Question

In a series $LC$ circuit, the applied voltage is $V_0$. If frequency is very low, then the voltage drop across the inductor $V_L$ and capacitor $V_C$ are

• A. $V_L=\frac{V_0}{2},V_C=\frac{V_0}{2}$
• B. $V_L=0,V_C=V_0$
• C. $V_L=V_0,V_C=0$
• D. $V_L=0,V_C=\frac{V_0}{2}$

Answer 1

The correct option is B $V_L=0,V_C=V_0$

For very low frequency,

$X_L=\omega L=2\pi fL\approx 0$ and

$X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}\to \infty$

So, capacitor acts like an open circuit and inductor behaves like a zero resistance wire.

So, $V_L=0$ and $V_C=V_{source}=V_0$

Hence, option $\left(B\right)$ is correct.