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Unformatted text preview: saldana (avs387) – hw15 – turner – (56705) 1 This printout should have 16 questions. Multiplechoice 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. 121 V 9 V X Y 3 . 2 Ω 3 . 9 Ω 3 . 6 Ω R 8 A Find the resistance R . Correct answer: 3 . 3 Ω. Explanation: E 1 E 2 X Y R 1 R 3 R 2 R I Let : E 1 = 121 V , E 2 = 9 V , R 1 = 3 . 2 Ω , R 2 = 3 . 6 Ω , R 3 = 3 . 9 Ω , and I = 8 A . From Ohm’s law, the total resistance of the circuit is R total = V I = E 1 − E 2 I = (121 V) − (9 V) (8 A) = 14 Ω . Therefore the resistance R is R = R total − R 1 − R 2 − R 3 = (14 Ω) − (3 . 2 Ω) − (3 . 6 Ω) − (3 . 9 Ω) = 3 . 3 Ω . 002 (part 2 of 3) 10.0 points Find the potential difference V XY = V Y − V X between points X and Y . Correct answer: 66 . 6 V. Explanation: The current in the circuit goes counter clockwise, so the potential difference between Y and X is V XY = E 2 + R 3 I + RI = (9 V) + (3 . 9 Ω + 3 . 3 Ω) (8 A) = 66 . 6 V , or = E 1 − R 1 I − R 2 I = (121 V) − (3 . 2 Ω + 3 . 6 Ω) (8 A) = 66 . 6 V . 003 (part 3 of 3) 10.0 points How much energy U E is dissipated by the 3 . 6 Ω resistor in 24 s? Correct answer: 5529 . 6 J. Explanation: Let : t = 24 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 = (8 A) 2 (3 . 6 Ω) (24 s) = 5529 . 6 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. 1 4 μ F 2 μ F 3 9 μ F 4 4 μ F 54 V S 2 S 1 a b c d saldana (avs387) – hw15 – turner – (56705) 2 Find the charge on the 14 μ F upperleft capacitor between points a and c . Correct answer: 556 . 302 μ C. Explanation: Let : C 1 = 14 μ F , C 2 = 20 μ F , C 3 = 39 μ F , C 4 = 44 μ F , and E B = 54 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 = (14 μ F) (39 μ F) 14 μ F + 39 μ F = 10 . 3019 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 = (20 μ F) (44 μ F) 20 μ F + 44 μ F = 13 . 75 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 = 10 . 3019 μ F + 13 . 75 μ F = 24 . 0519 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 = (10 . 3019 μ F) (54 V) = 556 . 302 μ C . C 2 and C 4 are in series, so Q 2 = Q 4 = Q l = C 24 E B = (13 . 75 μ F) (54 V) = 742 . 5 μ C . C 1 and C 3 are in series, so Q 3 = Q 1 = 556 . 302 μ C ....
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