Capacitors in circuits-solutions

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
C 1 = 9.1.052 ?C 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. 4.78 µF 1.15 µF 6.25 µF a b c d Q parallel = 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 µF + 8.59 µF = 13.37 µF . C 1 , C 23 , and C 4 are connected in series with equivalent capacitance 8.59 µF ( 11 V C ba = ( = 1 1 C 1 + C 23 + C1 1 ) −1 4 ) −1 1 1 Find the equivalent capacitance between points a and d. Correct answer: 0.905502 µF. Explanation: Let : C 1 = 1.15 µF , C 2 = 4.78 µF , C 3 = 8.59 µF , C 4 = 6.25 µF , and E B = 11 V . C 2 C 1 C 4 a b c d 1.15 µF + 13.37 µF + 6.25 µF = 0.905502 µF . 002 (part 2 of 3) 10.0 points Determine the charge on the 1.15 µF capacitor on the left-hand side of the circuit. Correct answer: 9.96052 µC. 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 µF) (11 V) C 3 E B For capacitors in series, = 9.96052 µC . 003 (part 3 of 3) 10.0 points Determine the charge on the 4.78 µF capacitor at the top center part of the circuit. 1 C series = V series = 1 C i V i , Correct answer: 3.56106 µC. Explanation: and the individual charges are the same. For parallel capacitors, The voltage drops across C 1 and C 4 are V 1 = Q 1 = 8.66133 V C parallel = C i 15 µF
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
C 4 = 9.6.052 ?C gilvin (jg47854) - 12. Capacitors in circuits - meyers - (21235) 2 and and their voltages add up to V , voltage of the battery V 4 = Q 4 = 1.59368 V . V 1 + V 2 = V 25 µF The remaining voltage is C 2 V 2 C 1 + V 2 = V C 2 V 2 + C 1 V 2 = V C 1 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 µF) (0.744991 V) = 3.56106 µC . 004 10.0 points A capacitor network is shown in the following figure. 4.14 µF 8 µF a b V 2 = V C 1 C 1 + C 2 = (18.6 V)(4.14 µF) 4.14 µF + 8 µF = 6.343 V . 005 (part 1 of 2) 10.0 points Consider the circuit 2 µF 3 µF c a b 5 µF 4 µF 100 V What is the equivalent capacitance for this 12.9 µF 18.6 V What is the voltage across the 8 µF upper right- hand capacitor? Correct answer: 6.343 V. Explanation: Let : C 1 = 4.14 µF, C 2 = 8 µF, C 3 = 12.9 µF, 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 , network? 1. C
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/29/2011 for the course PHYS 202 taught by Professor Doc during the Spring '11 term at North Texas.

Page1 / 13

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online