9/26/01
P121
CHAPTER 12.
PROBLEM SOLUTIONS
SOLUTION P12.1.
By conservation of energy, the instantaneous power consumed by each load when
summed together is equivalent to the total power consumed by the three phase load. Thus writing out
p
tot
(
t
)
=
p
AB
(
t
)
+
p
BC
(
t
)
+
p
CA
(
t
)
=
v
AB
(
t
)
i
AB
(
t
)
+
v
BC
(
t
)
i
BC
(
t
)
+
v
CA
(
t
)
i
CA
(
t
)
=
V
L
2
cos(
t
+
v
)
⋅
V
L
2
Z
cos(
t
+
i
)
+
V
L
2
cos(
t
+
v

120
o
)
⋅
V
L
2
Z
cos(
t
+
i

120
o
)
+
V
L
2
cos(
t
+
v
+
120
o
)
⋅
V
L
2
Z
cos(
t
+
i
+
120
o
)
=
V
L
2
Z
cos(
v

i
)
+
cos(2
t
+
v
+
i
)
( 29
+
V
L
2
Z
cos(
v

i
)
+
cos(2
t
+
v
+
i
+
120
o
)
( 29
+
V
L
2
Z
cos(
v

i
)
+
cos(2
t
+
v
+
i

120
o
)
( 29
=
3
V
L
2
Z
cos(
v

i
)
=
3
V
L
2
Z
×
pf
SOLUTION P12.2.
To justify the point of this problem we equate the following two equations:
(i)
For the 3 phase system:
P
loss
'
=
3
×
I
L
2
R
'
=
3
×
3
9
×
P
L
2
V
L
2
R
'
=
P
L
2
V
L
2
R
'
(ii)
For the single phase system:
P
loss
=
2
×
I
L
2
R
( 29
=
P
L
2
V
L
2
2
R
It follows that R' = 2R.
Since both systems have the same distance of transmission and the resistance of
a wire is inversely proportional to the cross sectional area, the condition R' = 2R implies that the cross
section A' of each wire in the threephase system need only be half of the area A of the wire in the single
phase system.
But there are two wires in the single phase system and three wires in the threephase
system.
Therefore the ratio of the materials used is:
material in 3

phase system
material in 1

phase system
=
3
A
'
2
A
=
3
2
×
1
2
=
75%
SOLUTION P12.3.
For row 1 of table 12.1, the impedance in (a) seen between each pair of terminal is
Z
ik
=
Z
∆
 2
Z
∆
=
2
Z
∆
3
.
In (b) the impedance seen between any two terminals is
Z
jk
=
Z
∆
3
+
Z
∆
3
=
2
Z
∆
3
.
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P122
In row 2, the impedance between any two terminals is
Z
jk
=
Z
Y
+
Z
Y
=
2
Z
Y
for (c), and
Z
ik
=
3
Z
Y
 6
Z
Y
=
2
Z
Y
for (d).
SOLUTION P12.4
.
Consider the
∆
Y relationship of the figures below (row 3 of table 12.1):
Let us consider the terminal pair (1,2):
(i)
For the Yconnected case,
Z
th
=
2
×
Z
∆
3
and
V
oc
=
V
p
3
∠

30
o
( 29

V
p
3
∠

150
o
( 29
=
V
p
3
∠
1
∠
30
o

1
∠ 
150
o
( 29
=
V
p
∠
(ii)
For the Dconnected case,
Z
th
=
Z
∆
/ / 2
Z
∆
( 29 =
2(
Z
∆
)
2
3
Z
∆
=
2
3
Z
∆
Now note that there is no load connected to the
∆
configuration.
Applying KVL to the indicated loop
implies that:
0
=
3
Z
∆
I
loop
+
V
p
∠
+
V
p
∠

120
o
+
V
p
∠
+
120
o
=
3
Z
∆
I
loop
Hence
I
loop
=
0 .
Finally,
V
oc
=
V
12
=
Z
∆
I
loop
+
V
p
∠
=
V
p
∠
.
Therefore, looking into terminals 12,
both the
∆
configuration and the Yconfiguration have the same Thevenin equivalent.
For terminal pairs
(13) and (23), the proof is virtually the same.
Hence this establishes the equivalence in row 3 of table
12.1.
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