Source current from parallel branch currents

Step 1: Write all currents as sine functions

• i₁ = 14.14 sin(ωt − 45°)
• i₂ = 28.3 cos(ωt − 60°) = 28.3 sin(ωt − 60° + 90°) = 28.3 sin(ωt + 30°)
• i₃ = 7.07 sin(ωt + 60°)

Step 2: Phasors (rms values, I = Im/√2) with voltage as reference

• I₁ = 10.0∠−45° = 7.07 − j7.07 A
• I₂ = 20.0∠30° = 17.32 + j10.00 A
• I₃ = 5.0∠60° = 2.50 + j4.33 A

Step 3: Source current

I = I₁ + I₂ + I₃ = (7.07 + 17.32 + 2.50) + j(−7.07 + 10.00 + 4.33) = 26.89 + j7.26 A = 27.86∠15.1° A (rms)

Peak value: Im = 27.86 × √2 = 39.4 A

i) Time expression

i = 39.4 sin(ωt + 15.1°) A

(Adding the peak values directly gives the same result: 38.04 + j10.27, magnitude 39.4 A, angle 15.1°.)

ii) Phasor diagram (effective values, V as reference)

            I3 = 5 A (60°)
           /
          /       I2 = 20 A (30°)
         /     /
        /   /        I = 27.86 A (15.1°)
       /  /    _ -
      / / _ -
  ---O-------------------------► V (0°)
      \
        \
          \  I1 = 10 A (−45°)

I₂ and I₃ lead V, and I₁ lags V by 45°. The resultant I leads V by about 15.1°, so the circuit is net capacitive.