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

<p>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:</p> <p>1. Resistance, $R = 90 \Omega$</p> <p>2. Supply voltage, $V_{\text{total}} = 120 \mathrm{~V}$</p> <p>3. Frequency, $f = 60 \mathrm{~Hz}$</p> <p>4. Voltage across the resistor, $V_R = 36 \mathrm{~V}$</p> <p>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:</p> <p>$I = \frac{V_R}{R} = \frac{36}{90} = 0.4 \mathrm{~A}$</p> <p>Next, we find the total impedance $Z$ of the series combination from the total supply voltage:</p> <p>$V_{\text{total}} = I \cdot Z$</p> <p>$Z = \frac{V_{\text{total}}}{I} = \frac{120}{0.4} = 300 \Omega$</p> <p>We know that the total impedance in a series circuit consisting of a resistor and an inductor is given by:</p> <p>$Z = \sqrt{R^2 + (X_L)^2}$</p> <p>where $X_L$ is the inductive reactance. Rearrange this to solve for $X_L$:</p> <p>$$X_L = \sqrt{Z^2 - R^2} = \sqrt{(300)^2 - (90)^2} = \sqrt{90000 - 8100} = \sqrt{81900} \approx 286 \Omega$$</p> <p>Now, we use the inductive reactance formula to find the inductance $L$:</p> <p>$X_L = 2\pi f L$</p> <p>$$L = \frac{X_L}{2\pi f} = \frac{286}{2 \pi \cdot 60} \approx \frac{286}{376.99} \approx 0.76 \mathrm{~H}$$</p> <p>Thus, the inductance of the coil is:</p> <p><strong>Option B: 0.76 H</strong></p>