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hw15-solutions

# hw15-solutions - marout(ssm979 hw15 turner(56725 This...

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marout (ssm979) – hw15 – turner – (56725) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points See the circuit below. 36 V 4 V X Y 1 . 9 Ω 1 . 6 Ω 1 . 9 Ω R 4 A Find the resistance R . Correct answer: 2 . 6 Ω. Explanation: E 1 E 2 X Y R 1 R 3 R 2 R I Let : E 1 = 36 V , E 2 = 4 V , R 1 = 1 . 9 Ω , R 2 = 1 . 9 Ω , R 3 = 1 . 6 Ω , and I = 4 A . From Ohm’s law, the total resistance of the circuit is R total = V I = E 1 − E 2 I = (36 V) (4 V) (4 A) = 8 Ω . Therefore the resistance R is R = R total R 1 R 2 R 3 = (8 Ω) (1 . 9 Ω) (1 . 9 Ω) (1 . 6 Ω) = 2 . 6 Ω . 002 (part 2 of 3) 10.0 points Find the potential difference V X Y = V Y V X between points X and Y . Correct answer: 20 . 8 V. Explanation: The current in the circuit goes counter- clockwise, so the potential difference between Y and X is V X Y = E 2 + R 3 I + R I = (4 V) + (1 . 6 Ω + 2 . 6 Ω) (4 A) = 20 . 8 V , or = E 1 R 1 I R 2 I = (36 V) (1 . 9 Ω + 1 . 9 Ω) (4 A) = 20 . 8 V . 003 (part 3 of 3) 10.0 points How much energy U E is dissipated by the 1 . 9 Ω resistor in 32 s ? Correct answer: 972 . 8 J. Explanation: Let : t = 32 s . The work done is W = P t = V I t = I 2 R t , so the energy dissipated is W = I 2 R 2 t = (4 A) 2 (1 . 9 Ω) (32 s) = 972 . 8 J . 004 (part 1 of 2) 10.0 points In the figure below consider the case where switch S 1 is closed and switch S 2 is open. 19 μ F 29 μ F 39 μ F 40 μ F 57 V S 2 S 1 a b c d

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marout (ssm979) – hw15 – turner – (56725) 2 Find the charge on the 19 μ F upper-left capacitor between points a and c . Correct answer: 728 . 224 μ C. Explanation: Let : C 1 = 19 μ F , C 2 = 29 μ F , C 3 = 39 μ F , C 4 = 40 μ F , and E B = 57 V . C 1 C 2 C 3 C 4 E B S 2 S 1 a b c d Redrawing the figure, we have E B C 1 C 2 C 3 C 4 b a c d C 1 and C 3 are in series, so C 13 = parenleftbigg 1 C 1 + 1 C 3 parenrightbigg 1 = C 1 C 3 C 1 + C 3 = (19 μ F) (39 μ F) 19 μ F + 39 μ F = 12 . 7759 μ F . C 2 and C 4 are in series, so C 24 = parenleftbigg 1 C 2 + 1 C 4 parenrightbigg 1 = C 2 C 4 C 2 + C 4 = (29 μ F) (40 μ F) 29 μ F + 40 μ F = 16 . 8116 μ F . Simplifying the circuit, we have E B C 13 C 24 a b C 13 and C 24 are parallel, so C ab = C 13 + C 24 = 12 . 7759 μ F + 16 . 8116 μ F = 29 . 5875 μ F . E B C ab a b C 1 and C 3 are in series, so Q 1 = Q 3 = Q 13 = C 13 E B = (12 . 7759 μ F) (57 V) = 728 . 224 μ C . C 2 and C 4 are in series, so Q 2 = Q 4 = Q l = C 24 E B = (16 . 8116 μ F) (57 V) = 958 . 261 μ C . C 1 and C 3 are in series, so Q 3 = Q 1 = 728 . 224 μ C . C 2 and C 4 are in series, so Q 4 = Q 2 = 958 . 261 μ C . 005 (part 2 of 2) 10.0 points
marout (ssm979) – hw15 – turner – (56725) 3 Now consider the case where switch S 2 is also closed. 19 μ F 29 μ F 39 μ F 40 μ F 57 V S 2 S 1 a b c d Find the charge on the 19 μ F upper-left capacitor between points a and c . Correct answer: 673 . 677 μ C. Explanation: A good rule of thumb is to eliminate junc- tions connected by zero capacitance. Switch S 2 is closed. E B C 1 C 2 C 3 C 4 b a c d When S 2 is closed, C 1 and C 2 are parallel, so C 12 = C 1 + C 2 = 19 μ F + 29 μ F = 48 μ F .

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