In the figure shown, after the switch \(‘S’\) is turned from position \(‘A’\) to position \(‘B’\), the energy dissipated in the circuit in terms of capacitance \(‘C’\) and total charge \(‘Q’\) is

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Answer & Solution

Correct Answer

(B) \(\frac{3}{8}\frac{{{Q^2}}}{C}\)

Detailed explanation

Before switch is shifted:

Energy stored,

\(\mathrm{U}_{1}=\frac{1}{2} \mathrm{CE}^{2}\)

and charge stored,

\(Q=C E\)

After switch is shifted:

\(V_{M}-V_{N}=\frac{Q}{4 C}=\frac{C E}{4 C}=\frac{E}{4}\)

Energy stored, \(U_{f}=\frac{1}{2}(4 \mathrm{C})\left(\frac{E}{4}\right)^{2}=\frac{1}{8} \mathrm{CE}^{2}\)

Energy dissipated \(=U_{i}-U_{f}=\frac{3}{8} C E^{2}\)

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