In a series LCR circuit R = 1 kΩ, C = 2 μF and the voltage across R is 100 V at resonant frequency (ω) \( 200\,\mathrm{rad\,s^{-1}} \). At resonant frequency, the voltage across L is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 250 V
At resonance \( X_L = X_C \). The same current flows through every element of a series circuit, so it can be found from the resistor.
\( \displaystyle I = \frac{V_R}{R} = \frac{100\,\mathrm{V}}{1000\,\Omega} = 0.1\,\mathrm{A} \)
\( \displaystyle X_L = X_C = \frac{1}{\omega C} = \frac{1}{200 \times 2\times10^{-6}} = 2500\,\Omega \)
\( \displaystyle V_L = I\,X_L = 0.1 \times 2500 = 250\,\mathrm{V} \)
The voltage across L (and across C) is larger than the voltage across R; this voltage magnification at resonance is why \( V_L \) can exceed \( V_R \).
Note: A similar question was also asked in CUET (UG) 2026 on 19-05-2026 (Slot 2).