806_Physics ProblemsTechnical Physics

806_Physics ProblemsTechnical Physics - 146 P28.42 Direct...

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146 Direct Current Circuits P28.42 Applying Kirchhoff’s loop rule, −+ = II I R gg p 75 0 0 . a f ej . Therefore, if I = 100 . A when I g = 150 m A , R I p g g = = × −× = 75 0 150 10 750 1 00 1 50 10 0113 3 3 . .. . A A a f af . FIG. P28.42 P28.43 Series Resistor Voltmeter VI R = :2 5 0 1 5 0 1 0 7 5 0 3 . + R s bg Solving, R s = 16 6 . k . FIG. P28.43 P28.44 (a) In Figure (a), the emf sees an equivalent resistance of 200 00 . I = = 60000 200 00 0030000 . . . V A 6.0000 V 20.000 180.00 180.00 180.00 A V A V (a) (b) (c) 20.000 20.000 FIG. P28.44 The terminal potential difference is ∆Ω R == = 0 030 000 180 00 5 400 0 . V a f . (b) In Figure (b), R eq =+ F H G I K J = 1 180 00 1 20 000 178 39 1 . . ΩΩ . The equivalent resistance across the emf is 178 39 0 500 00 20 000 198 89 . . ++ = . The ammeter reads I R = ε 0030167 . . V 198.89 A and the voltmeter reads R = 0 030 167 178 39 5 381 6 . V . (c) In Figure (c),
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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 .

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