Question

In a series $LC$ circuit, the applied voltage is $V_0$. If $\omega$ 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. $V_L=-V_C=\frac{V_0}{2}$

Answer 1

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

$X_L=\omega L$
If $\omega$ is very low, then $X_L=0$
$\therefore V_L=0$
or $V=V_C=V_0$