JEE Main 2025 — Alternating Current Question with Solution

$$\omega =\frac{1}{\sqrt{LC}}$$

For a series LCR circuit, the maximum current occurs at resonance, where the inductive reactance equals the capacitive reactance. Given the values:

Inductance: $$L=100\text{mH}=0.1\text{H}$$

Capacitance: $$C=25\text{nF}=25\times 10^{-9}\text{F}$$

we first calculate the product $$LC$$:

$$LC=0.1\times 25\times 10^{-9}=2.5\times 10^{-9}$$

Next, compute the square root of the product:

$$\sqrt{LC}=\sqrt{2.5\times 10^{-9}}$$

Recognize that:

$$\sqrt{2.5\times 10^{-9}}=\sqrt{2.5}\times \sqrt{10^{-9}}\approx 1.581\times 10^{-4.5}$$

Since $$10^{-4.5}=3.162\times 10^{-5}$$, we have:

$$\sqrt{LC}\approx 1.581\times 3.162\times 10^{-5}\approx 5.0\times 10^{-5}$$

Now, the angular frequency at resonance becomes:

$$\omega =\frac{1}{5.0\times 10^{-5}}=2.0\times 10^4\text{radians/s}$$

Thus, the angular frequency of the AC source for maximum current in the circuit is:

$$\omega =2\times 10^4\text{radians/s}$$