A three-phase, 600 V (line-to-line) delta-connected motor is drawing 10 A per phase. What is the approximate total power (in kW) consumed by the motor if its power factor is 0.85 and it has an efficiency of 90%?

Question: A three-phase, 600 V (line-to-line) delta-connected motor is drawing 10 A per phase. What is the approximate total power (in kW) consumed by the motor if its power factor is 0.85 and it has an efficiency of 90%?

Answer options:

• 8.8 kW
• 7.5 kW ✅ 8.3 kW
• 9.8 kW

Correct answer: 8.3 kW

Explanation: For a delta-connected motor, the line current is root-3 times the phase current. However, power calculation uses line values or phase values consistently. Total Apparent Power (S) = Root(3) * V_line * I_line. For a delta connection, I_line = Root(3) * I_phase. So, S = Root(3) * 600V * (Root(3) * 10A) = 3 * 600V * 10A = 18 kVA. Alternatively, using phase values, S = 3 * V_phase * I_phase. In delta, V_phase = V_line = 600V, so S = 3 * 600V * 10A = 18 kVA. Real Power (P) = S * Power Factor = 18 kVA * 0.85 = 15.3 kW. Since the question asks for power consumed by the motor, which implies input power, the efficiency is not applied to this calculation. The motor consumes 15.3 kW, but this option isn't available, suggesting a misinterpretation of 'per phase'. If '10 A per phase' refers to the line current (I_line = 10A), then P = Root(3) * V_line * I_line * PF = Root(3) * 600V * 10A * 0.85 = 1.732 * 600 * 10 * 0.85 = 8820 W or 8.82 kW. Checking the available options, 8.8 kW is the closest, meaning '10 A per phase' was intended as line current. Efficiency is applied to calculate output mechanical power, not input electrical power.