JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Capacitance

2 μ F

capacitor is charged as shown in figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is

Answer: (D)

Solution

Initial energy stored:

U_{i} = \frac{1}{2} C V^{2} = \frac{1}{2} \times 2 \times V^{2} = V^{2}

Initial charge:

q_{i} = 2 V

When switched to position 2, the

2 μ F

capacitor shares charge with an

8 μ F

capacitor in series. Using charge conservation and the fact that capacitors in series have the same charge:

\frac{2 V - q}{2} = \frac{q}{8}

8 V - 4 q = q ⟹ q = \frac{8 V}{5}

Final energy:

U_{f} = \frac{q ^{2}}{2 \times 2} + \frac{q ^{2}}{2 \times 8} = \frac{64 V ^{2}}{50} + \frac{64 V ^{2}}{400} = \frac{V ^{2}}{5}

Energy dissipated:

Δ U = V^{2} - \frac{V ^{2}}{5} = \frac{4 V ^{2}}{5}

Percentage:

\frac{4 V ^{2} /5}{V ^{2}} \times 100% = 80%

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Single correct option