Solution 12.70
8004001200PPP 21T =−=+=
Solution 12.71
(a) If
°∠= 0208
ab
V
,
°∠= 120–208
bc
V
,
°∠= 120208
ca
V
,
°∠
0208
ab
V
°∠−°∠=−= 75–708.1497.8316
BCCAcC III
(b)
W17.7398PPP
21T
=+=
Solution 12.72
From Problem 12.11,
V130220
AB °∠=V
and
A18030
aA
°∠=I
Solution 12.73
Consider the circuit as shown below.
°∠=+= 71.5762.3130j10Z
°∠
60–240
0120–24060–240
c=°∠−°∠+ZI
[ ]
[ ]
=°∠°∠== ).57101146.13)(60–240(ReReP *
111 IV
kW360.2
Z
Ib
Ia
I1
Z
Ic
Z
I2
Solution 12.74
Consider the circuit shown below.
For mesh 1,
In matrix form,
1
–2
208
I
ZZ
Z = 60 − j30 Ω
Z
Z
208∠0° V
+
−
I1
I2
208∠-60° V
−
+
Solution 12.75
V
12
V
Solution 12.76
If both appliances have the same power rating, P,
s
V
P
I=
P
Solution 12.77
A three-phase generator supplied 10 kVA at a power factor of 0.85 lagging. If 7.5 kW
are delivered to the load and line losses are 160 W per phase, what are the losses in the
generator?
Solution
Solution 12.78
°=θ→==θ 79.3185.0
60
51
cos 11
For each load,
Q
c
Solution 12.79
A balanced three–phase generator has an abc phase sequence with phase voltage
Van = 554.3∠0° V. The generator feeds an induction motor which may be represented by
a balanced Y-connected load with an impedance of 12 + j5 Ω per phase. Find the line
currents and the load voltages. Assume a line impedance of 2 Ω per phase.
Solution
Consider the per-phase equivalent circuit below.
Thus,
=°∠= 120–
ab
II
37.29∠–139.65° A
2 Ω
−
A
N
Ia
a
n
Solution 12.80
)7071.07071.0(8)]83.0sin(cos83.0[6
2
1
321
jSjSSSS −+++=++=
−