JEE Advanced 2024 Paper 1 · Q17 · LCR Series Resonance

Question

The circuit shown in the figure contains an inductor $L = 25\,	ext{mH}$, a capacitor $C_0 = 10\,\mu	ext{F}$, a resistor $R_0 = 5\,\Omega$ and an ideal $20\,	ext{V}$ battery. The circuit also contains two keys $K_1$ and $K_2$. Initially, both the keys are open and there is no charge on the capacitor. At an instant, key $K_1$ is closed and immediately after this the current in $R_0$ is found to be $I_1$. After a long time, the current attains a steady state value $I_2$. Thereafter, $K_2$ is closed and simultaneously $K_1$ is opened and the voltage across $C_0$ oscillates with amplitude $V_0$ and angular frequency $\omega_0$.

Match the quantities mentioned in List-I with their values in List-II and choose the correct option.

List-I:
(P) The value of $I_1$ in Ampere is
(Q) The value of $I_2$ in Ampere is
(R) The value of $\omega_0$ in kilo-radians/s is
(S) The value of $V_0$ in Volt is

List-II: (1) $0$; (2) $2$; (3) $4$; (4) $20$; (5) $200$.

[Figure: series RL loop with battery and key $K_1$, plus a parallel branch with capacitor $C_0$ and key $K_2$.]

SolveKar AI · Accepted Answer

Phase 1 — only $K_1$ closed (RL loop with battery). Inductor opposes sudden current change, so just after closing $K_1$: $I_1 = 0$. → P $	o$ (1).

Long-time steady state: $L$ acts as a wire, $I_2 = V/R_0 = 20/5 = 4\,	ext{A}$. → Q $	o$ (3).

Phase 2 — $K_2$ closed, $K_1$ opened: LC oscillation with stored magnetic energy converting to capacitor energy. $\omega_0 = 1/\sqrt{LC_0} = 1/\sqrt{(25	imes 10^{-3})(10	imes 10^{-6})} = 1/\sqrt{2.5	imes 10^{-7}} = 2	imes 10^3\,	ext{rad/s} = 2\,	ext{krad/s}$. → R $	o$ (2).

Energy conservation: $	frac{1}{2}LI_2^2 = 	frac{1}{2}C_0 V_0^2$ ⇒ $V_0 = I_2\sqrt{L/C_0} = 4\sqrt{(25	imes 10^{-3})/(10	imes 10^{-6})} = 4\sqrt{2500} = 4\cdot 50 = 200\,	ext{V}$. → S $	o$ (5). Mapping P→1, Q→3, R→2, S→5 = option (A).

JEE Advanced 2024 Paper 1 · Q17 · LCR Series Resonance

Solution