Inductor initial condition: An inductor, a 1 kΩ resistor, and a switch are in series across a 6 V DC source. Exactly at the moment the switch closes, what is the voltage across the inductor?

Introduction / Context:
Inductors oppose sudden changes in current. At the instant a DC source is applied, the current through an ideal inductor cannot change instantaneously, which determines the initial voltage across it.

Given Data / Assumptions:

• Series circuit: DC source 6 V, resistor 1 kΩ, ideal inductor.
• Switch just closed (t = 0+).
• Initial current is zero at the instant of closing.

Concept / Approach:

At t = 0+, the inductor current is still 0 A, so the resistor drop is i * R = 0 V. KVL requires the inductor to drop the entire source voltage initially. More formally, v_L = L * di/dt, which is initially equal to the source since di/dt is finite and the resistor drop is zero.

Step-by-Step Solution:

Verification / Alternative check:

As time increases, current rises toward 6 mA (6 V / 1 kΩ), and the inductor voltage decays to 0 V, consistent with first-order RL behavior.

Why Other Options Are Wrong:

0 V would imply instantaneous current rise. 12 V exceeds the source; no such surge exists in this ideal DC step. 4 V would require a simultaneous 2 V resistor drop at t = 0+, which contradicts i = 0 A.

Common Pitfalls:

Confusing capacitor and inductor initial conditions; an inductor initially behaves like an open circuit for DC steps.

Final Answer:

6 V