Two capacitors having capacitances `0.1muF " and "0.01muF` are joined in series. A potential difference of 22V is applied to the combination. If the capacitors are now joined in parallel, what will be the change in stored energy?

Explanation

If the capacitors are joined in series, their equivalent capacitance,
`C_s= (C_1xxC_2)/(C_1+C_2)= (0.1xx0.01)/(0.1+0.01)= (1)/(110) muF = (10^(-6))/(110)F`
Energy stored in the combination,
`E_1= (1)/(2)C_sV^2 = (1)/(2)xx(10^(-6))/(110)xx(22)^2 J = 22` erg
If the capacitors are joined in parallel, their equivalent capacitance,
`C_p = C_1+C_2= 0.1+0.01= 0.11 muF = 0.11xx10^(-6)F`
Energy stored in this combination,
`E_2 = (1)/(2)C_pV^2= (1)/(2)xx0.11xx10^(-6)xx(22)^2J = 266.2` erg
The change in stored energy `= E_2 - E_1 = 266.2 - 22= 244.2` erg
Stored energy increases by 244.2 erg.