In a series LCR circuit, the voltage across R is 200 V, \( C=5\,\mu\mathrm{F} \), \( R=2\,\mathrm{k}\Omega \). The resonant frequency \( \omega \) is 100 rad/s. At resonance, the voltage across L is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 200 V
The same current flows through every element of a series circuit. At resonance the inductive and capacitive reactances are equal, \( X_L = X_C = \frac{1}{\omega C} \), so the voltage across L can be found from the current and \( X_L \).
Current
\( \displaystyle I = \frac{V_R}{R} = \frac{200\,\mathrm{V}}{2000\,\Omega} = 0.1\,\mathrm{A} \)
Reactance at resonance
\( \displaystyle X_L = X_C = \frac{1}{\omega C} = \frac{1}{(100)(5\times10^{-6})}\,\Omega = 2000\,\Omega \)
Voltage across L
\( V_L = IX_L = 0.1\times2000 = 200\,\mathrm{V} \)
Shortcut: At resonance \( V_L = Q\,V_R \) with \( Q = \frac{1}{\omega_0CR} = \frac{1}{100\times5\times10^{-6}\times2000} = 1 \). So \( V_L = V_R = 200\,\mathrm{V} \) directly.