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

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:

Verification / Alternative check:

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: