p33_070

# p33_070 - V =(312 5 A(0 60 Ω = 1 9 × 10 2 V and P d...

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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 vo ltagedropisthen∆ V = I rms R =(3 . 125 A)(2)(0 . 30 Ω) = 1 . 9 V, and the rate of energy dissipation is P d = I 2 rms R =(3 . 125 A)(2)(0 . 60 Ω) = 5 . 9W . (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 .
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Unformatted text preview: 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 eFects the best choice among the three V t ’s above is V t = 80 kV....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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