Parallel RL phasor addition: In a parallel RL circuit with 3 A rms in the resistive branch and 3 A rms in the inductive branch, what is the total rms line current?

Introduction / Context:In AC analysis, currents in different branches can be out of phase. For a parallel RL, the current through the resistor is in phase with voltage, while the current through the inductor lags the voltage by 90°. Total current is found by vector (phasor) addition, not simple arithmetic sum.

Given Data / Assumptions:

• I_R = 3 A rms (in phase with V).
• I_L = 3 A rms (lags V by 90°).
• Parallel RL, both branches share the same applied voltage.

Concept / Approach:Because I_R and I_L are 90° apart, the magnitude of the total current is the square root of the sum of squares: I_total = sqrt(I_R^2 + I_L^2). This is basic orthogonal phasor addition.

Step-by-Step Solution:I_total = sqrt( I_R^2 + I_L^2 )I_total = sqrt( 3^2 + 3^2 )I_total = sqrt( 9 + 9 ) = sqrt( 18 )I_total ≈ 4.2426 A ≈ 4.24 A

Verification / Alternative check:Phasor diagram: one current on the real axis (3 A), one on the negative imaginary axis (3 A). The hypotenuse length is 3 * sqrt(2) ≈ 4.24 A. This confirms the computed value.

Why Other Options Are Wrong:

• 6 A: Simple scalar sum ignores the 90° phase difference.
• 424 mA: Off by a factor of 10.
• 42.4 A: Off by a factor of 10 in the other direction.

Common Pitfalls:Adding AC currents arithmetically when they are not in phase, or confusing series and parallel behavior. Always use phasor addition for orthogonal components.

Final Answer:4.24 A