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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, + = I I I R g g p 75 0 0 . a f e j . Therefore, if I = 1 00 . A when I g = 1 50 . mA , R I I I p g g = = × × = 75 0 1 50 10 75 0 1 00 1 50 10 0 113 3 3 . . . . . . A A A a f e j e j a f . FIG. P28.42 P28.43 Series Resistor Voltmeter V IR = : 25 0 1 50 10 75 0 3 . . . = × + R s b g 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 = = 6 000 0 200 00 0 030 000 . . . 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 V IR = = = 0 030 000 180 00 5 400 0 . . . A V b ga f . (b) In Figure (b), R eq = + F H G I K J = 1 180 00 1 20 000 178 39 1 . . . The equivalent resistance across the emf is 178 39 0 500 00 20 000 198 89 . . . . + + = . The ammeter reads I R = = = ε 6 000 0 0 030 167 . . V 198.89 A and the voltmeter reads V IR
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