QJEEM24Phb96f: JEE Main 2024 Physics - Class 12 JEEM24Phb96f Solution: When a coil is connected across a dc supply, it draws a current …

Let's first determine the resistance of the coil when connected to a DC supply. The current drawn by the coil in a DC circuit can be used to calculate its resistance using Ohm's law:

$R = \frac{V}{I}$

Given the DC supply voltage $V_{DC} = 20 \, \mathrm{V}$ and the current $I_{DC} = 5 \, \mathrm{A}$ , the resistance $R$ is:

$R = \frac{20 \, \mathrm{V}}{5 \, \mathrm{A}} = 4 \, \Omega$

Next, let's use the information given for the AC supply. When connected to an AC supply, the coil's impedance $Z$ can be determined using the given current. The total voltage and current in an AC circuit are related to the impedance by the formula:

$Z = \frac{V}{I}$

Given the AC supply voltage $V_{AC} = 20 \, \mathrm{V}$ and the current $I_{AC} = 4 \, \mathrm{A}$ , the impedance $Z$ is:

$Z = \frac{20 \, \mathrm{V}}{4 \, \mathrm{A}} = 5 \, \Omega$

The impedance $Z$ of the coil in an AC circuit is composed of both the resistance $R$ and the inductive reactance $X_L$ , related by:

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

We already know that $R = 4 \, \Omega$ . We can now solve for the inductive reactance $X_L$ :

$5 = \sqrt{4^2 + X_L^2}$

Squaring both sides of the equation:

$25 = 16 + X_L^2$

Solving for $X_L$ :

$X_L^2 = 25 - 16$

$X_L^2 = 9$

$X_L = \sqrt{9}$

$X_L = 3 \, \Omega$

The inductive reactance $X_L$ is also related to the inductance $L$ and the angular frequency $\omega$ by the formula:

$X_L = \omega L$

where $\omega = 2 \pi f$ . Given the frequency $f = 50 \, \mathrm{Hz}$ and using $\pi = 3$ , we find:

$\omega = 2 \times 3 \times 50$

$\omega = 300 \, \mathrm{rad/s}$

Now we can solve for the inductance $L$ :

$X_L = 300 L$

$3 = 300 L$

$L = \frac{3}{300}$

$L = 0.01 \, \mathrm{H}$

Since $1 \, \mathrm{H} = 1000 \, \mathrm{mH}$ , the self inductance of the coil is:

$L = 0.01 \, \mathrm{H} \times 1000 \, \mathrm{mH/H} = 10 \, \mathrm{mH}$

Therefore, the self inductance of the coil is $10 \, \mathrm{mH}$ .