38_Ch 18 College Physics ProblemCH18 Direct-Current Circuits

38_Ch 18 College Physics ProblemCH18 Direct-Current...

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Unformatted text preview: 136 CHAPTER 18 18.61 (a) Using the rules for combining resistors in series and parallel, the circuit reduces as shown below: 10.0 W 25.0 V –+ 10.0 W 25.0 V –+ I a 10.0 W 5.00 W 5.00 W 10.0 W 25.0 V –+ I b I1 10.0 W a I b 5.00 W 20.0 W I1 a 25.0 W Step 2 Step 1 b 2.94 W Step 3 From the figure of Step 3, observe that I= 25.0 V = 1.93 A 12.94 A ∆Vab = I ( 2.94 Ω ) = ( 1.93 A ) ( 2.94 Ω ) = 5.68 V and (b) From the figure of Step 1, observe that I1 = 18.62 ∆Vab 5.68 V = = 0.227 A 25.0 Ω 25.0 Ω (a) When the power supply is connected to points A and B, the circuit reduces as shown below to an equivalent resistance of Req = 0.099 9 Ω . 100 W R2 + R3 + 100 W = 111 W Req = 0.0999 W A R1 B R2 C R3 D A R1 = 0.100 W B I1 +– +– +– 5.00 V 5.00 V From the center figure above, observe that I R1 = I1 = and I R2 = I R3 = I100 = 5.00 V 5.00 V = 50.0 A 0.100 Ω 5.00 V = 0.045 0 A = 45.0 mA 111 Ω ...
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