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In Fig. E24.17, each capacitor has C = 4.00 µF and Vab =

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 17E Chapter 24

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 17E

In ?Fig. E24.17?, each capacitor has C = 4.00 µF and Vab = +28.0 V. Calculate (a) the charge on each capacitor; (b) the potential difference across each capacitor; (c) the potential difference between points a and d.

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Solution 17E If two capacitors with capacitances C and 1 are conn2cted in parallel, then the equivalent capacitance is given by C eq= C + C …..(1) 1 2 If two capacitors with capacitances C and C1are conne2ted in series, then the equivalent capacitance is given by C eq= C1C 2 …..(2) C 1 C2 Let us first calculate the equivalent capacitance of the given arrangement. The capacitors C a1d C are 2eries. So, their equivalent will be, C C C 12= 1 2 C1+C 2 4×4 C 12= 4+4F C = 2.0 F 12 Now, C 12is parallel with C 3 Hence their equivalent, C 123= 2.00 F + 4.00 F = 6.00 F Now, C 123 is in series with C4. Thus, their equivalent, C 1234= 6+4 F = 2.4 F Therefore, the equivalent capacitance is 2.4 F . The charge on this arrangement Q = C × V = 2.4 F × 28.0 V = 67.2 C 1234 ab Now, the charge magnitude is the same across each capacitor for series connection. Hence, the charge on C w4uld be 67.2 C . So, Q 4 67.2 C Q4 67.2 Potential difference across C 4 V 4 = C 4 = 4 V = 16.8 V Since C is in series with C , so the charge on the combination 123 would also be 123 4 67.2 C . Q 123= 67.2 C The 123 combination has a capacitance of 6 F . Hence, its voltage is, Q 123 67.2 V 123= 6F = 6.00V = 11.2 V Now, for parallel combination of capacitors, the voltage across each capacitor is the same, V 12 = V 3 = 11.2 V This is also the potential difference between the points a and d. (Answer of part c) The charge on the capacitor C ,3 = 31.2 V × 4 F = 44.8 C C a1d C are2in series. So, the charge on each of them is equal. Q 12= V 12× C 12= 11.2 V × 2.0 F = 22.4 C Q = Q = 22.4 C 1 2 Q1 22.4 Potential difference across C 1V 1 = C = 4 V = 5.6 V 1 Q 2 22.4 Potential difference across C 2V 2= C 2= 4 V = 5.6 V Summary Capacitor Charge Potential difference C 1 22.4 C 5.6 V C 2 22.4 C 5.6 V C 44.8 C 11.2 V 3 C 67.2 C 16.8 V 4

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Chapter 24, Problem 17E is Solved
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Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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In Fig. E24.17, each capacitor has C = 4.00 µF and Vab =