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?
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A6 V
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B0 V
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C12 V
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D4 V
Answer
Correct Answer: 6 V
Explanation
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:
At t = 0+: i = 0 A ⇒ v_R = i * R = 0 V.KVL: v_source = v_R + v_L ⇒ 6 V = 0 + v_L.Thus, v_L(0+) = 6 V.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