70.
(a) The rms current in the cable is I
rms
= P/V
t
= 250
× 10
3
W/(80
× 10
3
V) = 3.125 A. The rms
voltage d rop is then ∆V = I
rms
R = (3.125 A)(2)(0.30 Ω) = 1.9 V, andthe rate of energy dissipation
is P
d
= I
2
rms
R = (3.125 A)(2)(0.60 Ω) = 5.9 W.
(b) Now I
rms
= 250
× 10
3
W/(8.0
× 10
3
V) = 31.25 A, so ∆V = (31.25 A)(0.60 Ω) = 19 V and P
d
=
(3.125 A)
2
(0.60 Ω) = 5.9
× 10
2
W.
(c) Now I
rms
= 250
× 10
3
W/(0.80
× 10
3
V) = 312.5 A, so ∆V = (312.5 A)(0.60 Ω) = 1.9
× 10
2
V and
P
d
= (312.5 A)
2
(0.60 Ω) = 5.9
× 10
4
W. Both the rate of energy dissipation and the voltage drop
increase as V
t
decreases. Therefore, to minimize these effects the best choice among the three V
t
’s
above is V
t
= 80 kV.