JEE Advanced 2017 · Paper 1 · Physics Q6

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PhysicsMultiple correct+4 / -2Question 6 of 54

In the circuit shown, $L = 1\ \mu$ H, $C = 1\ \mu$ F and $R = 1$ kΩ. They are connected in series with an a.c. source $V = V_0\sin\omega t$ as shown. Which of the following options is/are correct?

Series circuit: an a.c. source V₀ sin ωt connected to an inductor L = 1 μH, a capacitor C = 1 μF and a resistor R = 1 kΩ in series.

The current will be in phase with the voltage if

\omega = 10^4\ \mathrm{rad\,s^{-1}}

The frequency at which the current will be in phase with the voltage is independent of

R

\omega \sim 0

the current flowing through the circuit becomes nearly zero

\omega \gg 10^6\ \mathrm{rad\,s^{-1}}

, the circuit behaves like a capacitor

Answer and worked solution

Official answer: B, C

Resonance.
$\omega_0=\frac1{\sqrt{LC}}=\frac1{\sqrt{10^{-6}\times10^{-6}}}=10^6\ \mathrm{rad\,s^{-1}}.$

(A) The current is in phase only at $10^6$ , not $10^4$ . False.
(B) $\omega_0$ does not involve $R$ . True.
(C) As $\omega\to0$ , $X_C\to\infty$ and the capacitor blocks the current. True.
(D) For $\omega\gg10^6$ , $X_L\gg X_C$ and the circuit is inductive. False.

(B), (C)

Topics:LCR circuits and resonance Reactance and impedance

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JEE Advanced 2017 · Paper 1, Physics question 6