Oklahoma Electrical Practice Exam

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Prepare for the Oklahoma Electrical Exam. Use flashcards and multiple choice questions, each accompanied by explanations and strategies. Be exam-ready!

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What is the full load current rating for a single phase, 1.5HP motor at 120V with 90% efficiency and 85% power factor?

10.5 amps
11.5 amps
12.2 amps
13.5 amps

The correct answer is: 12.2 amps

To determine the full load current rating for a single-phase, 1.5 HP motor at 120V with given efficiency and power factor, we can use the formula that relates horsepower to current, voltage, efficiency, and power factor. First, we need to convert horsepower to watts, since 1 HP is equivalent to approximately 746 watts. Therefore: 1.5 HP = 1.5 × 746 = 1119 watts. Next, we need to account for the efficiency of the motor. The input power can be calculated by dividing the output power (in watts) by the efficiency (expressed as a decimal): Input Power = Output Power / Efficiency = 1119 W / 0.90 = 1243.33 watts. Next, we need to consider the power factor to find the apparent power (in volt-amperes) that the motor draws: Real Power (P) = Apparent Power (S) × Power Factor (pf). This can be rearranged to find the apparent power: S = P / pf = 1243.33 W / 0.85 = 1460.39 volt-amperes. Finally, use the formula to find the full load current (I)