ch27-p049 - 49. The current in R2 is i. Let i1 be the...

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ε = iR i R ir 21 1 0. We apply the loop rule to the right-hand loop to obtain iR i i R V 11 1 0 −− = bg . The second equation yields i RR R i V V = + 1 1 We substitute this into the first equation to obtain ++ += Rr R i V V 1 0 b g This has the solution i R R R V VV 1 1 = + b gb g . The reading on the voltmeter is () ( ) ( ) ( ) ( ) 3 1 33 1 3.0V 5.0 10 250 300 100 250 5.0 10 250 5.0 10 1.12V. V RrRR R R ×Ω == + + +×Ω + = The current in the absence of the voltmeter can be obtained by taking the limit as R V becomes infinitely large. Then R RRr 1 12 30 250 250 300 100 115 = = = . . V V. b
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