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 :

Solution

Physics sir · 4 steps

The resistor's voltmeter reading gives the current, and the current gives the impedance.

1. Current in circuit. The same current flows through the resistor and the coil because they are in series. From the resistor, I = 36/90 = 0.4 A.
2. Total impedance. Z = V/I = 120/0.4 = 300 Ω. In an R-L circuit, R and X_L add like the sides of a right triangle, not directly.

   X_L = √(Z² - R²) = √(300² - 90²) = √81900 ≈ 286.2 Ω

3. Find L. X_L = 2π f L with f = 60 Hz, so ω = 2π × 60 ≈ 377 rad s⁻¹. Then L = 286.2/377 ≈ 0.759 H.
4. So. The inductance of the coil is about 0.76 H, option (b).

The idea to keep. In a series AC circuit, find the current from the pure resistor first. Then remember that R and X_L combine as Z = √(R² + X_L²), and voltages combine the same way.

Answer from the NTA final answer key. Solution by PrepWiser.