Problem 18: Two capacitors of capacitance 8 µF and 10 µF are connected in series to a source of P.D of 180 V. The capacitors are disconnected from the supply and are connected in parallel with each other. Find the new potential difference and charge on each capacitor.

Solution

The equivalent capacitance when the capacitors are connected in series with the battery is

As the capacitors are in series, therefore, charge on each capacitor is same and given as

Total charge on the system of capacitors = 800 + 800 = 1600 µC.

As the capacitors are now connected in parallel, their equivalent capacitance is given by
C_ep = 8 + 10 = 18 µF
Therefore, potential across the capacitors =

The charge on the individual capacitors in the parallel combination is,
Q₁ = 8 × 88.8 = 710.4 µC
Q₂ = 10 × 88.8 = 888 µC

(The sum of the charges on the capacitors should be 1600 micro C. The small difference comes from 1600/18).