos( v   i )
S
Impedance Z 
V Vrms Vrms


( v   i )
I I rms I rms
Power Triangle
Power triangle
Impedance triangle
Power triangle
Example 1
T he volt age and currentacrossa load are given as
v(t )  60 cos(t  10)

i (t )  1.5 cos(t  50)
Find (a) thecomplexand apparentpowers,(b) thereal and reactive
powers,and (c) thepower factorand theload impedance.
Sol :
(b) S  45   60  22.5  j 38.97
60
1.5
(a) Vrms 
  10 , I rms 
50
 P  22.5 W

2
2
Q  38.97 VAR

T hecomplexpower is
(c) pf  cos(60)  0.5 (leading)
*
S  VrmsI rms  45   60 VA
V 60  10
Z 
 40   60 
T heapparentpower  S  S  45 VA
I
1.550
Example 2
A load Z draws 12 - kVA at pf  0.856lagging froma 120- V rms
sinusoidal source.
Calculate: (a) theaverageand reactivepowersdelivered to theload,
(b) thepeak current,and (c) theload impedance.
S
10272 j 6204


Vrms
1200
Sol :
(a) pf  cos  0.856
   31.13
 S  12000VA
I
 P  S cos  10272W

Q  S sin   6204 VAR
T hepeak currentis
(b) T hecomplexpower is
S  P  jQ  10272 j 6204 VA
(c) Z 
 S  VrmsI *rms
*
rms
 10031.13
 I rms  100  31.13
I m  2 I rms  141.4 A
Vrms
1200

I rms 100  31.13
 1.231.13 
Conservation of AC Power
KCL gives I  I1  I 2
KVL gives V  V1  V2
T hecomplexpower supplied by
T hecomplexpower supplied by
thesource is
thesource is
1 * 1
S  VI  V (I1*  I *2 )
2
2
1
1
 VI1*  VI*2  S1  S 2
2
2
S1, 2 is thepower delivered to Z1,2.
1 * 1
S  VI  (V1  V2 )I *
2
2
1
1
 V1I *  V2 I *  S1  S 2
2
2
S1, 2 is thepower delivered to Z1,2.
Summary
• The total power supplied by the source equals
the total power delivered to the loads.
S  S1  S2      S N
• The complex, real, and reactive powers of the
sources equal the respective sums of the
complex, real, and reactive powers of the
individual loads.
Example
Find thereal power and reactive
power absorbed by : (a) thesource,
(b) theline, and (c) theload.
Sol :
T he totalimpedanceis
Z  (4  j 2)  (15  j10)
 19  j8  20.62  22.83
Vs
2200
I

Z 20.62  22.83
 10.6722.83
S line  VlineΙ *
 (47.7249.4)(10.67  22.83)
 455.4  j 227.7 VA (lagging)
(c) VL  (15  j10)I  192.38  10.87
(a) S s  Vs Ι *
S L  VL Ι *
 (2200)(10.67  22.83)
 2163.5  j 910.8 VA (leading)
(b) Vline  (4  j 2)I  47.7249.4
 (192.38  10.87)(10.67  22.83)
 1708 j1139 VA (leading)
Note that S s  S line  S L
Power Factor Correction
• It is the process of increasing the power factor without
altering the voltage or current to the original load.
pf
correction
Most loads
are inductive.
V
 I L   1
R  jL
I C  j C  V
IL 
I  I L  IC  I   2
pf Correction (Cont’d)
If we desire to increasepf fromcos1 to
cos 2 without alteringthereal power,
P1  S1 cos1  P2  S 2 cos 2  P
Q1  S1 sin 1  P1 tan1
Q2  S 2 sin  2  P tan 2
Applyingtheac power conservaton
i gives
QC  Q1  Q2  P (tan1  tan 2 )
2
2
Vrms
Vrms
2
But S  *  QC 
 CVrms
Z
XC
QC
P (tan1  tan 2 )
C 

2
2
Vrms
Vrms
Note that thereal power P is not affect edby
the pf correctionbecause PC is zero.
Example
Whenconnectedto a 120- V(rms),60 - Hz power line,
a load absorbs 4 - kW at 1 lagging power factorof 0.8.
Find thecapacitance necessaryto raise thepf to 0.95.
Sol :
When pf is raised to 0.95,
Vrms  120 V
  2 (60)  120 rad/s
We have 
 P  4000 W
 pf  0.8
pf  cos1  0.8  1  36.87
cos 2  0.95   2  18.19
P  S1 cos1  S1 
P
 5000VA
cos1
Q1  S1 sin 1  5000sin 36.87
 3000VAR
S2 
P
 4210.5 VA
cos 2
Q2  S1 sin  2  4210.5 sin 18.19
 1314.4 VAR
QC  Q1  Q2  3000 1314.4
 1685.6 VAR
QC
1685.6
C

 310.5 F
2
2
Vrms 120 120
Applications: Power Measurement
P
Low impedance
High impedance
Vm

V

 v
v(t )  Vm cos(t   v )  rms
2


I
 i(t )  I m cos(t   i )
 I rms  m  i

2
1
P  Vrms I rms cos( v   i )  Vm I m cos( v   i )
2
Example
Find the wattmeterreading.
Sol :

Vrms 

 I rms 

Vm
 v
2
Im
 i
2
1500
150
I rms 

(12  j10)  (8  j 6) 20  j 4
150(8  j 6)
Vrms  I rms (8  j 6) 
20  j 4
150(8  j 6) 150
*
S  VrmsI rms 
20  j 4 20  j 4
 423.7  j 324.6 VA
 P  Re(S)  423.7 W