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EE 212  Assignment 5  Solutions
1.
A balanced 3phase
∆
connected load is energized by a 3phase source that has linetoline
voltage of 208 V rms. and phase sequence abc. Per phase load impedance Z
L
= 12 + j16 ohms.
Find the line and phase voltages and currents, and show them in a phasor diagram. Determine
the real and reactive power delivered to the load.
Per phase equivalent circuit
:
Taking phase voltage V
ab
as the reference,
V
ab
= 208 /0
0
V
A
j
Z
V
I
ab
ab
0
0
13
.
53
4
.
10
16
12
0
208

∠
=
+
∠
=
=
Figure 2
Phase currents:
I
ab
=
10.4
∠
53.13
0
A
I
bc
=
10.4
∠
(53.13 – 120)
0
=
10.4
∠
173.13
0
A
I
ca
=
10.4
∠
(53.13 + 120)
0
=
10.4
∠
66.87
0
A
Line currents:
I
a
=
I
ab
 I
ca
=
10.4
∠
53.13
0
 10.4
∠
66.87
0
=
18.02
∠
83.1
0
A
I
b
=
I
bc
 I
ab
=
10.4
∠
173.13
0
 10.4
∠
53.13
0
=
18.02
∠
156.9
0
A
I
c
=
I
ca
 I
bc
=
10.4
∠
66.87
0
 10.4
∠
173.13
0
=
18.02
∠
36.9
0
A
Phase Voltages (Line Voltages are the same)
:
V
ab
=
208
∠
0
0
Volts
V
bc
=
208
∠
120
0
Volts
V
ca
=
208
∠
120
0
Volts
1
a
b
c
a
b
12+ j16
Ω
c
12+ j16
Ω
12+ j16
Ω
V
L
= 208 V
12+ j16
Ω
a
b
I
ab
V
ab
= 208 /0
0
V
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:
Power per phase:
P
ph
=
I
2
R
=
(10.4)
2
* 12
=
1297.92
W
Q
ph
=
I
2
X
=
(10.4)
2
* 16
=
1730.56
VAr
OR
P
ph
=
V
ab
* I
ab
Cos (53.13 )
=
(208) * (10.4) Cos(53.13 )
=
1297.92
W
Q
ph
=
V
ab
* I
ab
Sin (53.13 )
=
(208) * (10.4) Sin(53.13 )
=
1730.56
VAr
Total Power:
P
=
3 * 1297.92
=
3893.76 W
Q
=
3 * 1730.56
=
5191.67 VAr
2
I
ca
I
ab
I
bc
I
a
V
ab
V
ca
V
bc
30
0
53.13
0
I
ca
I
b
I
c
2.
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This note was uploaded on 09/25/2010 for the course EE 212 taught by Professor Gander during the Spring '10 term at University of Saskatchewan Management Area.
 Spring '10
 Gander
 Impedance, Volt

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