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 of the 6 µF capacitor is E.
Let V be the potential difference along the capacitor of capacitance 6 µF.

\[\text { Since } \frac{1}{2}C V^2 = E\]

\[ \therefore \frac{1}{2} \times 6 \times {10}^{- 6} \times V^2 = E\]

\[ \Rightarrow V^2 = \frac{E}{3} \times {10}^6 . . . (1)\]

Since potential is same for parallel connection, the potetial through 12 µF capacitor is also V. Hence, energy of 12 µF capacitor is

\[E_{12} = \frac{1}{2} \times 12 \times {10}^{- 6} \times V^2 = \frac{1}{2} \times 12 \times {10}^{- 6} \times \frac{E}{3} \times {10}^6 = 2E\]

(ii) Since charge remains constant in series, the charge on 6 µF and 12 µF capacitors combined will be equal to the charge on 3 µF capacitor. Using the formula, Q = CV, we can write

\[(6 + 12) \times {10}^{- 6} \times V = 3 \times {10}^{- 6} \times V'\]

\[\text { Using  (1) and squaring both sides, we get }\]

\[V '^2 = 12E \times {10}^6 \]

\[ \therefore E_3 = \frac{1}{2} \times 3 \times {10}^{- 6} \times 12E \times {10}^6 = 18E\]

(iii) Total energy drawn from battery is 

\[E_{total} = E + E_{12} + E_3 = E + 2E + 18E = 21E\]