Understanding Feeder Protection for Electric Motors

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Learn how to calculate the recommended feeder protection for various electric motors using the National Electrical Code guidelines, ensuring the safety and reliability of your electrical systems.

When you’re gearing up to tackle your HOLT Electricians exam, one of the burnished nuggets of knowledge you’ll want stashed away is how to calculate feeder protection for electric motors. The designations—like the 25-horsepower 208-volt three-phase motor and the trio of 3-horsepower 120-volt motors—can seem overwhelming at first. But once you break it down, it’s all quite manageable. So, let’s break it down, step by step!

What is Feeder Protection, Anyway?

First off, let’s toss the term "feeder protection" into the air. What does it really mean? In simple terms, feeder protection relates to the safeguards we put in place to protect motors and electrical systems from overloads or faults. Think of it like a guardian angel watching over your equipment, ensuring everything runs smoothly and safely.

The Calculation Craze

Now, let’s dive right into those calculations. We’re tasked with figuring out the recommended feeder protection for our friendly little duo of electric motors. To do this, we’ll follow some straightforward formulas. For the big fish—the 25-horsepower motor—here's the general formula we're working with:

[ I = \frac{HP \times 746}{\sqrt{3} \times V \times pf} ]

We're looking at a 25-horsepower motor operating at 208 volts. Let’s assume the power factor ( pf ) is 1, because why complicate things right now? Using that formula, we plug in the numbers:

[ I \approx \frac{25 \times 746}{\sqrt{3} \times 208} \approx 68.6 \text{ Amps} ]

Getting to Know the Smaller Motors

Of course, we also have three smaller companions—you guessed it: the 3-horsepower motors. Each operates at 120 volts, and we can use the same formula to find their full-load current:

[ I_{single} = \frac{3 \times 746}{120} \approx 18.65 \text{ Amps} ]

Multiplying that by three gives us:

[ I_{total} = 3 \times 18.65 \approx 55.95 \text{ Amps} ]

Add it All Up

Alright, now it’s time for the grand total! We’ll add the full-load current for all the motors together:

[ I_{total} = 68.6 + 55.95 \approx 124.55 \text{ Amps} ]

Before you start sweating the small stuff, remember that electrical codes dictate that we provide a safety margin. According to the National Electrical Code (NEC), it's smart to size up! That translates to recommending a feeder protection size.

Choosing the Right Feeder Protection

Let’s look at our options:

A. 200 ampere
B. 225 ampere
C. 250 ampere
D. 300 ampere

Based on our calculated total load, we lean towards 225 ampere as the optimal choice. This ensures you’re protected while leaving just enough wiggle room for safety concerns. Remember, it’s all about that cozy feeling of knowing your system is safeguarded from potential overload.

Wrapping It Up

So there you have it! When you take a methodical approach to calculating the current draw and applying NEC guidelines, tackling even the more complex problems on your HOLT Electricians exam can feel less like climbing a mountain and more like a pleasant stroll. Just don’t forget, those pesky calculations can save your bacon when working in the field. Stay sharp, and good luck with your studies!

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