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: (a) 0.91 H (b) 0.76 H (c) 2.86 H (d) 0.286 H

Answer: (b). I = VR/R = 36/90 = 0.4 A, so Z = (120)/(0.4) = 300 Ω. XL = √(Z² - R²) = √(300² - 90²) = √(81900) ≈ 286 Ω L = (XL)/(2π f) = (286)/(2π × 60) ≈ 0.76 H So the answer is (b) 0.76 H.

A coil of negligible resistance is connected in series with 90 Ω… | JEE Main 2024 (8 Apr, Shift 2)

Q25JEE Main 2024 · 8 APR · SHIFT 2Medium

RL and RC Series Circuits

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

Show answer and solution

Answer: (b)

$I = \frac{V_R}{R} = \frac{36}{90} = 0.4\text{ A}$ , so $Z = \frac{120}{0.4} = 300\ \Omega$ .
$X_L = \sqrt{Z^2 - R^2} = \sqrt{300^2 - 90^2} = \sqrt{81900} \approx 286\ \Omega$
$L = \frac{X_L}{2\pi f} = \frac{286}{2\pi \times 60} \approx 0.76\text{ H}$
So the answer is (b) 0.76 H.

Report a mistake

RL and RC Series Circuits question from JEE Main 2024 (8 Apr, Shift 2)

More questions on RL and RC Series Circuits