Susan Brown

[Susan Brown]

You are pretty much getting it. With the compressor off, the windings are in parallel, and have resistances of 4.4 ohms and 6.25 ohms. The condenser motor is over 1000. So, the equivalent resistance of the motor and the compressor will not be much affected by the condenser fan resistance.

If you do the 1/[(1/4.4) + (1/6.25)] formula you get Req of the windings of about 2.5 ohms.

We have an answer choice for that question that is close to that.

[Susan Brown]

^2 is another way to indicate “squared”. It’s not always easy to know how to do the superscript 2

So P=I^2xR is the same as P=IxIxR

[Susan Brown]

Hi There – I just reset you.

[Susan Brown]

Hi Giovanni,

I just took at look at your scores so far in the course. Looking good!

You are correct above. There’s just a small problem with your answer, which was “37.5 A”

A stands for Amps, which is current, not resistance.

37.5 is the correct value… you just used the wrong units.

[Susan Brown]

That was a glitch in the sharing/recording technology used for the webinar – it froze and didn’t show his “pointer” and what he was drawing until that little jump when it suddenly showed up. I think you can piece together what he was pointing out. There wasn’t a different slide or anything. But, if there’s something that isn’t clear, please ask another question and we’re glad to help.

[Susan Brown]

Hi Eric,

The mistake with that calculation is that it treats the circuit as if the loose connection is the only resistance in the circuit and thus all of the supply voltage is dropped across it. The element is still there, in series with the loose connection, and has a higher resistance, so most of the voltage will drop across it.

To use P=E^2/R, E has to be the voltage dropped across the R in question. You can calculate that, but here’s what I think is a more straightforward way (and the one we show in the video at the end of Unit 3).

1. Calculate the circuit current (which is determined by the voltage supply and the total resistance in the circuit)
2. Now that you know the current going through the loose connection, you can calculate the heat it generates by using P=I^2 x R.

[Susan Brown]

Hi Hassan – I had sent you an email about this. Did you receive it?

Also, for future reference, the Quiz and Exam Reset Request form is in the Campus Support menu.

[Susan Brown]

Question #1 – While each model has its own specific temperature that the ice maker needs to be at or below in order to function, we gave you a range of what these temps generally are. Based on that, which of the following might be one of these “cutoff” temperatures?

From Module 4, unit 1:

The key thing with ice makers is temperature. If the ice maker senses a temperature in its compartment above a certain amount, it will shut itself off until it senses the correct temperature. This will be some exact temperature, but which temperature it is varies by model. It generally falls somewhere between 8 – 17 degrees F.

It is not a trick question. Each refrigerator out there has a particular cutoff temperature. For example, my refrigerator might have a cutoff temperature of 13 degrees, and yours might have a cutoff temperature of 16 degrees.

We are expecting you to select the answer that is within that range we gave you.

(The Kleinert book is a useful reference, but we have found inaccuracies or outdated info here and there. Please give more weight to the information we give you in the course.)

I reset you so you can retake the quiz.

• This reply was modified 4 years ago by Susan Brown.

[Susan Brown]

Shaded pole is the answer.

[Susan Brown]

Haha – no worries! Most of us don’t deal with negative numbers that often – you can forget!

[Susan Brown]

Hi Jeffrey,

I’ll start with your last question – you are a student. I’m not sure what the significance of the “Spectator” designation is – I never pay attention to those! As long as you are able to post topics here, you are fine.

Our Appliantology Admin was out of town for a few days over the weekend, and I know he is now catching up on Membership requests. Did he communicate with you at all about your application? I notice you used a different email address than the one on your MST account.

[Susan Brown]

Hi Carl,

Yes! You are on the list for the next batch of Certificates to be printed. I expect that batch to be in the mail sometime within the next two weeks.

Great job!

[Susan Brown]

I assume you got my email last night that you passed the Exam, and thus your course?

Good job!

[Susan Brown]

Hmmm – a “Womanometer” sounds like one of the Samurai’s jokes. Which quiz is that on?

This is true: “the Samurai said that its important to use a meter with a high impedance input…”

We discuss that in Basic Electricity, Unit 7 in the “Loading vs. non-Loading meters” section. I don’t believe it is covered anywhere in Kleinert’s book.

[Susan Brown]

Hi Ben,

This is just one of those wording things. You have the right ideas.

“In order for something to have a voltage drop, it must have…” and current is the correct answer.

In other words, current flow is a necessary condition if there is to be any voltage drop.

It’s not saying that current flow through anything creates a voltage drop. As you point out, current through a closed switch won’t have voltage drop. But we’re not implying that with the question.

We’re not trying to be tricky, but to encourage thought. One reason we emphasize the distinction between voltage and voltage drop so much is that too many techs focus on voltage readings without paying attention to current/power. This mindset helps to avoid making that mistake. And that’s why this question is encouraging you to zero in on “current flow” as the most important causative factor of voltage drop.

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