39_Ch 18 College Physics ProblemCH18 Direct-Current Circuits

39_Ch 18 College Physics ProblemCH18 Direct-Current...

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Unformatted text preview: Direct-Current Circuits 137 (b) When the power supply is connected to points A and C, the circuit reduces as shown below to an equivalent resistance of Req = 1.09 Ω . 100 W R2 + 100 W = 110 W Req = 1.09 W A R1 B R2 C R3 D A R1 + R2 = 1.10 W C I1 +– +– 5.00 V +– 5.00 V 5.00 V From the center figure above, observe that I R1 = I R2 = I1 = I R3 = I100 = and 5.00 V = 4.55 A 1.10 Ω 5.00 V = 0.045 5 A = 45.5 mA 110 Ω (c) When the power supply is connected to points A and D, the circuit reduces as shown below to an equivalent resistance of Req = 9.99 Ω . 100 W 100 W Req = 9.99 W A R1 B R2 C R3 D A R1 + R2 + R3 = 11.1 W D I1 +– +– 5.00 V +– 5.00 V 5.00 V From the center figure above, observe that I R1 = I R2 = I R3 = I1 = and 18.63 I100 = 5.00 V = 0.450 A 11.1 Ω 5.00 V = 0.050 0 A = 50.0 mA 100 Ω In the circuit diagram at the right, note that all points labeled a are at the same potential and equivalent to each other. Also, all points labeled c are equivalent. To determine the voltmeter reading, go from point e to point d along the path ecd, keeping track of all changes in potential to find: ∆Ved = Vd − Ve = −4.50 V + 6.00 V = + 1.50 V a a a a I b A 10.0 W I2 5.00 W d V e 5.00 W I1 I3 6.00 W f c 6.00 V +– c +– c 4.50 V c ...
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