JEE Advanced 2023 Paper 1 · Q16 · LCR Series Resonance

Question

A series LCR circuit is connected to a $45\sin(\omega t)$ Volt source. The resonant angular frequency of the circuit is $10^5$ rad·s$^{-1}$ and current amplitude at resonance is $I_0$. When the angular frequency of the source is $\omega = 8 	imes 10^4$ rad·s$^{-1}$, the current amplitude in the circuit is $0.05\,I_0$. If $L = 50$ mH, match each entry in List-I with an appropriate value from List-II.
List-I:
(P) $I_0$ in mA
(Q) The quality factor of the circuit
(R) The bandwidth of the circuit in rad·s$^{-1}$
(S) The peak power dissipated at resonance in Watt
List-II: (1) $44.5$, (2) $18$, (3) $400$, (4) $2250$, (5) $500$

SolveKar AI · Accepted Answer

At resonance $\omega_r = 1/\sqrt{LC}$: $C = 1/(\omega_r^2 L) = 1/(10^{10}\cdot 0.05) = 2	imes 10^{-9}$ F. At $\omega = 8	imes 10^4$: $X_L = 4000\,\Omega$, $X_C = 6250\,\Omega$, $X = 2250\,\Omega$. From $0.05 = R/\sqrt{R^2+X^2}$: solving gives $R \approx 112.6\,\Omega$. (P) $I_0 = 45/R = 45/112.6 \approx 400$ mA — (3). (Q) $Q = \omega_r L/R = 10^5\cdot 0.05/112.6 \approx 44.5$ — (1). (R) Bandwidth $= R/L = 112.6/0.05 = 2252 \approx 2250$ rad/s — (4). (S) Peak power = $V_0^2/R = 2025/112.6 \approx 18$ W — (2).

JEE Advanced 2023 Paper 1 · Q16 · LCR Series Resonance

Solution