Parallel RL total current: A 3.3 kΩ resistor and a 120 mH inductor are connected in parallel across a 2 kHz, 12 V AC source. What is the total RMS current drawn from the source?

Introduction / Context:This problem blends branch calculations in a parallel RL circuit with phasor addition of currents that are not in phase. It is a practical exercise in using both magnitude and phase information to get the total current.

Given Data / Assumptions:

• R = 3.3 kΩ, L = 120 mH.
• Source: 12 V RMS at f = 2 kHz.
• Ideal inductor; line frequency steady state.

Concept / Approach:Calculate branch currents: IR = V / R (in phase with V), IL = V / XL (lagging by 90°) where XL = 2 * pi * f * L. The total current is the vector sum: I = sqrt(IR^2 + IL^2) because the two branches are orthogonal in phase for ideal RL.

Step-by-Step Solution:

Verification / Alternative check:Admittance method: Y = 1/R − j(1/XL) = 0.000303 − j0.000663 S, |Y| ≈ 0.000728 S ⇒ IT ≈ 12 * 0.000728 ≈ 8.74 mA (same result).

Why Other Options Are Wrong:

• 874 mA and 8.74 µA: Scale errors by factors of 100 or 1000.
• 874 µA: Off by exactly 10×.
• 3.64 mA: Resistor branch only; ignores inductive branch.

Common Pitfalls:

• Treating branch currents as algebraic sums instead of vector sums.
• Neglecting the 90° phase difference in an ideal inductor branch.

Final Answer:8.74 mA