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809_Physics ProblemsTechnical Physics

809_Physics ProblemsTechnical Physics - 149 Chapter 28...

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Chapter 28 149 P28.53 I R r = + ε , so P = = + I R R R r 2 2 2 ε a f or R r R + = F H G I K J a f 2 2 ε P . Let x ε 2 P , then R r xR + = a f 2 or R r x R r 2 2 2 0 + = a f . With r = 1 20 . , this becomes R x R 2 2 40 1 44 0 + = . . a f , which has solutions of R x x = ± 2 40 2 40 5 76 2 2 . . . a f a f . (a) With ε = 9 20 . V and P = 12 8 . W, x = 6 61 . : R = + ± = 4 21 4 21 5 76 2 3 84 2 . . . . a f or 0 375 . . (b) For ε = 9 20 . V and P = 21 2 . W, x = ε 2 3 99 P . R = + ± = ± 1 59 1 59 5 76 2 1 59 3 22 2 2 . . . . . a f . The equation for the load resistance yields a complex number, so there is no resistance that will extract 21.2 W from this battery. The maximum power output occurs when R r = = 1 20 . , and that maximum is: P max . = = ε 2 4 17 6 r W . P28.54 Using Kirchhoff’s loop rule for the closed loop, + = 12 0 2 00 4 00 0 . . . I I , so I = 2 00 . A V V b a = + = − 4 00 2 00 4 00 0 10 0 4 00 . . . . . V A V a fa f a fa f . Thus, V ab = 4 00 . V and point is at the higher potential a . P28.55 (a) R R eq = 3 I R = ε 3 P series = = ε ε I R 2 3 (b) R R R R R eq = + + = 1 1 1 1 3 b g b g b g I R = 3 ε P parallel
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