P28_039 - . × 10 3 Ω)(250 Ω) (300 Ω + 100 Ω)(250...

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39. The current in R 2 is i .L e t i 1 be the current in R 1 and take it to be downward. According to the junction rule the current in the voltmeter is i i 1 and it is downward. We apply the loop rule to the le ft-handlooptoobta in E− iR 2 i 1 R 1 ir =0 . We apply the loop rule to the right-hand loop to obtain i 1 R 1 ( i i 1 ) R V =0 . The second equation yields i = R 1 + R V R V i 1 . We substitute this into the Frst equation to obtain E− ( R 2 + r )( R 1 + R V ) R V i 1 + R 1 i 1 =0 . This has the solution i 1 = E R V ( R 2 + r )( R 1 + R V )+ R 1 R V . The reading on the voltmeter is i 1 R 1 = E R V R 1 ( R 2 + r )( R 1 + R V )+ R 1 R V
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Unformatted text preview: . × 10 3 Ω)(250 Ω) (300 Ω + 100 Ω)(250 Ω + 5 . × 10 3 Ω) + (250 Ω)(5 . × 10 3 Ω) = 1 . 12 V . The current in the absence of the voltmeter can be obtained by taking the limit as R V becomes inFnitely large. Then i 1 R 1 = E R 1 R 1 + R 2 + r = (3 . 0 V)(250 Ω) 250 Ω + 300 Ω + 100 Ω = 1 . 15 V . The fractional error is (1 . 12 − 1 . 15) / (1 . 15) = − . 030 , or − 3 . 0%....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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