The instantaneous power
p  t   P  P cos(2t)  Q sin(2t)
P 
VI
cos  V  I 
2
real (or average) power (Watts)
Which is the actual power absorb by the element
Examples Electric Heater , Electric Stove , oven Toasters, Iron …etc
Q  VI sin  V  I 
2
reactive power
Which is the reactive power absorb or deliver by the element
Reactive power represents energy stored in reactive elements
(inductors and capacitors). Its unit is Volt Ampere Reactive (VAR)
The instantaneous power
p  t   P  P cos(2t)  Q sin(2t)
P 
VI
cos  V  I 
2
Q 
VI
sin  V  I 
2
real (or average) power (Watts)
reactive power (Volt Ampere Reactive (VAR) )
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
Pˆ  P  jQ
were
Pˆ is the complex power
P is the average power
Q is the reactive power
Advantages of using complex power
P̂  P  jQ
 We can compute the average and reactive power from the complex power S
P  Real{Pˆ } VI cos(v  i )
2
Q  Ιmag{Pˆ } VI sin(v  i )
2
 complex power P̂ provide a geometric
interpretation
j
P̂  P  jQ  P̂ e
ˆ VA
|P|
Q
(reactive power) VAR

were
P̂ = P Q
2
2
Is called apparent power (VA)
P
( average power) Watts
V I cos(   ) 
 cos(   ) 
Q 
v
v
i   tan  
i   tan  tan(   ) 
  tan  
 v i 




P 
V
I
sin(



)
sin(



)
v
v
i 
i 


 =tan  
v  i
power factor angle
The geometric relations for a right triangle mean the four power triangle dimensions
( |P|
ˆ , P, Q,  ) can be determined if any two of the four are known
Power Calculations
P̂  P  jQ
 V I cos(v  i )  j V I sin(v  i )
2
2
j (v  i )
V
I
V
I
cos(v   )  j sin(v   ) 

e

i
i 
2
2
were
I*
Is the conjugate of the current phasor
I
I

V

Circuit
1
Pˆ  VI *
2
1 j (v )  j ( i )  1V I*
 Ve
Ie
2
2
In any circuit, conservation of complex power is achieved

P̂

0
i
allcircuit elements
This implies that in any circuit, conservation of average power and
Conservation of reactive power are achieved
 allcircuit elements Qi  0

allcircuit elements
Pi,AV  0
However, the apparent power (the magnitude of the complex power)
is not conserved
Ex:6.13 Determine the average and
reactive power delivered by the source.
10  30
The phasor current leaving the source is Î 
 1.86  98.2
2  j8  j3
The average power delivered by the source is:
1 
*
PAV,source  Re 10  30 1.86  98.2  

2 
1
 Re 10  30 1.8698.2  
2
 9.28cos  30  98.2  3.45W
The reactive power delivered by the source is:
Qsource 
1
Im 10  30 1.86  98.2  *
2

1
Im 10  30 1.8698.2  
2
 9.28sin  30  98.2  8.62 VAR
And the complex power delivered by the source is
P̂source  PAV,source  jQsource
 3.45  j8.62VA
Determine the average power and reactive power delivered to each element
Î  1.86  98.2
The voltage across the elements are:
ˆ  2Iˆ  2(1.86  98.2 )  3.71 98.2 V
V
R
ˆ  j8Iˆ  ( j8)(1.86  98.2 )  (890.0 )(1.86  98.2 )
V
L
 14.86 8.2 V
ˆ   j3Iˆ  ( j3)(1.86  98.2 )  (3  90.0 )(1.86  98.2 )
V
C
 5.57188.2 V
Thus the complex power delivered to each element is
1 ˆ ˆ
1
1
*
Pˆ R  V
I
*

(3.71


98.2)(1.86


98.2
)

(3.71  98.2)(1.8698.2 )
R
2
2
2
 3.450
 3.45  j0 VA
1 ˆ ˆ
1
Pˆ L  V
I
*

(14.86  8.2)(1.86  98.2 )*  13.7990  0  j13.79 VA
L
2
2
1 ˆ ˆ
1
 0  j5.17 VA
PˆC  V
I
*

(5.57  188.2)(1.86  98.2 )*  5.17  90
C
2
2
show that conservation of complex power, average
power, and reactive power is achieved.
Pˆsource
 PˆR
PˆL

 PˆC
3.45  j8.62  3.45  j0  0  j13.79  0  j5.17
PAV,source
3.45
Qsource
8.62
 PAV,R  PAV,L
 3.45  0
 QR  QL
 0  13.79
 PAV,C
 0
 QC
 5.17
6.6.1 Power Relations for the Resistor
The voltage and current are in phase so V  I  0
Average power is:
PAV,R
1 VR2 1 2

 IR R
2 R 2
Reactive power is zero for resistor
QR  0
6.6.1 Power Relations for the Inductor
ˆ  jLIˆ   L90  ˆI
V
L
L
L
The voltage leads the current by 90 so that
PAV,L
V  I  90
1
 VL I L cos 90  0
2
1
1
Q L  VL I L sin 90  VL I L
2
2
6.6.1 Power Relations for the Capacitor
The current leads the voltage by 90 so that V  I  90
ˆI  jCV
ˆ   C90  V
ˆ
C
C
C
ˆ  1 ˆI   j 1 ˆI   1   90
V
C
C
C

jC
C
 C
1
PAV,C  VC IC cos  90   0
2
1
1
QC  VC IC sin  90    VC IC
2
2
ˆ
 IC

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
pf  cos(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
6.6.2 Power Factor
ˆP  1 VI
ˆ ˆ *  PAV  jQ
2
PAV
VI

cos  V  I 
2
VI
Q
sin  V  I 
2
pf  cos  V  I 
PAV
1
 VI  pf
2
0  pf  1
Since
V  I  90
I  I  I
ˆ VA
|P|

Circuit
V V V

1
Pˆ  VI *
2
1
ZII *
2

P
( average power) Watts
 P  jQ V I cos(v  i )  j V I sin(v  i )
2
2
*
V 2
1 V  1 VV *
1
 V  

2  Z  2 Z*
2 Z*

Q
(reactive power) VAR

1
Z| I |2
2
pf  cos v  i 
I  I  I
I
+
V

ˆ VA
|P|

V V V
Circuit

P  VI
V2

R
Q
(reactive power) VAR

P
( average power) Watts
2
V
1
1
1
Pˆ  VI *  P  jQ  V I cos(v  i )  j V I sin(v  i ) 
 Z| I |2
2
2
2
2
2 Z*
IR
+
V
R

1
Pˆ  V I *  V R I R 
R 2
R R
2
+
V
L

1
Pˆ  V I *  V LI L
L 2
L L
2
 RI 2
IL
IC
+
V
C

1
Pˆ  V I *  V CI C
C 2
C C
2
1
RI R2
2
EX:6.15 Determine the average and reactive powers
delivered to the load impedance and the power factor
of the load
V̂load 
Îload 
5  j9  j2
1000  54.41  0.84 V
4  j6  5  j9  j2
1000
4  j6  5  j9  j2
Average power
OR P 
1 2
IR  R
2

P
 6.32  55.3
V
I
2
load load cos(v
A
 i )  (54.41)(6.32) cos(0.84  55.3 ) 100 W
2
1 2
1
2
Î L  5   6.32   5  100W
2
2
EX:6.15 Determine the average and reactive powers
delivered to the load impedance and the power factor
of the load
+
V
reactive

V̂load  54.41  0.84 V
Qload 
V
I
2
load load sin(v
IˆL  6.32  55.3
A
 i )  (54.41)(6.32) sin(0.84  55.3 )
2
140 VAR
I
 IˆL
I
| IˆL |  6.32
I
reactive
reactive
Qload  reactive reactive
OR
2
790
(j9 j2)
j7
1000  81.435.54
V

1000 
1000 
reactive 5 (j9 j2)
5 j7
8.654.46
V
V
EX:6.15 Determine the average and reactive powers
delivered to the load impedance and the power factor
of the load
+
V
reactive

IˆL  6.32  55.3
V̂load  54.41  0.84 V
Qload 
OR
V
I
2
load load sin(v
Qload 
V
A
 i ) 140 VAR
I
2
reactive reactive
Qload  (81.4)(6.32)
2
I
reactive
| IˆL |  6.32
V
reactive
 81.4 V
This could also be calculated from the complex power
delivered to the load
ˆP  1 V
ˆ ˆI*  1  54.41  0.84
load
load load
2
2
 100  j140VA
 6.32  55.3 
*
The power factor of the load is:
pf  cos  v  I   cos  0.84  55.3  0.581
The load is lagging because the current lags the voltage
A typical power distribution circuit
The consumer is charged for the average power consumed by the load
VI
VI
cos  V  I  
 pf
2
2
VI
The load requires a certain total apparent power
2
Ex 6.16 Suppose that the load
voltage figure is 170V
The line resistance is 0.1 ohm
The load requires 10KW of average power.
Examine the line losses for a load power factor of unity and for a power factor of 0.7 lagging.
The load current is obtained from
For unity power factor this is
For power factor of 0.7 I L 
PAV  12 VL IL  pf 
2 10KW 
IL 
 117.65A
170V 1
2 10KW 
 168.07A
170V  0.7 
The powers consumed in the line losses
PAV,line 
PAV,line 
 692.04W
1 2
I L R line  
2
1412.33W
 692.04W
1 2
I L R line  
2
1412.33W
720 W extra power to be generated if pf is 0.7 to supply the load
 unity pf 
 pf  0.7 
 unity pf 
 pf  0.7 
Power Factor Correction
Ex: 6.17: in Ex 6.16 determine the value of capacitor across the load to correct the
power factor to unity if power frequency is 60Hz.
1
The angle of the current is: I   cos  0.7   45.57 (Negative because lagging)
Thus the current into the load is: ÎL  168.07  45.57
The current through the added capacitor is: ÎC  jC 1700
Hence the total current ˆIline  ˆI L  ˆIC  168.07  45.57  jC 1700
 117.66  j120.02  j  2 60  C 1700
Solving this to cancel out the reactive component gives: C  1873F
C=120.02/(2*3.14*60*170)=0.001873F
Power Factor Correction
Ex: 6.17: in Ex 6.16 determine the value of capacitor across the load to correct
the power factor from 0.7 to unity if power frequency is 60Hz.
0.1 
ˆ  V 0o
V
S
S
+ ÎL
ˆ
Load
V
L
+

From Ex 6.16
ÎC
For power factor of 0.7
C

IL 
I   cos1  0.7   45.57
power factor 0.7 lagging
2 10KW 
 168.07A
170V
0.7

 
ÎL  168.07  45.57
ˆ
V
1700  jC 1700
The current through the added capacitor is: Î C  L

ZC
1 jC 
Hence the total current ˆI  ˆI  ˆI  168.07  45.57  jC 1700
line
L
C
 117.66  j120.02  j  2 60 C 170
cos(v  i ) 1
Unity power factor
v  i  0
Imaginary component of the line current is zero
C
120.02
 2 60  170
 1873 uF
i v  0
 j120.02  j  2 60 C 170  0
 R
 R TH
2
TH
ZL  R TH  jXTH
  XTH  XTH 
2

2
 R TH
 XTH  XTH

2
The average power delivered to the load is:
PAV ,load 
1 2
V TH2 R L
V TH2 RTH
1
I RL  1

2
2 (RTH  R L ) 2  (X TH  X L ) 2 2 (RTH  RTH ) 2  (X TH  X TH ) 2
V TH2
1 V TH2 RTH


2 (2RTH ) 2 8RTH
6.6.3 Maximum Power Transfer
Source-load Configuration
Determine the load impedance so that maximum average power is delivered to
that load.
Represent the source and the load impedances with real and imaginary parts:
Zˆ  R  jX
S
S
S
Zˆ L  R L  jX L
The load current is:
IˆL 
VˆS
VˆS

ˆ
ˆ
Z S  Z L ( RS  RL )  j ( X S  X L )
The average power delivered to the load is:
PAV ,load
V S2 R L
1 ˆ 2
1
 I L RL 
2
2 (R S  R L ) 2  (X S  X L ) 2
Since the reactance can be negative and to max value, we choose X L  X S
leaving
PAV ,load
1 V S2 R L

2 (R S  R L ) 2
Differentiate with respect to RL and set to zero to determine required RL which is
RL= RS
Hence:
Zˆ L  ZˆS*
In this case the load is matched to the source.
The max power delivered to the load becomes:
PAV ,load ,max
V S2

8R S
6.6.4 Superposition of Average Power
Average power computation when circuit contains more than one source
 i ' (t )  I 1 sin(1t  I 1 )
V s sin(1t  1 )   '
v (t ) V 1 sin(1t  V 1 )
 i '' (t )  I 2 sin(2t  I 2 )
I s sin(2t  2 )   ''
v (t ) V 2 sin(2t  V 2 )
 i ' (t )  I 1 sin(1t   I 1 )
V s sin(1t  1 )   '
v (t ) V 1 sin(1t  V 1 )
 i '' (t )  I 2 sin(2t   I 2 )
I s sin(2t  2 )   ''
v (t ) V 2 sin(2t  V 2 )
 i ' (t )  I 1 sin(1t  I 1 )
V s sin(1t  1 )   '
v (t ) V 1 sin(1t  V 1 )
 i '' (t )  I 2 sin(2t  I 2 )
I s sin(2t  2 )   ''
v (t ) V 2 sin(2t  V 2 )
The instantaneous power delivered to the element is
p (t )  v (t )i (t )  (v ' v '' )(i '  i '' )  (v 'i ' v ''i '' )  (v 'i '' v ''i ' )
Substituting
p (t )  [V 1 sin(1t  V 1 )I 1 sin(1t   I 1 ) V 2 sin(2t  V 2 )I 2 sin(2t   I 2 )]
 [V 1 sin(1t  V 1 )I 2 sin(2t   I 2 ) V 2 sin(2t  V 2 )I 1 sin(1t   I 1 )]
1
1
sin A sin B  cos( A  B )  cos(A  B )
Using the identity
p (t )  [V 1 sin(1t  V 1 )I 1 sin(1t   I 1 ) V 2 sin(2t  V 2 )I 2 sin(2t   I 2 )]
 [V 1 sin(1t  V 1 )I 2 sin(2t   I 2 ) V 2 sin(2t  V 2 )I 1 sin(1t   I 1 )]
1
2
1
2
Using the identity sin A sin B  cos(A  B )  cos(A  B )
p (t ) 
V 1I 1
V I
cos(V 1   I 1 )  2 2 cos(V 2   I 2 )
2
2
PAV 1
PAV
2
V 1I 1
V I
cos(21t  V 1   I 1 )  2 2 cos(22t  V 2   I 2 )
2
2
VI
VI
 1 2 cos[(1  2 )t  V 1   I 2 ]  1 2 cos[(1  2 )t  V 1   I 2 ]
2
2
V I
V I
 2 1 cos[(2  1 )t  V 2   I 1 ]  2 1 cos[(2  1 )t  V 2   I 1 ]
2
2

p (t ) 
V 1I 1
V I
cos(V 1   I 1 )  2 2 cos(V 2   I 2 )
2
2
PAV 1
PAV
Average powers delivered individually by the sources
2
V 1I 1
V I
cos(21t  V 1   I 1 )  2 2 cos(22t  V 2   I 2 )
2
2
VI
VI
 1 2 cos[(1  2 )t  V 1   I 2 ]  1 2 cos[(1  2 )t  V 1   I 2 ]
2
2
V I
V I
 2 1 cos[(2  1 )t  V 2   I 1 ]  2 1 cos[(2  1 )t  V 2   I 1 ]
2
2

Suppose that the two frequencies are integer multiples of some frequency 
and
2  m
as 1  n
The instantaneous power becomes
p (t ) 
V 1I 1
V I
cos(V 1   I 1 )  2 2 cos(V 2   I 2 )
2
2
PAV 1
PAV
2
V 1I 1
V I
cos(2n t  V 1   I 1 )  2 2 cos(2m t  V 2   I 2 )
2
2
VI
VI
 1 2 cos[(n  m )t  V 1   I 2 ]  1 2 cos[(n  m )t  V 1   I 2 ]
2
2
V I
V I
 2 1 cos[(m  n )t  V 2   I 1 ]  2 1 cos[(m  n )t  V 2   I 1 ]
2
2

Averaging the instantaneous over the common period T  2 / 
1 T
p (t )dt

0
T
PAV 1  PAV 2


V 1I 2
V 2I 1
P

P

cos(



)

cos(V 2   I 1 )
V1
I2
 AV 1 AV 2
2
2
PAV 
if n  m
if n  m
THUS: we may superimpose the average powers delivered by sources of
different frequencies, but we may not, in general, apply superposition to
average power if the sources are of the same frequency.
where
V 1I 1
1
cos(V 1   I 1 )  Re(Vˆ1Iˆ1* )
2
2
V I
1
 2 2 cos(V 2   I 2 )  Re(Vˆ2 Iˆ2* )
2
2
PAV 1 
PAV 2
Ex 6.18: Determine the average power delivered
by the two sources of the circuit
Iˆ' 
1030
 2.357  15
2  j 4 1 j 1
Hence the average power delivered by the voltage source is
'
AV
P
1
1
'
ˆ
 Re(1030 I *)   10  2.357  cos(30  15 )  8.333 W
2
2
This can be confirmed from average powers delivered to the two resistors
'
'
'
PAV
 PAV

P
,2 
AV ,1
1 ˆ' 2
1 ˆ' 2
 I 2  I 1  8.333 W
2
2
By current division:
Iˆx'' 
1 j
2
3
3  60  0.589  154.33
2
2  j 6 1 j
3
2 j6
Iˆy'' 
3  60  3.101  49.08
2
2  j 6 1 j
3
''
The voltage across the current source is
Vˆ  (2  j 6)Iˆx''  3.727  82.77
Hence the average power delivered by the current source is
''
PAV

1
1
Re(Vˆ '' 360 )   3.727  3  cos(82.77  60 )  5.154 W
2
2
This may be again confirmed by computing the average power delivered to the
1 ˆ'' 2
1 ˆ'' 2
''
''
''
Two resistors:
PAV  PAV ,2  PAV ,1  I x 2  I y 1  5.154 W
2
2
Since frequencies are not the same, total average power delivered is the sum
of average powers delivered individually by each source
EX 6.19: Determine the average power delivered
by the two sources
Since both sources have the same frequency, we can’t use superposition.
So we include both sources in one phasor circuit. The total average power
delivered by the sources is equal to the average power delivered to the resistor
We use superposition on the phasor circuit to find the current across the resistor
Iˆ' 
100
 3.536  45
2 j 4 j 2
Iˆ''  
j2
5  60  3.536  15
2 j 4 j 2
Iˆ  Iˆ'  Iˆ''  3.536 45  3.53615  6.831 30
Iˆ  Iˆ'  Iˆ''  3.536 45  3.53615  6.831 30
The phasor current is:
Hence the average power delivered to the resistor is PAV
1 ˆ2
 I  2  46.66 W
2
Note that we may not superimpose average powers delivered to the resistors by the
individual sources
1 ˆ' 2
1 ˆ '' 2
2
I
2
2
I
 2  25  46.66
We can compute this total average power by directly computing the average power
delivered by the sources from the phasor circuit
The voltage across the current source is
Vˆ  100  (2  j 4)Iˆ  22.88  132.63
The average power delivered by voltage source is
PAV , voltage source 
1
Re[(100 )Iˆ* ]  29.58 W
2
The average power delivered by the current source is
PAV , current source 
1
Re[Vˆ (560 )]  17.08 W
2
The total average power delivered by the sources is
PAV , source  29.58  17.08  46.66 W
6.6.5 Effective (RMS) Values of Periodic Waveforms
Sinusoidal waveform is one of more general periodic waveforms
Apply a periodic current source with period T on resistor R
i (t )  i (t  nT )
n  0,1, 2,3,...
The instantaneous power delivered to the resistor is
p (t )  i 2 (t )R
The average power delivered to the resistor is
PAV
1

T

T
0
1
i (t )R dt  R 
T
2

T
0

i 2 (t )dt 

2
I eff
Hence the average power delivered to the resistor by this periodic waveform
can be viewed as equivalent to that produced by a DC waveform whose value is
I eff
1

T

T
0
i 2 (t )dt
This is called the effective value of the waveform or the root-mean-square
RMS value of the waveform
Ex 6.20 Determine the RMS value of the current
waveform and the average power this would
deliver to 3 resistor
The RMS value of the waveform is
I rms
1 1

4dt  1.155 A

0
3
Hence the average power delivered to the resistor is
2
PAV  I rms
 3  4 W
RMS voltages and currents in phasor circuits
The sinusoid x t   X sin t    has a RMS value of
X rms
1

T

T
0
[X sin(t   )]2 dt 
X
 0.707X
2
Hence the average power delivered to a resistor by a sinusoidal voltage or current
2
1 V 2 V rms
1
waveform is
P


 I 2R  I 2 R
AV , R
2 R
R
rms
2
In general, the average power delivered to an element is
PAV
ˆ Iˆ* 

1
V
*
 Re(Vˆˆ
I *)  Re 
  Re(Vˆrms Iˆrms )
2
 2 2
Therefore, if sinusoidal voltages and currents are specified in their RMS values
rather than their peak values, the factor ½ is removed from all average-power
expressions. However, the time-domain expressions require a magnitude multiplied
by square root of 2 X sin(t   )  2X sin(t   )  X sin(t   )
rms
rms
Since X is the peak value of the waveform. Common household voltage are specified
as 120V. This is the RMS value of the peak of 170V.
Ex 6.21 Determine the average power delivered by
the source and the time-domain current i(t)
Phasor circuit with rms rather than peak
The phasor current is
7.0730
Iˆrms 
 1.25  15
4 j 4
Hence the average power delivered by the source is
1
*
PAV , source  ReVˆrms Iˆrms
 Re[(7.0730 )(1.2515 )]  6.25 W  Re[(1030 )(1.25 215 )]
2
The time-domain current is
i (t )  2I rms cos(2t 15 )  1.77cos(2t 15 ) A
Commercial Power Distribution
a
a
+


Van =Vp0

c



n
Vbn =Vp120
b

b
Vcn =Vp120

c
a
a


+
Van =Vp0
Van =Vp0
Vbn =Vp120

c



n
Vbn =Vp120
b 

b
Vp
is rms value of the voltage
The peak is
2 Vp
Time domain representation
Vcn =Vp120

Vcn =Vp120
v an (t )  2V p sin t V
c
v bn (t )  2V p sin(t  120 ) V
v cn (t )  2V p sin(t  120 ) V
a
a
Im
+


Van =Vp0

c



n
Vbn =Vp120
b 

Van =Vp0
Re
b
Vcn =Vp120

Vcn =Vp120
Vbn =Vp120
c
Im
Vcn =Vp120
Van =Vp0
Van  Vbn +Vcn= 0
Re
V p0 V p120 V p120  0
Vbn =Vp120
Im
Im
Im
Vcn
Van
Van
Van
Re
Re
Re
Vbn
Van  Vbn
Van  Vbn
Vbn
Vbn
a
a
+
+


Van =Vp0




Vbn =Vp120
b 
c
n

b
Vab = Van  Vnb = Van  V
bn

 V p0  V p120
Vcn =Vp120

3
1


2
2
0
1
 V p cos(0 ) + j V p sin(0 )   V p cos(120 ) + j V p sin(120 ) 

 

3 Vp
V 3

j p 
2
2

3V p30
c
3 3 Vp
V 3 
j p
2
2

3V p 


3  j 1 
2
2 
Im
Using Vectors
Vcn =Vp120
Van =Vp0
Re
Vab = Van  Vbn
Vbn =Vp120
Im
Vbn
Vab = Van  Vbn 
3V p
3V p30
Vbn
line voltage is 3 phase voltage
30
Vp
Re
Van
Im
Vab 
Vcn
30
3V p30
Re
Van
Vbn
line voltage is 3 phase voltage
a
a
+


Van =Vp0

c



n
Vbn =Vp120
b 

b
Vcn =Vp120

c
+
Vbc = Vbn  Vnc

Vbc = Vbn  Vnc = Vbn  Vcn  V p 120 V p120  3V p 90
Vca = Vcn  Vna = Vcn  Van
 V p120 V p0
 3V p150
Vca 
3V p150
Im
Vab 
Vcn
30
30
90
o
Re
Van
Vbn
Vbc 
3V p30
3V p 90
6.9.1 Wye-Connected Loads
We are going to investigate the transmission of power from
three phase generator to two types of load:
I – WYE connected load
V p 0
ˆ
Ia 
Zˆ L
V p   120
ˆ
Ib 
Zˆ L
V p 120
ˆ
Ic 
Zˆ L
ˆI  Iˆ  Iˆ  Vp 0  Vp   120  Vp 120  0
a
b
c
Zˆ L
Hence the current returning through the neutral wire is zero,
and the neutral may be removed
Zˆ L  Z L  Z L
Power calculation
For a balanced wye-connected load, the voltages across the individual
loads are the respective phase voltages whether the neutral is connected
or not.
The power delivered to the individual loads is three times the power
delivered to an individual load, because the individual loads are identical.
PAV ,total

V p 0
*
ˆ
 3Re(V p 0 I a )  3Re V p 0

ˆ*
Z
L

V p2

cos  Z L
  3
ZL

No ½ factor in power expression because values are rms
PAV ,total  3Re(V p 0 Iˆa* )

V p 0
 3Re V p 0

ˆ*
Z
L

V p2
3
cos  Z L
ZL
Since the line voltage is more accessible than the phase voltage
Power in terms of line-to-line voltages and load current gives
PAV ,total
V p2
3
cos  Z L  3V PV P cos   3 V P I L cos  Z
ZL
L
ZL
Z
L
VL
3
I L cos  Z L  3 V L I L cos  Z L
3



Example 6.24
Consider a balanced, wye-connected load
where each load impedance is
Zˆ L  50  j50 
the phase voltages are 120 V.
Determine the total average power delivered
to the load.
The line currents are
1200
Iˆa 
 1.7  45 A
50  j 50
120  120
Iˆb 
 1.7  165 A
50  j 50
120120
Iˆc 
 1.775 A
50  j 50
Hence, the average power delivered to each load is
PAV  Re[(1200 )(1.745 )]  120 1.7  cos(45 )  144 W
The total average power delivered to the load is
PAV , total  3 144  432 W
Example 6.25
If the line voltage of a balanced, wye-connected load is 208 V
and the total average power delivered to the load is 900 W,
determine each load if their power factors are 0.8 leading.
900 W  3 V L I L cos  Z L
Thus
IL 
900
 3.12 A
3(208)(0.8)
The phase voltage is Vp 
VL
 120 V
3
Thus the magnitude of the individual load impedance is Z L 
Vp
IL
 38.43 
Since the power factor is 0.8 leading (current leads voltage; voltage lags current)
Z   cos1 0.8  36.87 .
L
Thus the individual loads are
Zˆ L  38.43  36.87
 30.74  j 23.06 
Example 6.25
If the line voltage of a balanced, wye-connected load is
208 V and the total average power delivered to the load
is 900 W, determine each load if their power factors are
0.8 leading.
900 W  3 V L I L cos  Z L
Thus
IL 
900
 3.12 A
3(208)(0.8)
The phase voltage is
Vp 
VL
 120 V
3
Thus the magnitude of the individual load impedance is
ZL 
Vp
IL
 38.43 
Since the power factor is 0.8 leading (current leads voltage; voltage lags current)
Z   cos1 0.8  36.87 .
L
Thus the individual loads are
Zˆ L  38.43  36.87
 30.74  j 23.06 