JEE Main 2022 · 29 Jul · Shift 1 · Physics Q43

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PhysicsSingle correct+4 / -1Question 43 of 90

An alternating emf $E=440\sin100\pi t$ is applied to a circuit containing an inductance of $\dfrac{\sqrt2}\pi$ H. If an a.c. ammeter is connected in the circuit, its reading will be :

4.4 A

1.55 A

2.2 A

3.11 A

Answer and worked solution

Official answer: C

Reactance $X_L=100\pi\cdot\frac{\sqrt2}\pi=100\sqrt2\ \Omega$ .

rms current $\dfrac{440/\sqrt2}{100\sqrt2}=2.2$ A.

(C)

Topics:Reactance and impedance AC basics and RMS values

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JEE Main 2022 · 29 Jul · Shift 1, Physics question 43