Question

A series circuit consists of a 0.054 µF capacitor, a 0.140 µF capacitor, and a 410 V battery. Find the charge for the following situations.

(a) on each of the capacitors
µC (0.054 µF capacitor)
µC (0.140 µF capacitor)
(b) on each of the capacitors if they are reconnected in parallel across the battery
µC (0.054 µF capacitor)
µC (0.140 µF capacitor)

Answer 1

A.

when Capacitors are connected in Series,

Ceq = C1*C2/(C1 + C2)

Ceq = 0.054*0.140/(0.054 + 0.140)

Ceq = 0.039 uF

Now using equation

Qeq = Ceq*V

Qeq = 0.039*10^-6*410 = 1.60*10^-5 C

Qeq = 16.0*10^-6 C = 16 uC

Since Capacitors are connected in Series, So charge on each capacitors will be same

Charge on 0.054 uF capacitor = 16.0 uC

Charge on 0.140 uF capacitor = 16.0 uC

B.

When connected in parallel,

Ceq = C1 + C2

Ceq = 0.054 + 0.140 = 0.194 uF

Qeq = 0.194*10^-6*410 = 79.54*10^-6 C

In parallel circuit voltage on each capacitor wil be same, So

Charge on 0.054 uF capacitor will be

Q1 = C1*V1 = 0.054*10^-6*410 = 22.14*10^-6 C = 22.14 uC

Q2 = C2*V2 = 0.140*10^-6*410 = 57.4*10^-6 C = 57.4 uC