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