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JEE Mains · Physics · STD 12 -Alternating current
  • JEE Mains 2026
  • Easy
  • Numerical

A series LCR circuit with \(R = 20\ \Omega\), \(L = 1.6\text{ H}\) and \(C = 40\ \mu\text{F}\) is connected to a variable frequency a.c. source. The inductive reactance at resonant frequency is _______ \(\Omega\).

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(D) 200

Step-by-step Solution

Detailed explanation

Given \(L = 1.6\text{ H}\) and \(C = 40\ \mu\text{F} = 40 \times 10^{-6}\text{ F}\).

At resonance, the angular frequency is \(\omega_r = \dfrac{1}{\sqrt{LC}}\).

The inductive reactance at resonance is \(X_L = \omega_r L = \dfrac{L}{\sqrt{LC}} = \sqrt{\dfrac{L}{C}}\).

Substituting the given values:

\(X_L = \sqrt{\dfrac{1.6}{40 \times 10^{-6}}}\)

\(X_L = \sqrt{\dfrac{1.6}{4 \times 10^{-5}}}\)

\(X_L = \sqrt{4 \times 10^4}\)

\(X_L = 200\ \Omega\)

Answer: \(200\)
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