ε−−−=iRi Rir2110. We apply the loop rule to the right-hand loop to obtain iRi i RV1110−−=bg. The second equation yields iRRRiVV=+11We substitute this into the first equation to obtain −+++=RrRiVV10bgThis has the solution iRRRVVV11=+bgbg.The reading on the voltmeter is ()()( ) ( ) ( )313313.0V5.0 102503001002505.0 102505.0 101.12V.VRrRRRR×ΩΩ==+Ω+ΩΩ+×Ω+Ω=The current in the absence of the voltmeter can be obtained by taking the limit as RVbecomes infinitely large. Then RRRr11230250250300100115===..VV.b
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