Question

Discuss the behaviour of an inductor connected to (i) a dc source, and (ii) a high frequency ac source.

Hint:

Inductors oppose changes in current. Since DC is constant, it passes through effortlessly after an initial transient state. High-frequency AC changes direction incredibly rapidly, creating a massive back-EMF, causing the inductor to act as a barrier.

Answer 1

Concept: The inductive reactance \(X_L\) represents the effective opposition offered by an ideal inductor to the flow of an alternating current. It is mathematically quantified using the formula: \[ X_L = \omega L = 2 \pi f L \] Where \(f\) represents the operating frequency of the applied voltage source and \(L\) is the self-inductance of the inductor.

Step 1: Behavior when connected to a steady Direct Current (DC) source.
For a continuous steady-state Direct Current (DC) source, the current does not change with time, meaning the frequency is exactly zero (\(f = 0\)). Substituting this into our operational formula: \[ X_L = 2 \pi (0) L = 0 \] Since the inductive reactance is zero, an ideal inductor offers zero resistance to a steady DC signal. Once steady state is reached, it behaves exactly like a simple zero-resistance conducting wire (short circuit).

Step 2: Behavior when connected to a high-frequency Alternating Current (AC) source.
For an Alternating Current (AC) source operating at a very high frequency, \(f\) becomes extremely large (\(f \rightarrow \infty\)). Since \(X_L \propto f\), the inductive reactance escalates to a massive value: \[ X_L = 2 \pi f L \gg 0 \] As a direct consequence of this high reactance, the inductor presents a very high opposition to high-frequency AC signals. It effectively chokes or blocks high-frequency currents, acting nearly as an open circuit for high frequencies.