Susan Brown

[Susan Brown]

First of all, the best way to know how to do this problem is to re-watch the video at the end of Unit 3. We step you through it, although the resistances are a little different than in the midterm.

Let me know if there’s a step that you don’t understand, and I’m happy to elaborate or explain in a different way. You could tell me the time at which you don’t understand something (for example, “I didn’t follow what he did at the 5:30 point of the video.”)

As far as voltage goes, if you are talking about calculating circuit current, then the value for “E” is the source voltage. You know that by looking at the drawing of the circuit (is it L1-N or L1-L2?).

[Susan Brown]

Hi Brandon,

First note that I moved your topic the the correct Forum. I also renamed the topic because we prefer people talk more generally about the concept, rather than ask for help on a quiz or exam question.

We’ve stepped some other students through this already. Please see the topic below:

https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/calculating-equivalent-resistance/

You should also know the rule of thumb, which we give in unit 5. Do you know what it is? It was one of the quiz questions, for that unit, too.

[Susan Brown]

Glad you figured it out!

[Susan Brown]

Great question. You are asking why the equivalent resistance of loads in parallel is less than the smallest individual resistance.

This is all from the perspective of the power supply. Be Zen-like and become the power supply, which has constant voltage, and think about the affect that having multiple paths for current to flow (parallel circuits vs. just one series circuit) has on the number of electrons you can push out (current). You can push a lot more out when there are multiple branches, even though each branch has a load.

The best way to think this through is to draw out a couple of circuits on paper and play with the calculations.

The first one has two loads in parallel, R1 and R2. Assume a 120vac circuit, so L1 and N. Assign easy values to R1 and R2, say 10 ohms and 20 ohms. Then calculate the current flowing through each load. (I = E/R, and remember in parallel circuits each load gets the full 120 vac.) The total current draw from L1 would be the sum of those two different currents.

Now, let’s think about the “equivalent resistance” of this scenario. The equivalent resistance is taking the resistance of loads in parallel and theoretically combining them into a single load. So, do the math and come up with the equivalent resistance based on the loads above. Now use I = E/R to calculate the current draw from L1. Should be the same as what you came up with before.

Since current is inversely proportional to resistance, for the same amount of current to flow through our theoretical one load that was flowing through the two branches/loads in parallel, the single load, or equivalent load, would have to have a lower resistance.

Does that help?

[Susan Brown]

Hi Bill,

Each quiz and exam is set to automatically allow two attempts, regardless of your score on the first one, but it is up to you if you need to use the second attempt. So, in this case, you’ve already scored 100%, so you can just move on to the next unit. But if you take a quiz later and don’t score 100%, you can study some more, ask a question in the forum, etc., then retake the quiz to improve your score.

Does that answer your question?

[Susan Brown]

Yes – that’s it! I’m going to hide parts of your answers here so we don’t just give it away to others who read this.

[Susan Brown]

You’re welcome! If you are ever having trouble with any of the course material, please post a question. We like to answer them!

[Susan Brown]

Sure!

So – you see that the V for each load would have to be the voltage drop for that particular load, not the supply voltage. That can be calculated, but it’s faster to just use the P=I²x R formula.

[Susan Brown]

Hi Gregory,

P=V²/R can work, but what would you use for V for each load?

Remember – the supply voltage is the total amount being dropped over all loads in the circuit.

P.S. let me know if you need a reset on a quiz!

[Susan Brown]

Hi Dave,

Some good troubleshooting going on there!

Sounds like you were experiencing what we call the “hum-click Cha Cha”

Overheating can definitely cause internal damage in the compressor – the varnish on the windings can break down (which can allow a current leak to ground), the oil in the compressor can burn/break down, etc.

If you’re certain you used the correctly-sized OEM relay and overload and it still won’t run, then, yes, by process of elimination the compressor is bad.

There are ways you can do some more detailed testing on the compressor, if you want to play around with it, but it’s not diagnostically necessary.

[Susan Brown]

It’s fine to ask!

We don’t have a target date yet for that. Will be next year sometime, though. It takes awhile to put new courses together!

[Susan Brown]

In our opinion, you can’t charge for not being able to diagnose something correctly/fully.

Most of the time you will be able to do an accurate diagnosis – especially as you gain experience. But occasionally you may end up not diagnosing it correctly and lose money on the parts. Sometimes this situation happens through no fault of your own – repairing one problem reveals another that you had no way of testing until the first fault was fixed, for example – but that’s one of the costs of doing business. It shouldn’t happen too often, but it’s important that your customers are given accurate quotes up front.

Do you charge a diagnostic fee?

[Susan Brown]

Part of the challenge of this question is the funky way some of the lines are drawn. They have no meaning to the electrons! Everything comes down to is it a wire/switch or a load? Connections are not a load. Make sense?

[Susan Brown]

Glad it helped!

In a series circuit, the current is the same throughout the circuit. So the current is calculated from the *total* resistance and the voltage supply. In this case, 37 ohms and 240vac (using I = E/R)

Once you have that current, you can then calculate the heat generated by an individual load using P = I²R

[Susan Brown]

Hi John,

I’ll give you a (big) hint: the water that results from defrosting the evaporator is called “condensate”.

I know it’s referred to in the last video in that unit.

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