A series circuit has a resistor R=40…

A series circuit has a resistor R=40 Ω, an ideal inductor L=0.04 H and an ideal capacitor C=100 muF, connected to a source of constant rms voltage…

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A series circuit has a resistor

R=40\ \Omega

, an ideal inductor

L=0.04\ \mathrm{H}

and an ideal capacitor

C=100\ \mu\mathrm{F}

, connected to a source of constant rms voltage

100\ \mathrm{V}

whose angular frequency can be varied. Match List-I with List-II.

List-I
(P) rms current at the resonant frequency
(Q) Quality factor

\omega_0L/R

of the circuit at resonance
(R) Impedance of the circuit at

\omega=250\ \mathrm{rad\,s^{-1}}

(S) Average power dissipated in the circuit at

\omega=250\ \mathrm{rad\,s^{-1}}

List-II
(1)

0.5

(2)

160\ \mathrm{W}

(3)

2.5\ \mathrm{A}

(4)

50\ \Omega

(5)

2.0\ \mathrm{A}

Show answer and solution

Answer: (A) P→3, Q→1, R→4, S→2

Resonance occurs at

\omega_0=\dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{0.04\times10^{-4}}}=500\ \mathrm{rad\,s^{-1}}

, where

X_L=X_C

and

Z=R

. (P)

I=V/R=100/40=2.5\ \mathrm{A}

, which is (3). (Q)

Q=\dfrac{\omega_0L}{R}=\dfrac{500\times0.04}{40}=0.5

, which is (1). At

\omega=250\ \mathrm{rad\,s^{-1}}

X_L=250\times0.04=10\ \Omega

and

X_C=\dfrac{1}{250\times10^{-4}}=40\ \Omega

, so the net reactance is

-30\ \Omega

and

Z=\sqrt{40^2+30^2}=50\ \Omega

. (R) is (4). (S) The current is

100/50=2\ \mathrm{A}

, so the average power is

I_{rms}^2R=4\times40=160\ \mathrm{W}

, which is (2). The value

2.0\ \mathrm{A}

in (5) is a distractor (current, not power). At

250\ \mathrm{rad\,s^{-1}}

the circuit is capacitive since

X_C>X_L

, and its current is lower than the resonant maximum. Hence P→3, Q→1, R→4, S→2.

Original question written in the JEE Advanced 2024 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake