Susan Brown

[Susan Brown]

Hi Colin,

It is possible, but not something we advise doing since it’s not always saving that much money and it can be tricky to determine the exact state of all the various components. The best practice is to replace the board.

[Susan Brown]

Hi Brian,

Your original answer (down) is the correct one! I wonder if you accidentally looked at the wrong question/answer the first time you took it? I double-checked the quiz and it still is set up correctly.

But, your original thinking is correct, that’s the important part!

[Susan Brown]

And by the way, we have lots of students who are completely new to the field! You’ll do fine – it may take a little more work or repeating of some lessons than for those who already have some experience, but if you put in the work you’ll learn what you need to!

And feel free to ask questions when you need to! Just please start topics in the Forum for that particular course and unit. Thanks!

[Susan Brown]

Have you watched the video at the end of Unit 3? He steps you through it.

You are wanting to calculate the heat produced by a loose connection. Heat is a type of work, measured in watts (as described in the unit), and is also described as “power”. (You have to get used to different words meaning the same thing sometimes!)

Using the various Ohm’s Law equations there are actually a couple of ways you could calculate the answer to this problem, but it will always be done in two steps. The most straightforward way is to calculate the circuit current (because the current is always the same throughout a series circuit), then you can use that to calculate the watts of heat generated by the loose connection, knowing the resistance of that connection.

With that in mind, please re-watch the video.

[Susan Brown]

Hi Colin,

Even though it comes early in the course, the Basic Electricity module is often the most challenging one for our students! There’s a lot to comprehend, so don’t get discouraged. When we were developing this course I myself was trying to learn this stuff, and kept asking Scott questions until it started to click in place for me.

I encourage you to spend your study time right now going over each unit quiz and finding any questions/answers that you don’t fully understand, and use the forum to figure it out.

You might want to start with looking at previous conversations in the Forums – we’ve given lots of help to others that might be enlightening to you. For example here’s one CLICK.

And then start separate topics to ask remaining questions that you have (for example, the one you put at the bottom of your post). Read this if you need help on how to start a new topic.

So – please ask as many questions as you need to! That is what we are here for 🙂

[Susan Brown]

Hi Phil,

You always have to make sure you are using Ohm’s Law in the context of the realities of a circuit.

Let’s take a very simple circuit: one with a single load and 120vac supply. Let’s say the resistance of the load is 10 ohms.

If we use P= V^2/R we get 120×120/10 = 1440 watts.

If we want to use P= I^2xR then we first calculate I from the given V and R.
I = V/R = 12 amps.
Then 12x12x10 = 1440 watts.

So, that’s pretty simple when you are dealing with a single load. But how about if there are 2 loads? Let’s take my circuit above and say there are 2 loads, one is 2 ohms and the other 8 ohms of resistance.

The total power produced by the entire circuit will be identical to the calculations above, because the total resistance in the circuit still equals 10 ohms.

But if you want to calculate the power produced by each load individually, you have to think about what is different. We know the resistance of each load, and we know the total circuit current (because the current in a series circuit is the same at any point in the circuit).

So, you could use the P=I^2xR for each load to get 288 watts and 1152 watts, respectively.

But, how would you use P=V^2/R to do this? The mistake a lot of people make is to use the source voltage in this calculation, which would give you strange numbers for the watts produced (7200 and 1800, with the higher resistance load producing less heat, which doesn’t make sense).

So, now you have to think about voltage drop. The 2 loads combined will drop the 120vac, with each individual load dropping a portion of that voltage in direct proportion to its resistance.

So if you wanted to calculate power using the V^2/R equation, you’d first have to calculate the voltage drop across each load using V=IxR.

Try that and see if you get the same numbers as calculated above.

Does that answer your question?

[Susan Brown]

I just sent you an email about this!

[Susan Brown]

A wiggy (loading meter) can be used with either AC or DC, but only in situations where loads are being powered by higher voltages (e.g., 120v) from a robust power source.

In appliances this is almost always AC voltage. Very rare occasions will you have 120v DC powering a load.

[Susan Brown]

Hi Todd,

These questions are checking to see if you understand some of the basic things that have been taught so far about series and parallel circuits and can apply them to some basic scenarios.

The most basic concept is that for current to flow, you need a ____ circuit.

What goes in that blank?

Can current flow through a burned-out light bulb?

Two of the questions involve a series circuit, and one involves a parallel circuit. So you need to know the different behaviors of current and voltage in those. That info is presented in the first presentation in Unit 5. There’s also good info in the videos in Unit 4.

[Susan Brown]

Hi Jason,

Thanks for asking!

When we initially planned out the Academy, we thought that we would quickly move on to appliance-specific courses after Fundamentals, so some of our language reflects that. However, we ended up putting a lot of effort instead into creating more and better content for the Fundamentals course, as well as creating the Advanced Schematics course. We realized through experience and talking with techs, that this kind of material was the most critically needed.

Here’s what we have planned for our future courses:
A course on Data Communications and Connectivity
A Laundry course (washers and dryers)
A Cooking Appliances course

We’re in the initial planning phases, so I can’t give you an expected time frame yet.

Thanks for your interest!

[Susan Brown]

Yes, so that means that that load is out of the picture.

Try redrawing the circuits yourself, knowing what you know about the affect of the shunt, and do the “Zen” trick he talked about during the webinar. (There’s also a video in unit 5 on this.)

BTW – Scott doesn’t record the Q and A sessions.

[Susan Brown]

Hi Troy,

The detector switch being opened or closed does have a dramatic impact on these circuits. Do you recall at Office Hours the other night when Scott drew out a diagram almost exactly the same as this one? He then redrew the diagram, taking into account the fact that the detector is closed, which, as you say, makes a shunt. It was then very clear how the remaining loads were arranged with respect to Line and Neutral. (I’m having to be just a little bit vague here so I don’t just give away the answer! But I want to help you think through it.)

In a 120 vac circuit there’s always just one source for voltage and current, L1. But if a load is in series with parallel circuits, the current can flow through the parallel branches and then merge to go through the load in the series part of the circuit. In order to calculate voltage drops across loads in this scenario, you’d first calculate the equivalent resistance of the parallel circuits. Then use that equivalent resistance and the resistance of the series load to calculate the voltage drops like you normally do for a series circuit.

The challenge of this question on the midterm is mostly one of seeing exactly what is going on in this schematic. Are the loads all in parallel, or are some in series? Once you get a clear picture of that, then determining the voltage drops is straightforward, assuming you understand the basic principles of voltage in parallel and series circuits.

Has this helped at all?

[Susan Brown]

Hi Ed,

Thanks for the kind words! We’re always so glad to hear that people are getting a lot out of the courses.

You’ve made an excellent suggestion that we intend to implement. It is a bit of a production to produce downloadable copies of the presentation slides, so it may take a little time, but we’re going to put that on our project list and make those available as soon as possible. We’ll announce it in the newsletter once we’ve got them available.

thanks!

[Susan Brown]

Hi Sal,

You will be surprised to hear that this is not always the case! There are times you will find it either necessary or optimal to have the customer assist in some way.

For the most part a solo technician can do what he needs to do without assistance. But some tests can be incredibly tedious, if not impossible, to do by yourself, and it will facilitate the service call by asking for help. There are obviously some situations where you wouldn’t want to involve the customer, such as a test involving live voltage.

Some customers actually enjoy being part of the process. Others get in the way and are distracting. This is all part of the “art” component of the “art and science” of appliance repair – learning to read the customer and if/when to involve them in a repair.

I would encourage you to sign up at Appliantology! These are the type of conversations you can have with other techs, and learn from their experience. If you haven’t seen it, Scott gave a webinar on using Appliantology this week:
https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/appliantology-is-your-key-to-appliance-repair-service-call-success/

[Susan Brown]

Following up:

Those two characteristics listed above are not two types of EEPS. For two points to be EEPs, they have to meet BOTH of these criteria.

Understanding EEPs also requires that you have the meaning of voltage measurements clear in your mind. Voltage measurements are always done with respect to some other point. That’s why some people refer to voltage as being a “difference in potential”

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