Chapter 28155*P28.68(a)With the switch closed, current exists in a simple seriescircuit as shown. The capacitors carry no current. For R2we haveP=IR22IR==⋅=224018 5..VA7000 V AmA .The potential difference across R1and C1is∆VIR×=−12185 104000741..A VVejbg.The charge on C1FIG. P28.68(a)QCV×=−16300 10222∆CVVCafµ.The potential difference across R2and C2is∆ΩR×=−227000130. AV.The charge on C2×=−26600 10778∆CVVCaf.The battery emf isIRI RReq=+=×+=−122204bgA4000VAV.(b)In equilibrium after the switch has been opened, no currentexists. The potential difference across each resistor is zero. Thefull 204 V appears across both capacitors. The new charge C2×=−266 00 102041 222∆CVVCaffor a change of 1 222778444C CCµµ−=.FIG. P28.68(b)*P28.69The battery current is150 45 14 4213+++=afmAmA.(a)The resistor with highest resistance is thatcarrying 4 mA. Doubling its resistance willreduce the current it carries to 2 mA. Thenthe total current isFIG. P28.69150 45 14 2211=mAmA, nearly the same as before. The ratio is 2112130991=(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 .