Question

In the following arrangement of capacitors, the energy stored in the 6 µF capacitor is E. Find the value of the following:

(i) Energy stored in 12 µF capacitor.

(ii) Energy stored in 3 µF capacitor.

(iii) Total energy drawn from the battery

Answer 1

(i) Given that energy stored in 6 µF is E.

Let V be the voltage across 6 µF capacitor

Also, 6 µF and 12 µF capacitors are in parallel.

Therefore, voltage across 12 µF = Voltage across 6 µF capacitor

Energy stored in 12μF = \(\cfrac12\times12\left(\sqrt{\frac{E}3}\right)^2\) = 2E

(ii) Since charge remains constant in series. Sum of charge on 6 µF capacitor and 12 µF capacitor is equal to charge on 3 µF capacitor.

Using Q = CV,

Charge on 3 µF capacitor = (6 + 12) × V = 18 × V

Energy stored in 3 F capacitor

= 18 E

(iii) Total energy drawn from battery = E + 2E + 18E = 21E