A coil of negligible resistance is connected in series with 90 Ω resistor across 120 V , 60 Hz supply. A voltmeter reads 36 V across resistance. Inductance of the coil is :

Physics · Alternating Current

JEE Main 2024 — 8 April, Shift 2 — Question 42

A coil of negligible resistance is connected in series with $90 \Omega$ resistor across $120 \mathrm{~V}, 60 \mathrm{~Hz}$ supply. A voltmeter reads 36 V across resistance. Inductance of the coil is :

Option A:
0.76 H
Correct
Option B:
2.86 H
Option C:
0.286 H
Option D:
0.91 H

Answer: A

Step-by-step solution

$36=\mathrm{I}_{\text {rms }} \mathrm{R}$

$36=\frac{120}{\sqrt{\mathrm{X}_{\mathrm{L}}^{2}+\mathrm{R}^{2}}} \times \mathrm{R}$ $\mathrm{R}=90 \Omega \Rightarrow 36=\frac{120 \times 90}{\sqrt{\mathrm{X}_{\mathrm{L}}^{2}+90^{2}}}$