HOLT Electricians Practice Exam 2026 - Free Electrician Certification Practice Questions and Study Guide

Question: 1 / 400

What size dual-element fuse is required for a 5 horsepower, 208-volt, 3-phase motor with a service factor of 1.16?

15 ampere

20 ampere

To determine the appropriate size of the dual-element fuse for a 5 horsepower, 208-volt, 3-phase motor with a service factor of 1.16, it's essential to calculate the full-load current of the motor using the formula:

Full-load current (A) = (Horsepower × 746) / (Voltage × √3 × Efficiency)

Since efficiency is not provided, it is common practice to consider a standard efficiency of about 0.9 for industrial motors. By substituting the values, we can proceed with the calculation:

1. Convert horsepower to watts:

5 HP × 746 W/HP = 3730 W

2. Using the formula for 3-phase equipment:

Full-load current = (3730 W) / (208 V × √3)

≈ (3730 W) / (208 V × 1.732)

≈ (3730) / (360.6)

≈ 10.34 A

3. Next, to account for the service factor, the full-load current is multiplied by the service factor of 1.16:

Adjusted current = 10.34 A × 1.16 ≈

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25 ampere

30 ampere

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