Q). In the following arrangement of capacitors, the energy stored in the $6 \mu F$ capacitor is E. Find the value of the following:
(i) Energy stored in $12 \mu F$ capacitor.
(ii) Energy stored in $3 \mu F$ capacitor.
(iii) Total energy drawn from the battery.

Hello Abid dear, Given E is the energy stored in 6 u F
Recall the formula E = 1/2 * q^2 / C
Let Q be the charge stored in C = 6 u F
Or Q^2 = 12 E * 10^-6 ---(1) since C = 6 x 10^-6 F
So 12 u F would have 2 Q charge as 6 uF and 12 uF are at the same potential {recall Q = C V}
Now energy in 12 uF = 1/2 * (2Q)^2 / (12 * 10^-6)  = [2 * 10^6 /12] * Q^2 = 1/6 * 12 E = 2 E
Hence total energy in this combination E + 2 E = 3 E
Now charge on 3 uF would be the same as that of the combination 6 uF and 12 uF
So it will be 3 Q
Now energy = 1/2 * (3Q)^2 / C 
E' = 1/2 * 9 * 12 E / 3 = 18 E
(i) Energy stored in 12 u F = 2 E
(ii) Energy stored in 3 uF = 18 E
(iii) Total energy drawn from the battery = 21 E