alleryStart learning →The IMPEDANCE of a pure inductor at angular frequency $\omega$ is
A$\dfrac{1}{j\omega L}$
B$L / \omega$
C$\omega L$
D$j\omega L$
Answer & Solution
Correct answer: D. $j\omega L$
1. Inductor impedance: $Z_L = j\omega L$.
2. Magnitude: $|Z_L| = \omega L$. Phase: $+90^\circ$ (current LAGS voltage).
3. As $\omega \to 0$ (DC): $|Z_L| \to 0$ — short circuit (matches DC steady state).
4. As $\omega \to \infty$: $|Z_L| \to \infty$ — open circuit (opposes rapid changes).
5. Option A is the CAPACITOR formula. Options B, C miss the $j$.
_Source: Tony Kuphaldt, "Lessons in Electric Circuits — AC", Vol II, Ch 3 (Inductive reactance)._
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