What is the amp-load for a three-phase, 0.5-hp, 230-volt motor with an efficiency of 92% and a power factor of 80%?

What is the amp-load for a three-phase, 0.5-hp, 230-volt motor with an efficiency of 92% and a power factor of 80%?

• A. 1.27 amps
• B. 1.19 amps
• C. 2.55 amps
• D. 4.41 amps

Answer: (A)

Three-phase power ties current to both voltage and power factor, with input power related to current by P_in = √3 × V_L × I × PF. Since the motor’s 0.5 hp is the mechanical output, the electrical input is P_in = P_out / efficiency. Convert horsepower to watts: 0.5 hp × 746 W/hp = 373 W. Then P_in = 373 / 0.92 ≈ 405.4 W. Using V = 230 V and PF = 0.80, the current is I = P_in / (√3 × V × PF) ≈ 405.4 / (1.732 × 230 × 0.80) ≈ 1.27 A. So the amp-load is about 1.27 A. If you skip efficiency or misapply the three-phase formula, you’d get noticeably different values, which is why applying both efficiency and PF correctly gives the right result.

Three-phase power ties current to both voltage and power factor, with input power related to current by P_in = √3 × V_L × I × PF. Since the motor’s 0.5 hp is the mechanical output, the electrical input is P_in = P_out / efficiency. Convert horsepower to watts: 0.5 hp × 746 W/hp = 373 W. Then P_in = 373 / 0.92 ≈ 405.4 W. Using V = 230 V and PF = 0.80, the current is I = P_in / (√3 × V × PF) ≈ 405.4 / (1.732 × 230 × 0.80) ≈ 1.27 A. So the amp-load is about 1.27 A. If you skip efficiency or misapply the three-phase formula, you’d get noticeably different values, which is why applying both efficiency and PF correctly gives the right result.