12. Capacitors in circuits-solutions

12. Capacitors in circuits-solutions - gilvin(jg47854 –...

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Unformatted text preview: gilvin (jg47854) – 12. Capacitors in circuits – meyers – (21235) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This homwork is due Tuesday,March 22, at midnight Tucson time. 001 (part 1 of 3) 10.0 points Consider the group of capacitors shown in the figure. 11 V 4 . 78 μ F 8 . 59 μ F 1 . 15 μ F 6 . 25 μ F d a b c Find the equivalent capacitance between points a and d . Correct answer: 0 . 905502 μ F. Explanation: Let : C 1 = 1 . 15 μ F , C 2 = 4 . 78 μ F , C 3 = 8 . 59 μ F , C 4 = 6 . 25 μ F , and E B = 11 V . E B C 2 C 3 C 1 C 4 d a b c For capacitors in series, 1 C series = summationdisplay 1 C i V series = summationdisplay V i , and the individual charges are the same. For parallel capacitors, C parallel = summationdisplay C i Q parallel = summationdisplay Q i , and the individual voltages are the same. C 2 and C 3 are connected parallel with equivalent capacitance, so C 23 = C 2 + C 3 = 4 . 78 μ F + 8 . 59 μ F = 13 . 37 μ F . C 1 , C 23 , and C 4 are connected in series with equivalent capacitance C ba = parenleftbigg 1 C 1 + 1 C 23 + 1 C 4 parenrightbigg − 1 = parenleftbigg 1 1 . 15 μ F + 1 13 . 37 μ F + 1 6 . 25 μ F parenrightbigg − 1 = . 905502 μ F . 002 (part 2 of 3) 10.0 points Determine the charge on the 1 . 15 μ F capacitor on the left-hand side of the circuit. Correct answer: 9 . 96052 μ C. Explanation: Capacitors in series have the same charge on them. Using this together with the defini- tion of capacitance, we see that Q 1 = Q 4 = C ba V total = (0 . 905502 μ F) (11 V) = 9 . 96052 μ C . 003 (part 3 of 3) 10.0 points Determine the charge on the 4 . 78 μ F capacitor at the top center part of the circuit. Correct answer: 3 . 56106 μ C. Explanation: The voltage drops across C 1 and C 4 are V 1 = Q 1 C 1 = 9 . 96052 μ C 1 . 15 μ F = 8 . 66133 V gilvin (jg47854) – 12. Capacitors in circuits – meyers – (21235) 2 and V 4 = Q 4 C 4 = 9 . 96052 μ C 6 . 25 μ F = 1 . 59368 V . The remaining voltage is V remain = V total- V 1- V 4 = 11 V- 8 . 66133 V- 1 . 59368 V = 0 . 744991 V . This remaining voltage is the same across the parallel capacitors, so Q 2 = C 2 V remain = (4 . 78 μ F) (0 . 744991 V) = 3 . 56106 μ C . 004 10.0 points A capacitor network is shown in the following figure. 18 . 6 V 4 . 14 μ F 8 μ F 12 . 9 μ F a b What is the voltage across the 8 μ F upper right-hand capacitor? Correct answer: 6 . 343 V. Explanation: Let : C 1 = 4 . 14 μ F , C 2 = 8 μ F , C 3 = 12 . 9 μ F , and V = 18 . 6 V . 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 = (18 . 6 V)(4 . 14 μ F) 4 . 14 μ F + 8 μ F = 6 . 343 V ....
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12. Capacitors in circuits-solutions - gilvin(jg47854 –...

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