QJEEM24Phdb38: JEE Main 2024 Physics - Class 12 JEEM24Phdb38 Solution: A coil of negligible resistance is connected in series with…

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 $L$ :

$X_L = 2\pi f L$

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

Thus, the inductance of the coil is:

Option B: 0.76 H