797_Physics ProblemsTechnical Physics

797_Physics ProblemsTechnical Physics - Chapter 28 P28.15...

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Chapter 28 137 P28.15 R R I V R p s s = + F H G I K J = = + + = = = = 1 3 00 1 1 00 0 750 2 00 0 750 4 00 6 75 18 0 2 67 1 . . . . . . . . . V 6.75 A battery a f P = I R 2 : P 2 2 2 67 2 00 = . . A a f a f P 2 14 2 = . W in 2.00 P P P 4 2 2 4 2 4 3 1 3 3 2 3 2 1 1 1 2 2 67 4 00 28 4 2 67 2 00 5 33 2 67 4 00 10 67 18 0 2 00 2 00 3 00 1 33 2 00 1 00 4 00 = = = = = = = = = = = = = = = = . . . . . . . . . . . . . . . . . A A W in 4.00 A V, A V V V V W in 3.00 V W in 1.00 a f a f a fa f a fa f b g b g a f b g a f V V V V V V V V R V R p FIG. P28.15 P28.16 Denoting the two resistors as x and y , x y + = 690, and 1 150 1 1 = + x y 1 150 1 1 690 690 690 690 103 500 0 690 690 414 000 2 470 220 2 2 = + = + + = = ± = = x x x x x x x x x x y a f a f a f *P28.17 A certain quantity of energy E int time = P a f is required to raise the temperature of the water to 100 ° C. For the power delivered to the heaters we have P = = I V V R a f 2 where V a f is a constant.
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