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794_Physics ProblemsTechnical Physics

# 794_Physics ProblemsTechnical Physics - 134 Direct Current...

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134 Direct Current Circuits P28.4 (a) Here ε = + I R r a f , so I R r = + = + = ε 12 6 0 080 0 2 48 . . . V 5.00 A b g . Then, V IR = = = 2 48 5 00 12 4 . . . A V a fa f . (b) Let I 1 and I 2 be the currents flowing through the battery and the headlights, respectively. Then, I I 1 2 35 0 = + . A , and ε = I r I r 1 2 0 so ε = + + = I I 2 2 35 0 0 080 0 5 00 12 6 . . . . A V b gb g a f giving I 2 1 93 = . A. Thus, V 2 1 93 5 00 9 65 = = . . . A V a fa f . FIG. P28.4 Section 28.2 Resistors in Series and Parallel P28.5 V I R R = = 1 1 1 2 00 . A a f and V I R R R = + = + 2 1 2 1 1 60 3 00 b g a fb g . . A Therefore, 2 00 1 60 3 00 1 1 . . . A A a f a fb g R R = + or R 1 12 0 = . . P28.6 (a) R p = + = 1 1 7 00 1 10 0 4 12 . . . b g b g R R R R s = + + = + + = 1 2 3 4 00 4 12 9 00 17 1 . . . . (b) V IR = 34 0 17 1 . . V = I a f I = 1 99 . A for 4 00 . , 9 00 . resistors. Applying V IR = , 1 99 4 12 8 18 . . . A V a fa f Ω = 8 18 7 00 . . V = I
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