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815_Physics ProblemsTechnical Physics

# 815_Physics ProblemsTechnical Physics - Chapter 28*P28.68(a...

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Chapter 28 155 *P28.68 (a) With the switch closed, current exists in a simple series circuit as shown. The capacitors carry no current. For R 2 we have P = I R 2 2 I R = = = P 2 2 40 18 5 . . V A 7 000 V A mA . The potential difference across R 1 and C 1 is V IR = = × = 1 2 1 85 10 4 000 74 1 . . A V A V e j b g . The charge on C 1 FIG. P28.68(a) Q C V = = × = 1 6 3 00 10 74 1 222 . . C V V C e j a f µ . The potential difference across R 2 and C 2 is V IR = = × = 2 2 1 85 10 7 000 130 . A V e j b g . The charge on C 2 Q C V = = × = 2 6 6 00 10 130 778 . C V V C e j a f µ . The battery emf is IR I R R eq = + = × + = 1 2 2 1 85 10 7 000 204 b g b g . A 4 000 V A V . (b) In equilibrium after the switch has been opened, no current exists. The potential difference across each resistor is zero. The full 204 V appears across both capacitors. The new charge C 2 Q C V = = × = 2 6 6 00 10 204 1 222 . C V V C e j a f µ for a change of 1 222 778 444 C C C µ µ µ = . FIG. P28.68(b) *P28.69 The battery current is 150 45 14 4 213 + + + = a f mA mA . (a) The resistor with highest resistance is that carrying 4 mA. Doubling its resistance will reduce the current it carries to 2 mA. Then the total current is FIG. P28.69 150 45 14 2 211 + + + = a f mA mA , nearly the same as before. The ratio is 211
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