35_Ch 18 College Physics ProblemCH18 Direct-Current Circuits

35_Ch 18 College Physics ProblemCH18 Direct-Current...

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Direct-Current Circuits 133 Substitution of equation (4) into (3) yields 1 1 140 310 450 I I += or 2 11 310 450 140 0 II Solving this quadratic equation gives two possible values for the current . These are . Then, from 1 I 1 A and 0.452 = 1 1.0 A = 2 1 20 W = R I , we find two possible values for the resistance R . These are 20 or 98 RR =Ω = 18.57 When connected in series, the equivalent resistance is 12 eq n RRR Rn R = ++ . Thus, the current is ( ) ( ) se q IV R V
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