JEE Main 2022 · 24 Jun · Shift 1 · Physics Q54

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PhysicsInteger+4 / -1Question 54 of 90

As shown in the figure an inductor of inductance 200 mH is connected to an AC source of emf 220 V and frequency 50 Hz. The instantaneous voltage of the source is 0 V when the peak value of current is $\dfrac{\sqrt a}\pi$ A. The value of $a$ is ______.

A purely inductive AC circuit: an inductor L = 200 mH connected across an AC source.

Your answer

Answer and worked solution

Official answer: 242

Reactance. $X_L=2\pi fL=2\pi(50)(0.2)=20\pi\ \Omega$ .

Peak current. In a pure inductor the current lags by $90°$ , so it peaks when the voltage is zero. $I_0=\dfrac{220\sqrt2}{20\pi}=\dfrac{11\sqrt2}\pi=\dfrac{\sqrt{242}}\pi$ A.

So $a=242$ .

Answer: 242

Topics:Reactance and impedance

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JEE Main 2022 · 24 Jun · Shift 1, Physics question 54