JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Alternating Current

For an LCR series circuit, the current is in phase with the voltage at resonance, when $\omega L=\frac{1}{\omega C}$, giving $\omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{10^{-6}\times 10^{-6}}}=10^6$ rad/s. This resonance frequency is independent of $R$, confirming option (B). At $\omega =10^4$ rad/s, the circuit is not at resonance, so option (A) is incorrect. At $\omega \sim 0$, the capacitive reactance $X_C=\frac{1}{\omega C}\to \infty$, making the impedance very large and the current nearly zero, confirming option (C). At $\omega \gg 10^6$ rad/s, the inductive reactance $X_L=\omega L$ dominates, so the circuit behaves like an inductor, not a capacitor, making option (D) incorrect.