Delta load phase currents, power, line currents

From the figure: a delta load supplied from lines a, b and c at 100 V. The three arms are:

• Arm a–c: 8 Ω resistor in series with a 6 Ω inductive reactance, Z_ac = 8 + j6 = 10∠36.87° Ω
• Arm a–b: 6 Ω resistor in series with an 8 Ω inductive reactance, Z_ab = 6 + j8 = 10∠53.13° Ω
• Arm c–b: 4 Ω resistor in series with a 3 Ω capacitive reactance, Z_cb = 4 − j3 = 5∠−36.87° Ω

Assumptions: balanced 100 V line voltages with sequence a → b → c: V_ab = 100∠0°, V_bc = 100∠−120°, V_ca = 100∠120° V.

(i) Phase currents

• I_ab = 100/(6 + j8) = 10 A ∠−53.13° = 6.00 − j8.00
• I_bc = 100∠−120°/(4 − j3) = 20 A ∠−83.13° = 2.39 − j19.86
• I_ca = 100∠120°/(8 + j6) = 10 A ∠83.13° = 1.20 + j9.93

(ii) Total power

• P_ab = 10² × 6 = 600 W
• P_bc = 20² × 4 = 1600 W
• P_ca = 10² × 8 = 800 W

Total power = 3000 W = 3 kW (reactive power per arm: 800 VAR, −1200 VAR and 600 VAR, so the net is 200 VAR lagging)

(iii) Line currents

• I_a = I_ab − I_ca = 4.80 − j17.93 = 18.56 A (∠−75.0°)
• I_b = I_bc − I_ab = −3.61 − j11.86 = 12.39 A (∠−106.9°)
• I_c = I_ca − I_bc = −1.19 + j29.79 = 29.81 A (∠92.3°)

(Their sum is zero.) The line current magnitudes depend on the phase sequence assumed, whereas the phase currents and total power do not.