An alternating emf E = 440 sin 100π t is applied to a circuit containing an inductance of (√(2))/(π) H. If an a.c. ammeter is connected in the circuit, its reading will be: (a) 4.4 A (b) 1.55 A (c) 2.2 A (d) 3.11 A

Answer: (c). XL = ω L = 100π × (√(2))/(π) = 100√(2) Ω Erms = (440)/(√(2)) Ammeter reads Irms = ErmsXL = (440/√(2))/(100√(2)) = (440)/(200) = 2.2 A So the answer is (c) 2.2 A.

An alternating emf E = 440 sin 100π t is applied to a circuit… | JEE Main 2022 (29 Jul, Shift 1)

Q46JEE Main 2022 · 29 JUL · SHIFT 1Easy

Pure R, L and C Circuits

An alternating emf $E = 440 \sin 100\pi t$ is applied to a circuit containing an inductance of $\frac{\sqrt{2}}{\pi}\text{ H}$ . If an a.c. ammeter is connected in the circuit, its reading will be:

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Answer: (c)

$X_L = \omega L = 100\pi \times \frac{\sqrt{2}}{\pi} = 100\sqrt{2}\ \Omega$
$E_{rms} = \frac{440}{\sqrt{2}}$
Ammeter reads $I_{rms} = \frac{E_{rms}}{X_L} = \frac{440/\sqrt{2}}{100\sqrt{2}} = \frac{440}{200} = 2.2$ A
So the answer is (c) 2.2 A.

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Pure R, L and C Circuits question from JEE Main 2022 (29 Jul, Shift 1)

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