JEE Advanced 2023 · Paper 1 · Physics Q16

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A series LCR circuit is connected to a $45\sin(\omega t)$ Volt source. The resonant angular frequency of the circuit is $10^5\ \mathrm{rad\,s^{-1}}$ and current amplitude at resonance is $I_0$ . When the angular frequency of the source is $\omega = 8\times10^4\ \mathrm{rad\,s^{-1}}$ , 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 and choose the correct option.

List-I

P
$I_0$ in mA
Q
The quality factor of the circuit
R
The bandwidth of the circuit in $\mathrm{rad\,s^{-1}}$
S
The peak power dissipated at resonance in Watt

List-II

1
$44.4$
2
$18$
3
$400$
4
$2250$
5
$500$

P → 2; Q → 3; R → 5; S → 1

P → 3; Q → 1; R → 4; S → 2

P → 4; Q → 5; R → 3; S → 1

P → 4; Q → 2; R → 1; S → 5

Answer and worked solution

Official answer: B

Capacitance. $C=\dfrac{1}{\omega_0^2L}=\dfrac1{10^{10}\times0.05}=2\times10^{-9}$ F.

Resistance. At $\omega=8\times10^4$ :
$X=\omega L-\frac1{\omega C}=4000-6250=-2250\ \Omega.$
Since $I=0.05I_0$ , $Z=20R$ :
$R^2+2250^2=400R^2\implies R=\frac{2250}{\sqrt{399}}\approx112.5\ \Omega.$

List-I values.

(P) $I_0=45/112.5=0.4$ A $=400$ mA → 3.
(Q) $Q=\dfrac{\omega_0L}{R}=\dfrac{5000}{112.5}=44.4$ → 1.
(R) Bandwidth $=\dfrac RL=2250$ rad/s → 4.
(S) Peak power $=V_0I_0=45\times0.4=18$ W → 2.

P→3, Q→1, R→4, S→2.

Topics:LCR circuits and resonance Power in AC circuits

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