12.3 Solutions
561
12.22 Total resistance of the cable
R = 0.7 × 30 = 21
Voltage V = 100 kV = 10
5 V
Power P = 10
4 kW = 10
7 W
Current I =
P
V
=
10 7
10 5 = 100 A
Power dissipated = i
2 R = (100)
2
× 21 = 2.1 × 10
5 W
Fractional power loss =
2.1 × 10 5
10 7
= 0.021 or 2.1%
12.23 Let each of the three resistances be r . In the series arrangement the effective
resistance, R 1 = 3r :
P 1 =
ξ 2
R 1
=
ξ 2
3r
= 10
∴
ξ 2
r
= 30
(1)
In the parallel arrangement the effective resistance R 2 = r/3:
P 2 =
ξ 2
R 2
=
3ξ 2
r
= 3 × 30 = 90 W
where we have used (1).
12.24 If the heater resistance is R 0 ,
R 0 =
V 2
P
=
(100) 2
1000
= 10
The combined resistance of R 0 and R in parallel is
10R
R + 10
. As this is in
series with 10 , the effective resistance of the circuit
R e = 10 +
10R
R + 10
=
20R + 100
R + 10
If P is the power of the heater
R e =
20R + 100
R + 10
=
V 2
P =
(100) 2
62.5
Solving for R, we find R = 5 .