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Unformatted text preview: wilson (kw22588) – HW8 – Mackie – (20224) 1 This printout should have 24 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. warning: this includes capacitors with di electrics, which is material from before break. 001 (part 1 of 7) 10.0 points A capacitor network is shown in the following figure. 19 . 5 V 5 . 6 μ F 4 . 3 μ F 13 . 3 μ F a b What is effective capacitance C ab of the entire capacitor network? Correct answer: 15 . 7323 μ F. Explanation: Let : C 1 = 5 . 6 μ F , C 2 = 4 . 3 μ F , C 3 = 13 . 3 μ F , and E B = 19 . 5 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 . 6 μ F) (4 . 3 μ F) 5 . 6 μ F + 4 . 3 μ F + 13 . 3 μ F = 15 . 7323 μ F . 002 (part 2 of 7) 10.0 points What is the voltage across the 4 . 3 μ F upper righthand capacitor? Correct answer: 11 . 0303 V. Explanation: Since C 1 and C 2 are in series they carry the same charge C 1 V 1 = C 2 V 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 . 5 V)(5 . 6 μ F) 5 . 6 μ F + 4 . 3 μ F = 11 . 0303 V . 003 (part 3 of 7) 10.0 points If a dielectric of constant 2 . 39 is inserted in the 4 . 3 μ F top righthand capacitor (when the battery is connected), what is the elec tric potential across the 5 . 6 μ F top lefthand capacitor? Correct answer: 12 . 6221 V. Explanation: Let : κ = 2 . 39 . 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 . 39)(19 . 5 V)(4 . 3 μ F) 5 . 6 μ F + (2 . 39)(4 . 3 μ F) = 12 . 6221 V . wilson (kw22588) – HW8 – Mackie – (20224) 2 004 (part 4 of 7) 10.0 points If the battery is disconnected and then the dielectric is removed, what is the charge on 5 . 6 μ F top lefthand capacitor? Correct answer: 51 . 0255 μ C. Explanation: Immediately before the battery was discon nected the charges on the capacitors had been Q ′ 3 = C 3 V = (13 . 3 μ F)(19 . 5 V) = 259 . 35 μ C Q ′ 1 = Q ′ 2 = C ′ 12 V = (3 . 62482 μ F)(19 . 5 V) = 70 . 6839 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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This note was uploaded on 02/17/2011 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

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