QJEEM25Ph4875: JEE Main 2025 Physics - Class 12 JEEM25Ph4875 Solution: In a series LCR circuit, a resistor of , a capacitor of 25 nF…

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

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:

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