New York State Master Electrician Practice Exam

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What is the required fuse size for a 100HP DC motor at 220V with 90% efficiency?

  1. 400 A

  2. 500 A

  3. 600 A

  4. 700 A

The correct answer is: 600 A

To determine the required fuse size for a 100HP DC motor operating at 220V with 90% efficiency, it’s essential to first convert the motor's horsepower to watts. 1 horsepower is equivalent to approximately 746 watts. Therefore, a 100HP motor translates to: 100 HP × 746 W/HP = 74,600 watts. Since the motor has an efficiency of 90%, the actual power that it consumes from the power supply is calculated by accounting for the efficiency rate: Power input = Output Power / Efficiency Power input = 74,600 W / 0.90 = 82,888.89 watts. Next, we convert this power consumption into current using the formula for power in a DC circuit: Power (watts) = Voltage (volts) × Current (amps) Rearranging this gives us: Current (amps) = Power (watts) / Voltage (volts) Substituting in the values: Current = 82,888.89 W / 220 V = 376.77 A. Fuses should be selected with a safety margin, typically around 125% of the calculated continuous current. This ensures that the fuse will not blow under normal operating conditions

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