Easy INTEGER +4 / -1 PYQ · JEE Mains 2025

In a series LCR circuit, a resistor of $300 \Omega$, a capacitor of 25 nF and an inductor of 100 mH are used. For maximum current in the circuit, the angular frequency of the ac source is _________ $\times 10^4$ radians $\mathrm{s}^{-1}$

Answer (integer) 2

Solution

<p>$\omega = \frac{1}{\sqrt{LC}}$</p> <p>For a series LCR circuit, the maximum current occurs at resonance, where the inductive reactance equals the capacitive reactance. Given the values:</p> <p><p>Inductance: $L = 100 \, \text{mH} = 0.1 \, \text{H}$</p></p> <p><p>Capacitance: $C = 25 \, \text{nF} = 25 \times 10^{-9} \, \text{F}$</p></p> <p>we first calculate the product $LC$:</p> <p>$LC = 0.1 \times 25 \times 10^{-9} = 2.5 \times 10^{-9}$</p> <p>Next, compute the square root of the product:</p> <p>$\sqrt{LC} = \sqrt{2.5 \times 10^{-9}}$</p> <p>Recognize that:</p> <p>$$ \sqrt{2.5 \times 10^{-9}} = \sqrt{2.5} \times \sqrt{10^{-9}} \approx 1.581 \times 10^{-4.5} $$</p> <p>Since $10^{-4.5} = 3.162 \times 10^{-5}$, we have:</p> <p>$\sqrt{LC} \approx 1.581 \times 3.162 \times 10^{-5} \approx 5.0 \times 10^{-5}$</p> <p>Now, the angular frequency at resonance becomes:</p> <p>$\omega = \frac{1}{5.0 \times 10^{-5}} = 2.0 \times 10^{4} \, \text{radians/s}$</p> <p>Thus, the angular frequency of the AC source for maximum current in the circuit is:</p> <p>$\omega = 2 \times 10^4 \, \text{radians/s}$</p>

About this question

Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C

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