815_Physics ProblemsTechnical Physics

815_Physics ProblemsTechnical Physics - Chapter 28 *P28.68...

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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 = IR 2 2 I R == = 2 240 18 5 . . VA 7000 V A mA . The potential difference across R 1 and C 1 is VI R × = 1 2 185 10 4000 741 .. A V V ej bg . The charge on C 1 FIG. P28.68(a) QCV × = 1 6 300 10 222 CV V C a f µ . The potential difference across R 2 and C 2 is ∆Ω R × = 2 2 7000 130 . A V . The charge on C 2 × = 2 6 600 10 778 C V V C a f . The battery emf is IR I R R eq =+ = × + = 12 2 204 b g A 4 0 0 0 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 × = 2 6 6 00 10 204 1 222 C V V C a f for a change of 1 222 778 444 C C C µµ −= . FIG. P28.68(b) *P28.69 The battery current is 150 45 14 4 213 +++ = af 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 = mA mA, nearly the same as before. The ratio is 211 213 0991 = (b) The resistor with least resistance carries 150 mA. Doubling its resistance changes this current
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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