For a single-phase motor, which expression correctly gives current from horsepower, voltage, efficiency, and power factor?

For a single-phase motor, which expression correctly gives current from horsepower, voltage, efficiency, and power factor?

• A. I = HP × 746 / Volts × Efficiency × Power Factor
• B. I = HP × 746 × Volts / (Efficiency × Power Factor)
• C. I = Volts × Efficiency × Power Factor / (HP × 746)
• D. I = (HP + 746) / (Volts × Efficiency × Power Factor)

Answer: (A)

The current is found from how electrical power, efficiency, and power factor relate to the motor’s mechanical output. A motor’s mechanical power is HP × 746 watts. Because efficiency equals output power divided by input power, the electrical input power is P_in = (HP × 746) / efficiency. Real power also equals P_in = V × I × pf, so solving for current gives I = P_in / (V × pf) = (HP × 746) / (V × efficiency × pf). So the correct expression is I = (HP × 746) / (Volts × Efficiency × Power Factor). The other forms misplace factors or mix terms incorrectly—for example, putting Volts in the numerator or adding HP and 746 together—so they don’t reflect how input power, voltage, and pf determine current.

The current is found from how electrical power, efficiency, and power factor relate to the motor’s mechanical output. A motor’s mechanical power is HP × 746 watts. Because efficiency equals output power divided by input power, the electrical input power is P_in = (HP × 746) / efficiency. Real power also equals P_in = V × I × pf, so solving for current gives I = P_in / (V × pf) = (HP × 746) / (V × efficiency × pf).

So the correct expression is I = (HP × 746) / (Volts × Efficiency × Power Factor).

The other forms misplace factors or mix terms incorrectly—for example, putting Volts in the numerator or adding HP and 746 together—so they don’t reflect how input power, voltage, and pf determine current.