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Physhw 8-solutions (1)

# Physhw 8-solutions (1) - billing(cab4763 hw 8...

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billing (cab4763) – hw 8 – opyrchal – (121104) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of3)10.0points 22 . 9 Ω 4 . 29 Ω 18 . 1 Ω 27 . 9 V 13 . 95 V Find the current through the 18 . 1 Ω (lower) resistor. Correct answer: 0 . 743836 A. Explanation: r 1 r 2 R E 1 E 2 I 1 I 2 I 3 Let : E 1 = 27 . 9 V , E 1 = 13 . 95 V , r 1 = 22 . 9 Ω , r 2 = 4 . 29 Ω , and R = 18 . 1 Ω . Assuming currents I 1 , I 2 , and I 3 in the direction show, we get I 3 = I 1 + I 2 . Applying Kirchhoff’s loop rule, we can get two equations. E 1 = I 1 r 1 + I 3 R (1) E 2 = I 2 r 2 + I 3 R = ( I 3 - I 1 ) r 2 + I 3 R = - I 1 r 2 + I 3 ( R + r 2 ) , (2) Multiplying Eq. (1) by r 2 , Eq. (2) by r 1 , E 1 r 2 = I 1 r 1 r 2 + r 2 I 3 R E 2 r 1 = - I 1 r 1 r 2 + I 3 r 1 ( R + r 2 ) Adding, E 1 r 2 + E 2 r 1 = I 3 [ r 2 R + r 1 ( R + r 2 )] I 3 = E 1 r 2 + E 2 r 1 r 2 R + r 1 ( R + r 2 ) = (27 . 9 V) (4 . 29 Ω) + (13 . 95 V) (22 . 9 Ω) (4 . 29 Ω) (18 . 1 Ω) + (22 . 9 Ω) (18 . 1 Ω + 4 . 29 Ω) = 0 . 743836 A . 002(part2of3)10.0points Determine the current in the 22 . 9 Ω (upper) resistor. Correct answer: 0 . 630418 A. Explanation: From (1), get I 1 = E 1 - I 3 R r 1 = 27 . 9 V - (0 . 743836 A) (18 . 1 Ω) 22 . 9 Ω = 0 . 630418 A . 003(part3of3)10.0points Determine the current in the 4 . 29 Ω (middle) resistor. Correct answer: 0 . 113419 A. Explanation: From (2), get I 2 = E 2 - I 3 R r 2 = 13 . 95 V - (0 . 743836 A) (18 . 1 Ω) 4 . 29 Ω = 0 . 113419 A .

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billing (cab4763) – hw 8 – opyrchal – (121104) 2 004(part1of2)10.0points The circuit has been connected as shown in the figure for a “long” time. 12 V S 11 μ F 20 Ω 28 Ω 4 Ω 44 Ω What is the magnitude of the electric po- tential across the capacitor? Correct answer: 4 V. Explanation: E S 1 C t b a b I t R 1 I t R 2 I b R 3 I b R 4 Let : R 1 = 20 Ω , R 2 = 28 Ω , R 3 = 4 Ω , R 4 = 44 Ω , and C = 11 μ F . After a “long time” implies that the ca- pacitor C is fully charged and therefore the capacitor acts as an open circuit with no cur- rent flowing to it. The equivalent circuit is I t R 1 I t R 2 R 3 I b I b R 4 a b R t = R 1 + R 2 = 20 Ω + 28 Ω = 48 Ω R b = R 3 + R 4 = 4 Ω + 44 Ω = 48 Ω I t = E R t = 12 V 48 Ω = 0 . 25 A I b = E R b = 12 V 48 Ω = 0 . 25 A Across R 1 E 1 = I t R 1 = (0 . 25 A) (20 Ω) = 5 V .
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Physhw 8-solutions (1) - billing(cab4763 hw 8...

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