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Unformatted text preview: Reilly (mar3978) – oldhomework 11 – turner – (56725) 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 3 . 5 μ C when charged by a potential differ ence of 2 . 22 V . a) Find its capacitance. Correct answer: 1 . 57658 × 10 − 6 F. Explanation: Let : Q = 3 . 5 μ C and Δ V 1 = 2 . 22 V . The capacitance is given by C = Q Δ V = 3 . 5 × 10 − 6 C 2 . 22 V = 1 . 57658 × 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 . 45 V battery? Correct answer: 1 . 65738 × 10 − 6 J. Explanation: Let : Δ V 2 = 1 . 45 V U electric = 1 2 C (Δ V ) 2 = 1 2 (1 . 57658 × 10 − 6 F) (1 . 45 V) 2 = 1 . 65738 × 10 − 6 J 003 (part 1 of 7) 10.0 points A capacitor network is shown in the following figure. 11 V 4 . 96 μ F 9 . 7 μ F 11 . 9 μ F a b What is effective capacitance C ab of the entire capacitor network? Correct answer: 15 . 1819 μ F. Explanation: Let : C 1 = 4 . 96 μ F , C 2 = 9 . 7 μ F , C 3 = 11 . 9 μ F , and E B = 11 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 = (4 . 96 μ F) (9 . 7 μ F) 4 . 96 μ F + 9 . 7 μ F + 11 . 9 μ F = 15 . 1819 μ F . 004 (part 2 of 7) 10.0 points What is the voltage across the 9 . 7 μ F upper righthand capacitor? Correct answer: 3 . 72169 V. Explanation: Since C 1 and C 2 are in series they carry the same charge C 1 V 1 = C 2 V 2 , Reilly (mar3978) – oldhomework 11 – turner – (56725) 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 = (11 V)(4 . 96 μ F) 4 . 96 μ F + 9 . 7 μ F = 3 . 72169 V . 005 (part 3 of 7) 10.0 points If a dielectric of constant 2 . 65 is inserted in the 9 . 7 μ F top righthand capacitor (when the battery is connected), what is the electric potential across the 4 . 96 μ F top lefthand capacitor? Correct answer: 9 . 22077 V. Explanation: Let : κ = 2 . 65 . When the dielectric is inserted, the capaci tance formerly C 2 becomes C ′ 2 = κC 2 , and the new voltage across C 1 is V ′ 1 = V C ′ 2 C 1 + C ′ 2 = κV C 2 C 1 + κC 2 = (2 . 65)(11 V)(9 . 7 μ F) 4 . 96 μ F + (2 . 65)(9 . 7 μ F) = 9 . 22077 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 4 . 96 μ F top lefthand capacitor? Correct answer: 38 . 1831 μ C. Explanation: Immediately before the battery was discon nected the charges on the capacitors had been Q ′ 3 = C 3 V = (11 . 9 μ F)(11 V) = 130 . 9 μ C Q ′ 1 = Q ′ 2 = C ′ 12 V = (4 . 15773 μ F)(11 V) = 45 . 735 μ C ....
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This note was uploaded on 12/01/2010 for the course PHY 60230 taught by Professor Turner,j during the Spring '10 term at University of Texas.
 Spring '10
 TURNER,J

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