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:

Engineering

Physics

Circuit Analysis Having Capacitor Only

Question

$\frac{5}{8} \frac{Q^{2}}{C}$

$\frac{3}{4} \frac{Q^{2}}{C}$

$\frac{3}{8} \frac{Q^{2}}{C}$

$\frac{1}{8} \frac{Q^{2}}{C}$

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Before switch is shifted:

Energy stored,

$U_{i} = \frac{1}{2} {CE}^{2}$

and charge stored,

Q = CE

After switch is shifted:

$V_{M} - V_{N} = \frac{Q}{4 C} = \frac{CE}{4 C} = \frac{E}{4}$

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

$Energy dissipated = U_{i} - U_{f} = \frac{3}{8} {CE}^{2}$