Question

In Fig. 25-31, a 20.0 V battery is connected across capacitors of capacitances C_1 = C_6 = 3.00 mu F and C_3 = C_5 = 2.00C_2 = 2.00 C_4 = 4.00 mu F. What are (a) the equivalent capacitance C_eq of the capacitors and (b) the charge stored by C_eq the What are (c) V_1 and (d) q_1 of capacitor 1, (e) V_2 and (f) q_2 of capacitor 2, and (g) V_3 and (h) q_3 of capacitor 3?

Answer 1

(a) C3 and C5 are in series and are equivalent to C35 [1/4 1/4 uF 2.00 uF C35. C2 and C4 are in parallel and their equivalent capacitance is C2345 -C2+ C35 + C4 (2.00 + 2.00 + 2.00uF-6.00 uF. This is now in series with the lower parallel combination of C and Cs. These add to give C1,6 6.00 μF. C 16 and C2345 are and series and add reciprocally to give deq-(1/64 1/6T F-3.00 F (b) By our definition for capacitance, Ce4 qCeaV 60.0 uC (c) Capacitors in series have the same charges on them, so C16q/Vi6 and C2345 -q/V2345 Since C16-C2345-3.00 μΕ the potential drop across the bottom and top sets of capacitors is the same and must add up to 20.0 V. The potential drop across G and C% (and also across C2 and C4, and the series combination of C, and Cs) is 10.0 V. (d) The charge carried by C1 is q C1V (3.00 uF) 10.0 V) 30.0 uC (e) The potential difference across C2 is given by V - Vi- (20.0 V - 10.0 V)- 10.0 V. (f) The charge carried by C2 is q,-C2V,-(2.00 μF)( 10.0 V) = 20.0 μC. (g) Since the potential difference is divided equally between C3 and the other C5 (since both have the same capacitance) capacitor connected in series with it, the potential difference across it is half the total ofy, or y,-W2-5.00 y (h) q3 -C3V3 (4.00 uF) (5.00 V) 20.0 uC.