Skip to main content

Star load (8+j6) line, phase current

Short answer

Impedance 8+j6 Ω = 10∠36.87° Ω, phase voltage 230.9 V. Phase current = line current = 23.09 A. Total power = 3 × 23.09² × 8 = 12.8 kW.

A balanced star-connected load of (8 + j6)Ω per phase is connected to a balanced 3 phase 400V, 50Hz supply. Find the line current, phase current and total power consumed. Take RYB phase sequence.

Given a balanced star load Z_ph = 8 + j6 = 10∠36.87° Ω, V_L = 400 V, 50 Hz, RYB sequence.

  • V_ph = 400/√3 = 230.9 V
  • Phase voltages: V_R = 230.9∠0°, V_Y = 230.9∠−120°, V_B = 230.9∠+120° V

Currents

I_R = V_R/Z = 23.09∠−36.87° A

I_Y = 23.09∠−156.87° A

I_B = 23.09∠+83.13° A

  • Phase current = line current = 23.09 A (for star)

Power

  • Power factor = 8/10 = 0.8 lagging
  • Total power P = 3 I_ph² R = 3 × 23.09² × 8 = 12 800 W = 12.8 kW

Check: √3 V_L I_L cos φ = 1.732 × 400 × 23.09 × 0.8 = 12.8 kW. Also Q = 9.6 kVAR, S = 16 kVA.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Basic Electrical Engineering chapter 8 questions →