Question

Consider the circuit shown in the figure below, where C1 = 8.00 µF, C2 = 7.00 µF, and ?V = 22.0 V. Capacitor C1 is first charged by closing switch S1. Switch S1 is then opened, and the charged capacitor is connected to the uncharged capacitor by closing S2. (a) Calculate the initial charge acquired by C1. µC (b) Calculate the final charge on each capacitor. q1 = µC q2 = µC

Answer 1

(a)

After switch S1 is closed ,charge on the capacitoe C1

Q1 = C1*V

Q1 = 8*10^-6 *22

Q1 = 176 uC

(b)

After switch S2 closed and S1 is open position ,connection between capacitor C1 and C2 act as parallel .

The equivalent capacitor is

Ceq = C1+C2= 8*10^-6 + 7*10^-6 = 15*10^-6 F

As the second capacitor is uncharged , the total charge on the two capacitor is equal to charge on the C1 ,

Q = Ceq *V

V = Q /Ceq

V = 176*10^-6 / 15*10^-6

V = 11.73 V

Now charge on capacitor C1

Q1 = C1*V

Q1 = 8*10^-6 *11.73

Q1= 93.86*10^-6 C = 93.86 uC

Charge on capacitor C2

Q2 = C2*V

Q2 = 7*10^-6 *11.73

Q2 = 82.11 uC