Q.An applied e.m.f. signal consists of superposition of a d.c. source and an a.c. source of high frequency. The circuit consists of an inductor L and a capacitor C in series. Show that d.c. signal appears across C and a.c. signal appears across L.

Reactances $X_L=\omega L$ and $X_C=1/(\omega C)$ behave oppositely with frequency — this naturally separates d.c. onto C and high-frequency a.c. onto L.

Let the applied signal be $v = V_{dc} + V_0\sin\omega t$ (a d.c. term plus a high-frequency a.c. term), applied across an inductor $L$ and capacitor $C$ connected in series.

For the d.c. component ($\omega = 0$):

• Inductive reactance: $X_L = \omega L = 0$ — an ideal inductor behaves like a plain connecting wire for steady d.c.; it drops no voltage.
• Capacitive reactance: $X_C = \dfrac{1}{\omega C} \to \infty$ — in the steady state a capacitor carries no d.c. current at all (it charges up and then blocks further d.c. flow), so it behaves like an open circuit for d.c., and the entire d.c. voltage appears across C.

For the high-frequency a.c. component ($\omega$ large):

• Inductive reactance: $X_L = \omega L$ becomes very large.
• Capacitive reactance: $X_C = \dfrac{1}{\omega C}$ becomes very small. …