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Unformatted text preview: howard (cah3459) – oldhomework 11 – turner – (56705) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A parallelplate capacitor has a charge of 5 . 1 μ C when charged by a potential differ ence of 2 . 89 V . a) Find its capacitance. Correct answer: 1 . 76471 × 10 6 F. Explanation: Let : Q = 5 . 1 μ C and Δ V 1 = 2 . 89 V . The capacitance is given by C = Q Δ V = 5 . 1 × 10 6 C 2 . 89 V = 1 . 76471 × 10 6 F . 002 (part 2 of 2) 10.0 points b) How much electrical potential energy is stored when this capacitor is connected to a 1 . 86 V battery? Correct answer: 3 . 05259 × 10 6 J. Explanation: Let : Δ V 2 = 1 . 86 V U electric = 1 2 C (Δ V ) 2 = 1 2 (1 . 76471 × 10 6 F) (1 . 86 V) 2 = 3 . 05259 × 10 6 J 003 (part 1 of 7) 10.0 points A capacitor network is shown in the following figure. 19 . 1 V 5 . 85 μ F 7 . 8 μ F 11 . 6 μ F a b What is effective capacitance C ab of the entire capacitor network? Correct answer: 14 . 9429 μ F. Explanation: Let : C 1 = 5 . 85 μ F , C 2 = 7 . 8 μ F , C 3 = 11 . 6 μ F , and E B = 19 . 1 V . E B C 1 C 2 C 3 a b C 1 and C 2 are in series with each other, and they are together are parallel with C 3 . So C ab = C 1 C 2 C 1 + C 2 + C 3 = (5 . 85 μ F) (7 . 8 μ F) 5 . 85 μ F + 7 . 8 μ F + 11 . 6 μ F = 14 . 9429 μ F . 004 (part 2 of 7) 10.0 points What is the voltage across the 7 . 8 μ F upper righthand capacitor? Correct answer: 8 . 18571 V. Explanation: Since C 1 and C 2 are in series they carry the same charge C 1 V 1 = C 2 V 2 , howard (cah3459) – oldhomework 11 – turner – (56705) 2 and their voltages add up to V , voltage of the battery V 1 + V 2 = V C 2 V 2 C 1 + V 2 = V C 2 V 2 + C 1 V 2 = V C 1 V 2 = V C 1 C 1 + C 2 = (19 . 1 V)(5 . 85 μ F) 5 . 85 μ F + 7 . 8 μ F = 8 . 18571 V . 005 (part 3 of 7) 10.0 points If a dielectric of constant 4 . 93 is inserted in the 7 . 8 μ F top righthand capacitor (when the battery is connected), what is the electric potential across the 5 . 85 μ F top lefthand capacitor? Correct answer: 16 . 578 V. Explanation: Let : κ = 4 . 93 . When the dielectric is inserted, the capaci tance formerly C 2 becomes C 2 = κ C 2 , and the new voltage across C 1 is V 1 = V C 2 C 1 + C 2 = κ V C 2 C 1 + κ C 2 = (4 . 93)(19 . 1 V)(7 . 8 μ F) 5 . 85 μ F + (4 . 93)(7 . 8 μ F) = 16 . 578 V . 006 (part 4 of 7) 10.0 points If the battery is disconnected and then the dielectric is removed, what is the charge on 5 . 85 μ F top lefthand capacitor? Correct answer: 71 . 2607 μ C. Explanation: Immediately before the battery was discon nected the charges on the capacitors had been Q 3 = C 3 V = (11 . 6 μ F)(19 . 1 V) = 221 . 56 μ C Q 1 = Q 2 = C 12 V = (5 . 07755 μ F)(19 . 1 V) = 96 . 9813 μ C ....
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This note was uploaded on 10/11/2010 for the course PHY 56715 taught by Professor Turner during the Spring '10 term at University of Texas.
 Spring '10
 Turner

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