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Resultant of two alternating currents

Short answer

Adding phasors: I1=7∠0° A, I2=10∠90° A gives resultant peak magnitude √(7²+10²)=12.21 A at angle 55.0°. Resultant current equation: i=12.21 sin(ωt+55.0°) A. RMS value: 12.21/√2=8.63 A.

Two alternating currents represented by the equations i1 = 7sinωt and i2 = 10sin(ωt + π/2) are fed into a common conductor. Find the equation for the resultant current and its RMS value.

Given i₁ = 7 sin ωt and i₂ = 10 sin(ωt + π/2) = 10 cos ωt.

Phasor addition (peak values)

  • I₁ = 7∠0° = 7 + j0
  • I₂ = 10∠90° = 0 + j10
  • I = I₁ + I₂ = 7 + j10

Magnitude: Im = √(7² + 10²) = √149 = 12.21 A

Phase angle: φ = tan⁻¹(10/7) = 55.0° = 0.960 rad

Resultant current: i = 12.21 sin(ωt + 55.0°) A

RMS value

I_rms = Im / √2 = 12.21 / 1.4142 = 8.63 A

Check: I₁rms = 4.95 A, I₂rms = 7.07 A, and √(4.95² + 7.07²) = 8.63 A (the two currents are in quadrature).

✓ Answer checked by Claude Opus 5.5 (high effort)

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