JEE Main 2024 — Alternating Current Question with Solution

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:

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

$$\omega =300rad/s$$

Now we can solve for the inductance $$L$$:

$$X_L=300L$$

$$3=300L$$

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

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

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

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

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