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E E 2415
Lecture 9
Phasor Circuit Analysis, Effective
Value and Complex Power: Watts,
VAR’s and Volt-Amperes
Effective Value of a Sinusoid (1/3)
i(t)
Average Power:
10 
Effective Value of a Sinusoid (2/3)
Effective Value of a Sinusoid (3/3)
In our example:
Also:
The effective value is also called the Root Mean
Square value or rms value.
R-C Circuit Example (1/6)
40 
vs(t)
+ vR - +
i
vc
-
88.42 F
Capacitive Reactance
R-C Circuit Example (2/6)
40 
120
~
+ VR - +
~
~
I
VC
-
Using rms phasor for voltage source.
-j30 
R-C Circuit Example (3/6)
Calculate Real Power:
And Reactive Power:
Apparent power is the product of voltage and
current of the source.
Also:
R-C Circuit Example (4/6)
Power Factor is the ratio of real power to
apparent power:
Power Factor is also the Cosine of the angle
between the load voltage and the load current:
If the load current leads the load voltage, the
power factor is leading; if it lags the load voltage,
the power factor is lagging.
R-C Circuit Example (5/6)
2.4A
~
I
36.87o
120V
~
V
Phasor Diagram of Voltage and Current
Current leads voltage.
R-C Circuit Example (6/6)
imag
230.4 W
288.0 VA
real
-172.8 VAR
The Power Triangle showing leading
power factor.
Calculating Complex Power (1/2)
~
I
jX
R
~
V
+
S  P  jQ  I
~
I
c
jd
2
-
 R  jX 
I  c  jd ,
2
I  I I
 I  c  jd
*
then I  I  c  d  I
S  P  jQ  I  I
*
*
*
2
 R  jX 
2
2
Calculating Complex Power (2/2)
V   R  jX  I
S  P  jQ  V  I
*
From now on, we use the above method
to calculate complex power.
Lagging Power Factor Example (1/4)
5
v(t)
i
22.97 mH
Lagging Power Factor Example (2/4)
5
1200°
~
I
j8.66 
Calculate complex power directly:
Lagging Power Factor Example (3/4)
Power Factor:
Power Factor is Lagging
imag
1200° V
60o
real
12-60° A
Phasor Diagram of Voltage and Current
Lagging Power Factor Example (4/4)
14
40
VA
imag
1247 VAR's
real
720 W
Power Triangle for lagging Power Factor
Phasor Power Example (1/4)
0.5 
~
I
j2 
+
40 
~
V
480
V(rms)
-j150 
j30 
-
(40  j30)( j150)
Zp 
  (56.25  j18.75)
40  j 30  j150
4800 V
I
 7.94  20.08 A
(56.75  j 20.75)
Phasor Power Example (2/4)
V  I  (56.25  j18.75)  471  1.65
S RLC  V  I *
 (471V   1.65)(7.94 A20.08 )
 3.55 kW  j1.18 kVAR
PF 
3.55
 0.949 or 94.9% Lagging
3.74
Phasor Power Example (3/4)
I RL
471V   1.65

 9.42 A  38.52
(40  j30)
*
S RL  V  I RL
 (471V   1.65)(9.42 A38.52)
 3.55 kW  j 2.66 kVAR
Capacitor VAR’s:
4712 V 2
Qc 
 1.48 kVAR
150 
Phasor Power Example (4/4)
2.66 kVAR
1.18 kVAR
Q
44
.
4
A
V
k
}1.48 kVAR
VA
k
3.74
3.55 kW
P
Impedance and Admittance
Z  R  jX
Impedance
1
Y   G  jB Admittance
Z
R
Conductance
G 2
R  X2
X
B 2
R  X2
Susceptance
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