809_Physics ProblemsTechnical Physics

# 809_Physics ProblemsTechnical Physics - 149 Chapter 28...

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Chapter 28 149 P28.53 I Rr = + ε , so P == + IR R 2 2 2 af or R += F H G I K J 2 2 . Let x 2 , then x R a f 2 or x R r 22 20 +− = . With r = 120 . , this becomes Rx R 2 240 144 0 = .. a f , which has solutions of R xx = −− ± 576 2 2 . a f a f . (a) With = 920 V and = 12 8 . W, x = 661 . : R = = 421 2 384 2 ... . or 0375 . (b) For = V and = 21 2 . W , x ≡= 2 399 . R = = ±− 159 2 322 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 , and that maximum is: max . 2 4 17 6 r W. P28.54 Using Kirchhoff’s loop rule for the closed loop, + −−= 12 0 2 00 4 00 0 . II , so I = 200 A VV ba −= + = 4 00 2 00 4 00 0 10 0 4 00 . V A V a f ΩΩ . Thus, V ab = 400 . V and point is at the higher potential a . P28.55 (a) RR eq = 3 I R = 3 series I R 2 3 (b) R RRR R eq = ++ = 1 111 3 bg I R = 3 parallel I R 3 2 (c) Nine times more power is converted in the parallel connection. *P28.56 (a) We model the generator as a constant-voltage power supply.
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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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