944_Physics ProblemsTechnical Physics

944_Physics ProblemsTechnical Physics - Chapter 33 P33.64...

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Chapter 33 285 P33.64 Suppose each of the 20 000 people uses an average power of 500 W. (This means 12 kWh per day, or $36 per 30 days at 10¢ per kWh). Suppose the transmission line is at 20 kV. Then I V rms rms W V A == P 20 000 500 20 000 10 3 bg af ~ . If the transmission line had been at 200 kV, the current would be only ~10 2 A. P33.65 R = 200 , L = 663 mH , C = 26 5 . F µ , ω = 377 1 s, V max . = 50 0 V L = 250 , 1 100 C F H G I K J = , ZR XX LC =+ = 2 2 250 (a) I V Z max max . . === 50 0 0200 V 250 A φ = F H G I K J tan . 1 36 8 R ( V leads I ) (b) VI R R , max V max . 40 0 at 0 (c) V I C C , max . max V 20 0 at =− ° 90 0 . ( I leads V ) (d) L L ,m a x . max V 50 0 at ° 90 0 . ( V leads I ) P33.66 L = 200 . H , C 10 0 10 6 F , R = 10 0 , vt t a f = 100sin (a) The resonant frequency 0 produces the maximum current and thus the maximum power delivery to the resistor. 0 6 11 200 100 10 224 × = LC .. ej rad s (b) = V R max . a f 2 2 2 100 2100 500 W (c) I V Z V RL C rms rms rms +− ∆∆
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