ch27-p096 - R 1 R 2 = 5, where R 1 = 100 Ω . We solve for...

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96. When connected in series, the rate at which electric energy dissipates is P s = ε 2 /( R 1 + R 2 ). When connected in parallel, the corresponding rate is P p = 2 ( R 1 + R 2 )/ R 1 R 2 . Letting P p / P s = 5, we get ( R 1 + R 2 ) 2 /
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Unformatted text preview: R 1 R 2 = 5, where R 1 = 100 Ω . We solve for R 2 : R 2 = 38 Ω or 260 Ω . (a) Thus, the smaller value of R 2 is 38 Ω. (b) The larger value of R 2 is 260 Ω....
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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