812_Physics ProblemsTechnical Physics

# 812_Physics ProblemsTechnical Physics - 152*P28.62 Direct...

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152 Direct Current Circuits *P28.62 (a) ∆∆ VV I R I R II I I IR R I RR R I IR I R IR I 12 1 1 2 2 121 11 2 1 21 2 1 2 2 2 1 2 == =+=+ = + = + + = I R 1 R 2 I 1 I 2 FIG. P28.62(a) (b) The power delivered to the pair is P IR IR I I R 1 2 2 2 2 2 bg . For minimum power we want to find I 1 such that d dI 1 0 = . d dI I I R 1 1 2 22 1 0 =+ = a f 2 0 −+ = I IR 1 2 = + This is the same condition as that found in part (a). P28.63 Let R m = measured value, R = actual value, I R = current through the resistor R I = current measured by the ammeter. (a) When using circuit (a), V 20 000 or R I I R =− L N M O Q P 20 000 1 . But since I V R m = and I V R R = , we have I I R R Rm = (a) (b) FIG. P28.63 and R R m m = 20 000 . (1) When
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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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