A 24 V peak sinusoidal source drives a 900 µH inductor at f = 22 kHz. What is the reactive power (inductive VAR) drawn by the inductor?

Introduction / Context:Reactive power quantifies the non-dissipative exchange of energy between reactive components and the source. For a pure inductor, average real power is zero, but reactive power Q is nonzero and depends on the RMS voltage and inductive reactance XL. This question requires converting peak to RMS and applying the correct formula for Q.

Given Data / Assumptions:

• Vp = 24 V (sinusoidal peak).
• f = 22 kHz, L = 900 µH.
• Ideal inductor (no resistance), sinusoidal steady state.

Concept / Approach:Reactive power for a purely inductive load is Q = Vrms^2 / XL, where XL = 2 * π * f * L. Also Vrms = Vp / sqrt(2) for a sine wave. Compute XL, then compute Q. Units are VAR (volt-ampere reactive).

Step-by-Step Solution:Vrms = 24 / sqrt(2) ≈ 16.97 V.XL = 2 * π * 22,000 * 900e-6 ≈ 124.4 Ω.Q = Vrms^2 / XL = (16.97^2) / 124.4 ≈ 288.0 / 124.4 ≈ 2.31 VAR.Rounded to two significant figures: Q ≈ 2.3 VAR.

Verification / Alternative check:Compute current magnitude: I = Vrms / XL ≈ 16.97 / 124.4 ≈ 0.136 A. Then Q = Vrms * I ≈ 16.97 * 0.136 ≈ 2.31 VAR, consistent with the first method.

Why Other Options Are Wrong:

• 4.6 VAR and 9.3 VAR: Would correspond to either double or quadruple the correct current/voltage relationships.
• 0 VAR: Only true for zero reactive components; a pure inductor at AC has nonzero Q.
• 1.1 VAR: Roughly half the correct value, inconsistent with calculations.

Common Pitfalls:

• Using peak voltage directly in Q = V^2 / XL; must use RMS.
• Confusing real power P with reactive power Q; for an ideal inductor, P = 0 while Q ≠ 0.

Final Answer:2.3 VAR