Parallel RL across 15 V AC: A 470 Ω resistor and an inductor with 125 Ω inductive reactance are connected in parallel across a 15 V (rms) source. What is the rms current through the inductor branch?

Introduction / Context:
This problem checks your ability to compute branch current in an AC parallel RL network using reactance. Because the branches are in parallel, each branch sees the full source voltage, and currents are determined independently by their branch impedances.

Given Data / Assumptions:

• AC source: 15 V (rms).
• Resistive branch: R = 470 Ω (not needed for the inductor current).
• Inductive branch: XL = 125 Ω.
• Ideal components; frequency is such that the given reactance applies.

Concept / Approach:
For any single reactive branch, the magnitude of branch current is I = V / |Z_branch|. For a pure inductor branch, |Z| = XL. Since the branches are in parallel, the inductor sees the full 15 V (rms).

Step-by-Step Solution:
Given: V_rms = 15 V, XL = 125 ΩI_L = V_rms / XLI_L = 15 / 125 = 0.12 AI_L = 120 mA (rms)

Verification / Alternative check:
If instead you computed total current, you would need phasor addition of the resistor and inductor currents. However, the question asks only for the inductor branch current, which is simply V / XL = 120 mA.

Why Other Options Are Wrong:

• 152 mA, 32 mA, 12 mA: Do not match V / XL for 15 V and 125 Ω.

Common Pitfalls:
Confusing parallel with series behavior, or accidentally using the resistor value or total impedance instead of the inductor’s reactance for the branch current.

Final Answer:
120 mA