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).

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

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.

During switching, heat generated in the circuit is zero

answer is B, D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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.