Two-phase generator with neutral: Two 120 V (rms) phases at 90° feed two 90 Ω resistors; a common neutral is present. What current flows in the neutral conductor?

Introduction / Context:
Neutral current in multi-phase systems depends on both magnitude and phase angle of the branch currents. For a two-phase system with 90° displacement (historical 2-φ systems), balanced resistive loads produce equal-magnitude currents that are orthogonal in phase, so their vector sum in the neutral is the diagonal of a right triangle.

Given Data / Assumptions:

• Two-phase source: V1 and V2 are 120 V (rms) each and 90° apart.
• Each phase drives a 90 Ω resistor (purely resistive).
• Common neutral returns the sum of the two phase currents.

Concept / Approach:

Current in each phase: I = V / R = 120 / 90 = 1.333… A. The two currents are separated by 90°. The neutral current magnitude equals the vector (phasor) sum: I_N = √(I^2 + I^2) = I * √2.

Step-by-Step Solution:

Verification / Alternative check:

Geometric interpretation: forming a right triangle with legs 1.333… A yields a hypotenuse ≈ 1.88 A. Power per phase = I^2 * R ≈ (1.333…^2 * 90) = 160 W; the system total ≈ 320 W, consistent.

Why Other Options Are Wrong:

1.33 A ignores vector addition. 2.66 A assumes simple magnitude sum. 1.77 A is a miscalculation of √2 scaling.

Common Pitfalls:

Using arithmetic sums instead of phasor sums; mixing degrees with radians; rounding errors.

Final Answer:

1.88 A