A potential difference of 240 V is applied to a seriesconnection of two capacito

Question

A potential difference of 240 V is applied to a seriesconnection of two capacitors of capacitances C1= 2.21 µF and C2 = 7.00 µF. Whatare (a) charge q1 and(b) potential difference V1 oncapacitor 1 and (c) q2and (d) V2 on capacitor2? The charged capacitors are then disconnected from eachother and from the battery. Then the capacitors are reconnectedwith plates of the same signs wired together (the batteryis not used). What now are (e) q1, (f) V1, (g) q2, and (h) V2? Suppose, instead, the capacitorscharged in part (a) are reconnected with plates ofopposite signs wired together. What now are(i) q1,(j) V1,(k) q2, and(l) V2? Give yourcharges in Coulombs.

Explanation / Answer

(a) charge q1 equivalent capactior: C=C1*C2/(C1+C2)=1.68 F q1=V*C=403C (b)potential difference V1 on capacitor1 V1*C1=V2*C2=q1 => V1=182.4V (c)q2=q1=403C (d) V2=q2/C2=57.6V the equivalent capacitor now is: C=C1+C2=9.21F q1/C1=q2/C2=total charges before/9.21 F =2*403C/9.21F (*) (e)=> q1=96.7C     (from (*)) (f)V1=q1/C1=43.7V    (from (*)) (g)q2=306.4C       (from (*) (h)V2=V1=43.7V   (having the same potential when connectingparallel) the capacitors charged in part (a) are reconnected with platesof opposite signs wired together => total charges now is zero   (-Q+Q=0) =>q1=q2=0 V1=V2=0

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