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=I R22IR==⋅=P22 4018 5..V A7 000 V AmA .The potential difference across R1and C1is∆VIR==×=−121 85104 00074 1..AVAVejbg.The charge on C1FIG. P28.68(a)QCV==×=−163 001074 1222∆..C VVCejafµ.The potential difference across R2and C2is∆ΩVIR==×=−221 85107 000130.AVejbg.The charge on C2QCV==×=−266 0010130778∆.C VVCejafµ.The battery emf isIRI RReq=+=×+=−1221 85107 000204bgbg.A4 000VAV.(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 C2QCV==×=−266 00102041 222∆.C VVCejafµfor a change of 1 222778444CCCµµµ−=.FIG. P28.68(b)*P28.69The battery current is15045144213+++=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. P28.6915045142211+++=afmAmA, nearly the same as before. The ratio is 211
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Resistor, Potential difference, Electrical resistance, Series and parallel circuits