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.