15_Ch 18 College Physics ProblemCH18 Direct-Current Circuits

15_Ch 18 College Physics ProblemCH18 Direct-Current...

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Direct-Current Circuits 113 18.17 We name the currents as shown. Using Kirchhoff’s loop rule on the rightmost loop gives 12 3 , , and II I ( ) () 3 2 5.00 1.00 I I −+ 12.0 V- 1.00+3.00 4.00 V 0 + = ( or ) 32 4.00 V = 2.00 3.00 + ( (1) Applying the loop rule to the leftmost loop yields ) ( ) 21 4.00 V+ 1.00 0 ( +5.00 8.00 +− = or ) ( ) 2.00 V = 4.00 3.00 123 += I (2) From Kirchhoff’s junction rule, (3) Solving equations (1), (2) and (3) simultaneously gives 3 =0.846 A, 1.31 A I = =0.462 A, and All currents are in the directions indicated by the arrows in the circuit diagram. ±²³³´ ± ± µ µ¶²³ · ¸²³³ · µ²³³ ± ± ± ± ¹²³³ ± µ²³³ ± º²³³ ± 18.18 Observe that the center branch of this circuit, that is the branch containing points a and b , is not a continuous conducting path, so no current can flow in this branch. The only
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