JEE Main 2024 — Alternating Current Question with Solution

To find the inductance of the coil, we need to analyze the given circuit and use the information provided. The circuit consists of a resistor and an inductor in series, connected to an AC supply. Here are the given values:

1. Resistance, $$R=90\Omega$$

2. Supply voltage, $$V_{\text{total}}=120\mathrm{V}$$

3. Frequency, $$f=60\mathrm{Hz}$$

4. Voltage across the resistor, $$V_R=36\mathrm{V}$$

First, we calculate the current through the resistor (which is the same as the current through the inductor, since they are in series) using Ohm's law:

$$I=\frac{V_R}{R}=\frac{36}{90}=0.4\mathrm{A}$$

Next, we find the total impedance $$Z$$ of the series combination from the total supply voltage:

$$V_{\text{total}}=I\cdot Z$$

$$Z=\frac{V_{\text{total}}}{I}=\frac{120}{0.4}=300\Omega$$

We know that the total impedance in a series circuit consisting of a resistor and an inductor is given by:

$$Z=\sqrt{R^2+(X_L{)}^2}$$

where $$X_L$$ is the inductive reactance. Rearrange this to solve for $$X_L$$:

$$X_L=\sqrt{Z^2-R^2}=\sqrt{(300{)}^2-(90{)}^2}=\sqrt{90000-8100}=\sqrt{81900}\approx 286\Omega$$

Now, we use the inductive reactance formula to find the inductance $$L$$:

$$X_L=2\pi fL$$

$$L=\frac{X_L}{2\pi f}=\frac{286}{2\pi \cdot 60}\approx \frac{286}{376.99}\approx 0.76\mathrm{H}$$

Thus, the inductance of the coil is:

Option B: 0.76 H