oldhomework 11-solutions

oldhomework 11-solutions - marout(ssm979 oldhomework 11...

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marout (ssm979) – oldhomework 11 – turner – (56725) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A parallel-plate capacitor has a charge of 7 . 2 μ C when charged by a potential differ- ence of 1 . 58 V . a) Find its capacitance. Correct answer: 4 . 55696 × 10 6 F. Explanation: Let : Q = 7 . 2 μ C and Δ V 1 = 1 . 58 V . The capacitance is given by C = Q Δ V = 7 . 2 × 10 6 C 1 . 58 V = 4 . 55696 × 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 . 47 V battery? Correct answer: 4 . 92357 × 10 6 J. Explanation: Let : Δ V 2 = 1 . 47 V U electric = 1 2 C V ) 2 = 1 2 (4 . 55696 × 10 6 F) (1 . 47 V) 2 = 4 . 92357 × 10 6 J 003 (part 1 of 7) 10.0 points A capacitor network is shown in the following figure. 10 . 8 V 3 . 03 μ F 6 μ F 13 . 5 μ F a b What is effective capacitance C ab of the entire capacitor network? Correct answer: 15 . 5133 μ F. Explanation: Let : C 1 = 3 . 03 μ F , C 2 = 6 μ F , C 3 = 13 . 5 μ F , and E B = 10 . 8 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 = (3 . 03 μ F) (6 μ F) 3 . 03 μ F + 6 μ F + 13 . 5 μ F = 15 . 5133 μ F . 004 (part 2 of 7) 10.0 points What is the voltage across the 6 μ F upper right-hand capacitor? Correct answer: 3 . 62392 V. Explanation: Since C 1 and C 2 are in series they carry the same charge C 1 V 1 = C 2 V 2 ,
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marout (ssm979) – 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 = (10 . 8 V)(3 . 03 μ F) 3 . 03 μ F + 6 μ F = 3 . 62392 V . 005 (part 3 of 7) 10.0 points If a dielectric of constant 3 . 11 is inserted in the 6 μ F top right-hand capacitor (when the battery is connected), what is the electric potential across the 3 . 03 μ F top left-hand capacitor? Correct answer: 9 . 29129 V. Explanation: Let : κ = 3 . 11 . 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 = (3 . 11)(10 . 8 V)(6 μ F) 3 . 03 μ F + (3 . 11)(6 μ F) = 9 . 29129 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 3 . 03 μ F top left-hand capacitor? Correct answer: 22 . 5753 μ C. Explanation: Immediately before the battery was discon- nected the charges on the capacitors had been Q 3 = C 3 V = (13 . 5 μ F)(10 . 8 V) = 145 . 8 μ C Q 1 = Q 2 = C 12 V = (2 . 60672 μ F)(10 . 8 V) = 28 . 1526 μ C . When we remove the dielectric, the sum of the charges stays the same, and the voltages on C 3 and on C 12 (where C 12 is the equivalent capacitance of C 1 and C 2 in series) are equal to each other Q ′′ 1 + Q ′′ 3 = Q 1 + Q 3 . (1) Since the potential drop across the top and bottom are equal, we have V ab = Q ′′ 1 C 12 = Q ′′ 3 C 3 , so Q ′′ 3 = C 3 C 12 Q ′′ 1 .
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