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Homework # 6
P5.41* This is a capacitive load because the reactance is negative.
( 15 ) 100 22. 5 kW
2 2
P = Irms R= =
( 15 ) ( 50 ) 11. 25 kVAR
2 2
Q = Irms X = − =−
D tan tan ( 0. 5 ) 26. 57
1 1 = − =
− −
P
Q
θ
power factor = cos( θ )= 89. 44 %
P5.44*
P5.47* Load A:
LoadB:
Source:
Powerfactor cos ( 20. 66 ) 0. 9357 93. 57 %leading
Apparentpower 10. 68 kVA
Q V sin 3. 770 kVAR
cos 10 kW
rms
D
D
D D
rms rms
rms
rms rms
V I
I
P V I
j j
θ
θ
I
tan 4. 843 kVAR
cos 0. 9 25. 84
10 kW
1 A
= =
−
A A A
A
Q P
P
θ
θ
D
cos( ) 12 kW
sin 9 kVAR
cos 0. 8 36. 87
15 kVA
1
−
B rms Brms B
B rms Brms B
B
rms Brms
P V I
Q V I
V I
θ
θ
θ
D
- 8462 84. 62 % lagging Apparentpower
Powerfactor
Apparentpower 26 kVA
- 84 kVAR
22 kW
2 2
s
s s
s A B
s A B
P
P Q
Q Q Q
P P P
P5.49* (a)
(b)
The capacitor must be rated for at least 387.3 kVAR. With the
capacitor in place, we have:
(c) The line current is smaller by a factor of 4 with the capacitor in
place, reducingI R
2 losses in the line by a factor of 16.
P5.52* (a) Zeroing the current source, we have:
Thus, the Thévenin impedance is
100 50 111. 8 26. 57 Ω
D
Z t = + j = ∠
Under open circuit conditions, there is zero voltage across the
inductance, the current flows through the resistance, and the Thévenin
voltage is
( )
( ) ( )
D
D
400 A
1 kV 0. 25
100 kW
cos
cos
cos 0. 25
I
θ
θ
θ
θ
rms
rms
rms rms
V
P
I
P V I
( )
1027 F
Q 0
sin 387. 3 kVAR
2 3
total
μ
ω
θ
=− × =
C
C
X
X
V
Q
Q Q
Q V I
C
C
rms C
load C
load rms rms
D 100 0
100 A
100 kW
I
rms
rms rms
I
P V I
P5.58*
P5.59*
P5.63* This is a positive sequence source. The phasor diagram is shown in Figure
5.40 in the book. Thus, we have:
D 0 3
V an = ∠
The impedance of a equivalent wye-connected load is
= ∆^ = 1. 667 − 0. 6667 Ω
j
Z
Z Y
The equivalent circuit for the a-phase of an equivalent wye-wye circuit is:
Thus, the line current is:
( ) ( )
- 36 kW
3V cos 3 440 14. 67 cos 0
- 67 Arms 30
3 3 440 762. 1 Vrms
Y
=
= = × × ×
= × = × =
L^ θ
Y L
L Y
P I
R
V
I
V V
+ ×
∆
−
4
D
D
Y
Y
Z Z
j
j
R j C
Z
( )
( )
3 ( ) 0. 5 20. 50 kW
3 5 68. 32 kW
2
2
aA
= × =
= × =
∆
line aArms
load ABrms
AB AB
AB
An an aA
Y
an
P I
P I
Z
j
j Z
D
D
D
D
V
I
V
V V I
V
I
