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.
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: \(200\)
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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