AC inductive reactance: how much current flows through a 0.02 H choke connected to a 12 V (rms), 100 Hz AC source?

Difficulty: Medium

Correct Answer: 0.955 A

Explanation:


Introduction / Context:
An inductor resists changes in current with a frequency-dependent reactance XL. In steady-state AC, the current magnitude through an ideal inductor is I = V / XL. This problem tests the ability to compute XL and the resulting rms current for given L, f, and V.


Given Data / Assumptions:

  • Inductance L = 0.02 H.
  • Frequency f = 100 Hz.
  • Applied voltage (rms) V = 12 V.
  • Ideal inductor; ignore winding resistance.


Concept / Approach:

  • Inductive reactance: XL = 2 * π * f * L.
  • Current magnitude: I = V / XL.


Step-by-Step Solution:

Compute XL = 2 * π * 100 * 0.02 = 2 * π * 2 = 4 * π ≈ 12.566 Ω.Compute I = 12 / 12.566 ≈ 0.955 A (rms).Therefore, the current is approximately 0.955 A.


Verification / Alternative check:

Sanity check: If L were halved, XL halves and current doubles; numbers scale as expected.


Why Other Options Are Wrong:

  • 0.02 A: Would imply XL = 600 Ω, inconsistent with given f and L.
  • 10 A and 2.02 A: Far too high; would require very small XL not supported by L and f.
  • None of the above: Incorrect because 0.955 A matches calculation.


Common Pitfalls:

  • Confusing 2 * π * f * L with 2 * π * L / f (incorrect).
  • Mixing peak and rms values; V given is rms.


Final Answer:

0.955 A

More Questions from Electronic Principles

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion