Susan Brown

[Susan Brown]

Hi Chris,

When I’m trying to understand this stuff, I always pull up that colorful Ohm’s law wheel and use the equations to show me the relationships between power, resistance, amps, and voltage. Math helps to keep our thinking straight!

One equation we use in the loose connection scenario is P = I² x R.

P in this case is the heat generated from the current going through something that has resistance. From this equation you can see that increased resistance can increase the heat generated, even if the current is a little lower.

Normally a wire has essentially zero resistance, which generates essentially no heat. But the loose wire connection creates a spot of resistance, which generates heat, which then causes the wire to deteriorate even further, which increases its resistance, and so on, until there is finally a failure at that point or somewhere else in the circuit.

This loose connection lowers the current in the entire circuit, and thus the power in the element. But that doesn’t make it easier for the circuit. The reason this puts stress on the circuitry is that the current has to flow for a longer period of time in order to get the oven up to temperature – beyond the design parameters.

Does that make sense?

[Susan Brown]

Hi John,

Sorry you are having a problem with one of the videos playing!

I just tried and it played fine for me. There may be something going on on your end.

If you haven’t already, please read the Video Playback section at the Student Resource page:
https://mastersamuraitech.com/appliance-repair-course-support/sta-resource-page/

Then please let me know if you get the video to play or not.

[Susan Brown]

You’re welcome!

Some students have difficulty with equations that have a “squared” number in them because they aren’t sure how to square a number on their calculators. The easiest way is to just multiply the number times itself.

For future reference, here’s info on how to start a topic the next time you have a question:

https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/please-read-how-to-post-a-topic-in-the-forum/

[Susan Brown]

Hi Jose,

First, note that I moved this question to a new topic. Your question should be a new topic with an appropriate title.

There is an equation you could have used to do the calculation in one step: R = E squared/P.

However, you can still get the correct answer by calculating current first, then calculate R from that and the voltage. Mathematically it works out the same either way.

[Susan Brown]

Try combining what you went over above on voltage drop with the information on supply voltage in parallel circuits in the first video in this unit, particularly starting at the 11 minute mark:
https://my.mastersamuraitech.com/module-3/basic-electricity-series-and-parallel-circuits/

[Susan Brown]

Question 4 involves a series circuit, and question 8 involves circuits in parallel.

There’s an important fact about voltage in parallel circuits that you need to account for here! Watch the presentation again in Module 3 Unit 5 starting at about the 11 minute mark, and see if that enlightens you!
https://my.mastersamuraitech.com/module-3/basic-electricity-series-and-parallel-circuits/

[Susan Brown]

The answer can vary depending on how you treat the 1/30, since in decimal form it is 0.03333….

But the slight difference caused by how many decimal places you use is trivial – around 19 ohms is certainly good enough!

The most important point is to see that the equivalent resistance is less than the smallest resistance value.

[Susan Brown]

Hi Alex,

You don’t need to do a calculation to get the correct answer – it’s pretty simple to eyeball it and get it right. However, I’ve updated the explanation on the quiz with math that is a little easier to explain. See if that helps.

[Susan Brown]

P.S. We’ve removed the statement about the power supply being verified good to simplify the scenario even further.

[Susan Brown]

We’re not asking questions in this exam that require you to know about the rest of the dryer circuit. The purpose of the midterm exam is to test your understanding of fundamental circuit concepts. As for troubleshooting problems using stock schematics on the entire appliance, you’ll be getting plenty of that in the troubleshooting module.

[Susan Brown]

It sounds like perhaps you are not clear on measuring voltage vs. voltage drop. (This is a concept that take a little while to grasp!)

When you take a voltage reading, that does not tell you that current is in fact flowing. It just tells you that there is a voltage difference between two points.

In the question it stated that there was zero volts across the heating element, which tells you that no work is being done, thus no current is flowing in that circuit (so, there isn’t a complete circuit).

It helped me to watch the video in Module 3 Unit 8 again when contemplating this scenario. Watch that and see if it helps. Let me know!

https://my.mastersamuraitech.com/module-3/basic-electricity-voltage-drop-and-load/

[Susan Brown]

It sounds like you are confusing measuring voltage potential vs. voltage drop.

That’s a hint… does that help clear it up? If not, we can discuss further!

[Susan Brown]

The question you are referring to is:
Question #19 – If the Defrost Heater and the Freezer Drain Heater are operating within specifications, what equivalent resistance would you measure from Pin 13 to Pin 7 in the diagram below:

So, we are talking about equivalent resistance.

What kind of circuit is shown in the diagram? Series or parallel?

(P.S. “Current drop” is not something that happens in a circuit – did you mean to say “voltage drop”? Make sure you have a clear understanding of current and voltage, and their characteristics in series and parallel circuits!)

[Susan Brown]

No, because these are specialty items that aren’t needed for most calls, especially when someone is starting out.

[Susan Brown]

There aren’t a lot of times that you simply can’t do a job without a second man. One that comes to mind is replacing the drum bearings on a front-load washer. But, there are some times when a second man can be instrumental to avoiding damage to the surroundings, when there is an unusual amount of disassembly or effort involved because of the particular setting of the appliance. This most often happens with laundry appliances, which can be stacked or in tight closets. You have to evaluate the setting and decide what the risks seem to be – do they have fancy floors or cabinetry or trim work that could get scratched or dented in the process of gaining access to the appliance? If so, you may have to schedule to come back with help.

We eventually acquired some specific equipment to help with these situations – an all-dolly for wall ovens and an air sled to help move appliances in difficult settings. This is pricey equipment, so we charge when they have to be used (essentially, they are the “second man”).

The vast majority of jobs you can do solo. But, it would be good to identify a person or two who you might be able to arrange to help you on occasional jobs. Someone who has a flexible schedule (self-employed handyman, for example). Or even just a friend. They don’t need to be skilled – you just need an extra set of hands, usually!

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