811_Physics ProblemsTechnical Physics

811_Physics ProblemsTechnical Physics - Chapter 28 P28.59...

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Chapter 28 151 P28.59 Let the two resistances be x and y. Then, Rxy I s s =+= = = P 22 225 900 W 5.00 A a f . yx =− . and R xy xy I p p = + == = 50 0 200 . . W 5.00 A af so xx . . . +− = a f 2 180 0 −+ = .. . Factoring the second equation, −− = 600 300 0 so x = or x = . Then, gives y = or y = . The two resistances are found to be and . x y x y FIG. P28.59 P28.60 Let the two resistances be x and y. Then, I s s =+= 2 and R xy I p p = + = 2 . From the first equation, y I x s 2 , and the second becomes xIx xI x I s s p 2 2 2 = ej or x I x I s sp 2 24 0 F H G I K J += PP . Using the quadratic formula, x I ss s p = ±− 2 2 4 2 . Then, y I x s 2 gives y I s p = 2 2 4 2 . The two resistances are s p I 2 2 4 2 and s p I 2 2 4 2 . x y x y FIG. P28.60 P28.61 (a) εε ε + = IR ch bg 12 0 40 0 4 00 2 00 0 300 0 300 6 00 6 00 0 . . . . . V A V +++−+ = a f a f R ;s o R = 440 (b)
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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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