Susan Brown

[Susan Brown]

Yep, every year that goes by, more and more people are “googling” rather than looking in phone books or other print media. It’s very important to have a good online presence, including a mobile-friendly website, since more and more people are searching using devices. It doesn’t have to be a complicated site – just make it easy for them to call you!

Google AdWords can be very effective, too.

[Susan Brown]

Hi Igor and Mark,

Very sorry for the delay in answering this post! We’ve just realized there is an occasional glitch in our software that sometimes prevents us being notified of a post – we are gonna get that fixed!

We do have a section in Mod 4 Unit 4 of the course on legal structures with info on the most commonly used one for small businesses, the LLC.

At this point we do not include any detailed info on growing a business to include multiple techs. That is a topic we are planning to address eventually, but our focus at the moment is developing more technical courses.

A current resource that I can think of if you have immediate questions about growing your business would be connecting with other business owners in the Tech forums at Appliantology.

Thanks for the question!

[Susan Brown]

Hi Igor,

Basically what we are saying is if you are a self-employed professional tech, are any of the items in that list things that shouldn’t be factored into your pricing so that you are compensated for them?

(And I’ve unblocked you.)

[Susan Brown]

Hi Walter,

I just wanted to chime in and say that we cover the type of questions you’re asking here – current, voltage, voltage drop, various types of circuits, electrical measurements, etc. – very thoroughly in the Basic Electricity Boot Camp (including a discussion on how “the path of least resistance” is an errant way of thinking when dealing with electrical circuits!).

The BEBC lays a firm foundation for folks to then be able to get the most out of the Advanced Schematics course. We’ve found that techs without a clear understanding of basic electricity can have difficulty really getting their troubleshooting skills where they want them, even if they take Advanced Schematics. Click here to see the topics covered.

Just wanted to throw that out there in case you were interested in taking the BEBC! If you are, I can message you privately to discuss it further. Lemme know!

🙂

[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?

[Author]

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