Spring 11 Phys 121 - HW 08

# Spring 11 Phys 121 - HW 08 - dzifanu(sd26397 hw 8...

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dzifanu (sd26397) – hw 8 – opyrchal – (121102) 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 34 . 1 Ω 4 . 2 Ω 25 Ω 38 . 9 V 19 . 45 V Find the current through the 25 Ω (lower) resistor. Correct answer: 0 . 750986 A. Explanation: r 1 r 2 R E 1 E 2 I 1 I 2 I 3 Let : E 1 = 38 . 9 V , E 1 = 19 . 45 V , r 1 = 34 . 1 Ω , r 2 = 4 . 2 Ω , and R = 25 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 ) = (38 . 9 V) (4 . 2 Ω) + (19 . 45 V) (34 . 1 Ω) (4 . 2 Ω) (25 Ω) + (34 . 1 Ω) (25 Ω + 4 . 2 Ω) = 0 . 750986 A . 002(part2of3)10.0points Determine the current in the 34 . 1 Ω (upper) resistor. Correct answer: 0 . 590186 A. Explanation: From (1), get I 1 = E 1 - I 3 R r 1 = 38 . 9 V - (0 . 750986 A) (25 Ω) 34 . 1 Ω = 0 . 590186 A . 003(part3of3)10.0points Determine the current in the 4 . 2 Ω (middle) resistor. Correct answer: 0 . 160799 A. Explanation: From (2), get I 2 = E 2 - I 3 R r 2 = 19 . 45 V - (0 . 750986 A) (25 Ω) 4 . 2 Ω = 0 . 160799 A .

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dzifanu (sd26397) – hw 8 – opyrchal – (121102) 2 004 10.0points In the figure below the battery has an emf of 19 V and an internal resistance of 1 Ω . Assume there is a steady current flowing in the circuit. 3 μ F 7 Ω 3 Ω 1 Ω 19 V Find the charge on the 3 μ F capacitor. Correct answer: 15 . 5455 μ C. Explanation: Let : R 1 = 7 Ω , R 2 = 3 Ω , r in = 1 Ω , V = 19 V , and C = 3 μ F . The equivalent resistance of the three resistors in series is R eq = R 1 + R 2 + r in = 7 Ω + 3 Ω + 1 Ω = 11 Ω , so the current in the circuit is I = V R eq , and the voltage across R 2 is V 2 = I R 2 = R 2 R eq V = 3 Ω 11 Ω (19 V) = 5 . 18182 V . Since R 2 and C are parallel, the potential difference across each is the same, and the charge on the capacitor is Q = C V 2 = (3 μ F) (5 . 18182 V) = 15 . 5455 μ C .
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