#
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.

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**