For three-phase systems, which formula gives current when horsepower is known?

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Multiple Choice

For three-phase systems, which formula gives current when horsepower is known?

Explanation:
In three-phase systems, the current is found from how electrical input power relates to mechanical output power through the three-phase power equation and efficiency. The mechanical power in horsepower converts to watts as HP × 746. The electrical input power must be higher by a factor of 1/η due to efficiency, so P_in = HP × 746 / η. Real power in a three-phase system is P_in = √3 × V × I × PF, where V is line voltage and PF is the power factor. Solving for the current gives I = P_in / (√3 × V × PF) = (HP × 746 / η) / (√3 × V × PF) = HP × 746 / (V × η × PF × √3). Since √3 ≈ 1.73, this becomes I = HP × 746 / (Volts × Efficiency × PF × 1.73). This matches the correct form. The other arrangements either place Volts in the numerator, omit dividing by efficiency or PF, or invert the relationship, which does not align with the fundamental power balance in a three-phase system.

In three-phase systems, the current is found from how electrical input power relates to mechanical output power through the three-phase power equation and efficiency. The mechanical power in horsepower converts to watts as HP × 746. The electrical input power must be higher by a factor of 1/η due to efficiency, so P_in = HP × 746 / η. Real power in a three-phase system is P_in = √3 × V × I × PF, where V is line voltage and PF is the power factor. Solving for the current gives I = P_in / (√3 × V × PF) = (HP × 746 / η) / (√3 × V × PF) = HP × 746 / (V × η × PF × √3). Since √3 ≈ 1.73, this becomes I = HP × 746 / (Volts × Efficiency × PF × 1.73).

This matches the correct form. The other arrangements either place Volts in the numerator, omit dividing by efficiency or PF, or invert the relationship, which does not align with the fundamental power balance in a three-phase system.

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