Susan Brown

[Susan Brown]

Hi Matt!

You can get help with this question over at our tech support site, Appliantology. We keep these Student Forums focused on questions specifically relating to the course material.

I can’t remember if you have a professional membership over there or not (if not, read here about our special offer for STA students!). That would give you access to tech-only forums, downloads, etc. It’s the perfect resource for asking tech questions. You can post in the Kitchen Forum without having a tech membership, however.

See you over at Appliantology!

[Susan Brown]

The answer to question 3 is figured out by studying the schematic and seeing which inputs (sensors) are connected to which boards.

I want to avoid giving out quiz answers on the forum, but I’ll just emphasize that to answer 9 and 10, you need to look at the schematic and the board pin-out diagram.

[Susan Brown]

Ah – that’s just a rounding difference!

1/30 is actually 0.033333….

If you happen to use a calculator that stores the values as you go along (rather than using the rounded-off 0.03), then you get 18.75.

FYI – in practical terms, for appliance repair, there’s no significant difference between 18.75 ohms and 20 ohms.

[Susan Brown]

Hi Brandon,

I’m glad you’re enjoying the course so far!

There’s another Forum question where I stepped through calculating equivalent resistance. The resistance numbers were different, but see if this helps:
https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/equivalent-resistance/

Also, the most important thing to remember is the basic rule of thumb: the equivalent resistance in a parallel circuit will be something less than the smallest resistance.

Let me know if you’re able to get the correct answer!

[Susan Brown]

As you discovered, your exam was graded shortly after you posted this! (Congratulations!)

It can take anywhere from 1 to 3 days, usually, for an exam to get graded. It happens to be our busy season right now, so sometimes it might take a little longer.

Thanks, and great job on Fundamentals!

We’ll be sending out your certificate soon.

[Susan Brown]

It may not be stated outright, but hopefully you can figure it out based on what you know so far. DC has polarity (+ or -), because it moves in one direction. AC is constantly switching directions. Can you see why measuring DC voltage will result in a positive or negative reading, depending on where you put your leads, whereas measuring AC will not? (It will just be a number with no + or – sign.)

Does that help you figure out the answer?

[Susan Brown]

Hi Dennis,

Thanks for asking questions!

First question – Sometimes it will be as easy as reading the label on a motor, but the main way that you’ll know what type of power a particular component is getting is when you do a load analysis using the schematic diagram for the appliance. You will be learning how to do this as you continue to go through Fundamentals, so be on the lookout for it!

As for your second question, appliance techs never need to measure DC current, that’s why we had this message above the video: “Here’s a quick video showing the use of a digital multi-meter, to help clarify the reading. The amp reading he does at the end is not something you’ll ever need to do so you can stop watching at about 3:30 into the video – I’ll show you in another video how to use a clamp-on ammeter (a picture of which is on page 95 of the Kleinert text).”

However, if you are asking just to try to understand the difference, it is because DC current moves in one direction only, so in order to read it, the meter has to physically be in the circuit. AC current, on the other hand, is constantly changing direction which creates a moving magnetic field that the clamp-on meter can read. Hopefully, you’ll be able to have a better feel for this after you have learned about transformers and motors, later in the course.

[Susan Brown]

Great! Glad to help.

[Susan Brown]

I just went to look at your quiz, and it was questions 6 and 7 that you missed. 6 is above, here is 7:

Question #7 – In order to calculate the voltage drop across a load in series with another load, we first need to calculate the ______ .

I’ll deal with this one first. The choices are total circuit resistance, current, or voltage. We can rule out total circuit voltage, because that is not calculated, it’s given.

To calculate voltage drop across a load, we need to know both the current in the circuit (which is constant throughout the circuit) and the resistance of the load, right? E=I x R.

The resistance of the load is what it is. In other words, it’s an intrinsic value of the load and must be specified on the tech sheet.

Current, however, has to be calculated knowing the circuit supply voltage and the total resistance of the circuit. I = E/R. Does that make sense? So, the first step in eventually being able to calculate the voltage drop across a particular load is to calculate the total circuit resistance so that you can then calculate the circuit current.

For Question 6, the possible answers are:
The current flow through each load will be proportional to their resistance.
The current flow through each load will be proportional to their voltage drop.
The voltage drop across each load will be proportional to their intrinsic resistance.
(Note – we clarified the wording just a bit, changing “them” to “each load”)

We are talking about each load in the series. The current is the same throughout the circuit, right? So do the first two answers make sense? The current flow in the entire circuit is a function of the TOTAL resistance and supply voltage, but does not vary through each load.

E = IxR. Current is constant, but each load has it’s own resistance, so the voltage drop for a load will be proportional to the resistance of that load.

Hope that helps! Let me know if you have any follow-up questions.

[Susan Brown]

Hi Tyler,

Don’t feel bad – this is supposed to be a challenging course! The struggle is just part of the learning process. And we’re here to help ya.

First, let me just verify that these are the two questions you are asking about:

Question #5: What are loads in series?
Question #6: What is the true statement about loads in series?

Are those them?

[Susan Brown]

It’s hard to write out equations here, but I’ll try to show it.

It’s 1/(1/58 + 1/320)

1/58= 0.017
1/320= 0.003
Add those together = 0.02
Then 1/0.02 = 50

I rounded the decimals a bit, which is fine. To do the fractions on your calculator, you just enter 1 and then divide by the denominator.

Is this what you needed?

[Susan Brown]

Hi Robert,

That particular question was set up so that you just needed to know the rule of thumb about equivalent resistance: that it will be something less than the smallest resistance in the parallel circuit.

You can calculate the exact value of the equiv. resistance by using the formula that is shown in the last presentation in that unit (starting at about 2 1/2 minutes in), but you didn’t need to for this question.

Does that answer your question?

[Susan Brown]

Hi John,

Since March, we’ve added 9 new presentations to Fundamentals, including the one you are asking about. We almost immediately decided that the Ohm’s Law presentation went beyond the scope of Fundamentals, which focuses on basics, and was advanced enough that we should include it among the presentations in a new course on Advanced Schematic Analysis and Troubleshooting. We left the presentation in the Fundamentals course for the time being while we built the ASAT course, but moved it over to the ASAT course last week when that course was launched.

So you were one of the few students who happened to catch it during that brief window that it appeared in Fundamentals and got a bonus lesson (and a preview of the ASAT). Lucky you! 🙂

We’ve noted this and all other updates in the Course Updates log at the site.

[Susan Brown]

Hi Glenn,

I’m not sure what the “ugh of ohms law” is that you mentioned – perhaps that was a typo? 🙂

To cover all the bases, I’ll go over the basic math functions that are in the Ohm’s law pie chart that first makes its colorful appearance in Unit 3 of Basic Electricity.

1. Multiplication: as in, E = I x R
2. Division: as in, R = E ÷ I (written in the pie chart as E / I, which is just another way to show a division problem)

Those two are easy to do on any calculator.

3. Squared: as in, P = I² x R. A number that is “squared” is simply that number times itself. For example, 2 squared is the same as 2 x 2 which is 4. So, if your calculator doesn’t have a squared function button (usually labelled as x²) just enter the number and multiply it by the same number.

4. Square root: as in, E = √ P x R 
A square root is the opposite of a square. You are basically saying, what number was squared to get this number? If you want to know the square root of 4, you ask “what number times itself will equal 4?” The answer is 2. For all but the simplest numbers, you’ll need a calculator that has a square root button to calculate this. If you are using the calculator on an iPhone, turn the phone sideways to get a lot of extra buttons, including the square root. It looks something like this: ²√ x 

I hope that helps! Lemme know if you have any other questions.

[Susan Brown]

That’s some wise advice, Smashy!

You can retake the quiz when you are ready.

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