In the circuit shown, the resistance of the connecting wires is negligibly small. Both the switches $S_{1} a n d S_{2}$ are closed simultaneously at $t = 0$ . The process of closing the switches (called “switching”) is finished in negligibly small amount of time. Mark the correct statement(s).

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During switching, heat generated in the circuit is zero

During switching, heat generated in the circuit is $\frac{3 C V_{0}^{2}}{2}$

After switching, maximum current through inductor is $5 V_{0} \sqrt{\frac{L}{C}}$

At $t = \frac{π}{2} \sqrt{L C}$ , magnetic energy stored in L is equal to total electric energy stored in the capacitors.

answer is B, D.

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Detailed Solution

Immediately after switching, redistribution of charge will take place on $C_{1} & C_{2}$ and they will come at common potential.
$V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}} = \frac{C V_{0} + g C V_{0}}{4 C} = \frac{5}{2} V_{0} ∴ Δ H = \frac{1}{2} \frac{C_{1} C_{2}}{C_{1} + C_{2}} {(V_{1} - V_{2})}^{2} = \frac{1}{2} \times \frac{3 c}{4} \times {(2 v_{0})}^{2} = \frac{3}{2} C V_{0}^{2}$

$t = \frac{T}{4} = \frac{π}{2} \sqrt{L C}$ ; magnetic energy stored in inductor will be max.