In a series resonant LCR circuit, the voltage across \(R = 1\,\mathrm{k\Omega}\) is \(100\,\mathrm{V}\) and \(C = 2\,\mu\mathrm{F}\). The resonant frequency \(\omega\) is \(200\,\mathrm{rad/s}\). At resonance, the voltage across \(L\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(250\,\mathrm{V}\)
At resonance \( X_L = X_C \), so the impedance is just R and the voltages across L and C are equal and opposite.
\( \displaystyle I = \frac{V_R}{R} = \frac{100\,\mathrm{V}}{1000\,\Omega} = 0.1\,\mathrm{A} \)
\( \displaystyle X_C = \frac{1}{\omega C} = \frac{1}{200\,\mathrm{rad\,s^{-1}} \times 2\times10^{-6}\,\mathrm{F}} = 2500\,\Omega \)
\( \displaystyle X_L = X_C = 2500\,\Omega \)
\( \displaystyle V_L = IX_L = 0.1\,\mathrm{A} \times 2500\,\Omega \)
\( \displaystyle V_L = 250\,\mathrm{V} \)
Note: A similar question was also asked in CUET (UG) 2025 on 13-05-2025 (Slot 1).