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796_Physics ProblemsTechnical Physics

# 796_Physics ProblemsTechnical Physics - 136 P28.11 Direct...

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136 Direct Current Circuits P28.11 (a) Since all the current in the circuit must pass through the series 100 resistor, P = I R 2 P max max = RI 2 so I R max . . = = = P 25 0 0 500 W 100 A R V R I eq eq = + + F H G I K J = = = 100 1 100 1 100 150 75 0 1 V max max . (b) P = = = I V 0 500 75 0 37 5 . . . A V W a fa f total power P P P 1 2 3 2 2 25 0 100 0 250 6 25 = = = = . . . W A W RI a fa f FIG. P28.11 P28.12 Using 2.00- , 3.00- , 4.00- resistors, there are 7 series, 4 parallel, and 6 mixed combinations: Series 2.00 6.00 3.00 7.00 4.00 9.00 5.00 Parallel Mixed 0.923 1.56 1.20 2.00 1.33 2.22 1.71 3.71 4.33 5.20 The resistors may be arranged in patterns: P28.13 The potential difference is the same across either combination. V IR I R = = + 3 1 1 1 500 c h so R R 1 1 500 3 + F H G I K J = 1 500 3 + = R and R = = 1 000 1 00 k . . FIG. P28.13 *P28.14 When S is open ,
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