In the circuit shown, L = 1 , C = 1 = 1kΩ . They are connected in...

The current will be in phase with voltage when phase angle is zero,

$∴ tan \text{ϕ} = \dfrac{\text{\omega}L - \dfrac{1}{\text{\omega}C}}{R} = 0$

Where $\text{\omega}, L, C, R$ are the angular frequency, inductance, capacitance and resistance respectively.

$⇒ \text{\omega} = {\text{\omega}}_{0} = \dfrac{1}{\sqrt{LC}}$

$⇒ {\text{\omega}}_{\text{ο}}$is independent of $R$

$⇒ ( resonantfrequency ) {\text{\omega}}_{0} = {10}^{6}rad / s$

Current in the circuit is given by,

${i}_{0} = \dfrac{{V}_{0}}{\sqrt{{R}^{2} + {\left(\text{\omega}L - \dfrac{1}{\text{\omega}C}\right)}^{2}}}$

When $\text{\omega} \to 0, {i}_{0} \to 0$

For $\text{\omega} ≫ {\text{\omega}}_{0}$ circuit behaves as inductor.