Susan Brown

[Susan Brown]

Great!

I’d say that choice is up to you – we don’t have a strong recommendation. In fact, these are two that you could even do simultaneously, if you felt like it, as they are very different.

Advanced Troubleshooting will give you more schematic practice.

W&D will give you a deep dive into the technology found in washers and dryers.

[Susan Brown]

Hi Don,

First of all, are you going to retake the Unit 2 quiz?

I always like to make sure students are looking at the correct light, since there is another one on the schematic that is labeled similarly.

Be sure you are looking at the LF Hot Indicator Light

In unit 4, Question 2, there are two correct answers out of the choices given – we want you to select both.

After I’m sure we’re looking at the same thing, I’ll try to help you with what you’re asking about the switch.

[Susan Brown]

Yes!

So, now you can think about power. One of the equations of power is P = I x E.

To use that in this scenario, I is the circuit current, but what number we use for E requires more thinking about.

If we use the source voltage, then we’ll calculate the heat generated by the entire circuit.

But if we only want to know the heat generated by one of the loads, we’d have to use the voltage drop for that load.

This tells us that the load with the higher resistance will generate more heat (power).

Getting back to Questions 7 and 8 on the quiz.

To use P = I x E, you’d have to calculate voltage drop across each load as well as the circuit current. Those are both doable, but there’s a little more straightforward way.

We showed you how to do it in a video back in unit 3, where there was a loose connection in series with an element. It’s a very similar scenario to this one. Rewatch how we did the calculations and see if you can follow it. Let me know if you have any questions about it, and if it helps you to see what to do on these questions.

[Susan Brown]

Correct.

The total resistance of the circuit (in this case, 5 + 32) will determine what that current is.

So, each load has the same current flowing through it.

Now, before we get back to power, what about voltage drop in these two loads?

As we taught you, when current goes through a resistance (a load), it will create a voltage drop across the load. The voltage drop will be proportional to the resistance (V = I x R) And the sum of the two voltage drops will equal the source voltage.

Which of the two loads will have a higher voltage drop?

[Susan Brown]

Okay – now we’re onto something.

In a series circuit, is the current the same throughout the circuit, or does it change at different points in the circuit?

(We covered this in unit 5)

[Susan Brown]

You got it!

[Susan Brown]

Well, we know the common mistakes students make on these questions, so we often put those answers as choices. Not to make the quiz unnecessarily difficult, but to reveal to the student that they need to improve their understanding.

For Questions 7 and 8, you’ve got two loads in series. The resistance of the first one is 5 ohms, the second one is 32 ohms.

First question I’d like you to answer:

Based on what we’ve taught, which one would you expect to generate more heat?

(Answer that, then I’ll give you a tip on where to look for more help on this.)

[Susan Brown]

I’ll get there and thank you

We’re glad to help you get there!

[Susan Brown]

Hi Michael,

The current in a series circuit is the same at every point in the circuit. It’s just the way electron movement works. If electrons are moving in a circuit (if there is voltage and a complete circuit), then they will all move at the same rate.

What that rate is is determined by the *total* resistance of the loads in the circuit.

So, a 120vac circuit with 20 ohms of total resistance in it will have a current of 6 amps (at every point in the circuit). That will be true whether it has one load with 20 ohms, or two loads of 10 ohms each.

Does that help?

[Susan Brown]

Hi Joe,

Here’s the paragraph from Mod 7, unit 1:

It’s also not uncommon to see switches test fine with your ohm meter but then not work right when they’re in a live circuit actually switching voltage. This is called “failing under load.” That’s why I personally prefer doing my troubleshooting on a live circuit and testing for voltages rather than checking for continuity. Continuity tests are okay, sometimes necessary, sometimes expedient. But the gold standard for testing components like switches is live voltage testing.

Here are the 3 answer choices for the quiz:

It’s not actuating properly.
It checks good on continuity but fails in a live circuit.
It’s having trouble carrying its weight.

[Susan Brown]

Hi Steve,

Good question! Appliance techs don’t need an Oscilloscope for their work in the field. The one you see in the video is primarily used for training/teaching purposes, rather than diagnostics.

[Susan Brown]

Hi Rudy,

Thanks for your comments.

We’ve had about 2000 students go through the course over the past few years, but are always open to student feedback. I’ll take a look at the items you’ve mentioned.

But I also wanted to point out that we have these Ask the Teacher Forums to account for the fact that everyone has different needs when it comes to training and learning. You do have two attempts available on each quiz, as you know, and we also have the reset program. So, if after taking a quiz for the first time there are questions that aren’t clear to you, just ask us to clarify!

[Susan Brown]

Good question, Matt.

There are two resistances in the circuit – the loose connection and the element.

To get the heat generated by one of the resistances only, the “E” in the equation P=IxE would have to the be voltage drop across that individual load, not the total voltage.

So, you’d have to calculate voltage drop across the loose connection.

This is do-able, but most students have an easier time of following the logic behind the technique shown above in V’s post and in the video. And it requires fewer steps.

Here’s how you could do it.

The total voltage dropped across the entire circuit is 240vac.

The resistances (in our video scenario) are 32 ohms and 5 ohms, so total is 37ohms

Voltage drop across loose connection is directly proportional to the resistance of that connection, so is
240 (5/37) = 32.4 volts

32.4 x 6.5 = 211 watts

so – that’s a little more complicated than just using P = I squared x R

[Susan Brown]

Yes – that is correct – good job!

FYI – I’ll need to hide part of your answer so we don’t give it away to other students.

[Susan Brown]

Good question!

You’ll get a feel for these numbers as you go along. Here are a few guidelines.

Ohms and Watts do not need to be reported with any decimal places – whole numbers are fine.

With the example you gave, of 1350 watts vs. 1352 watts, those are essentially the same. When you’re talking about overall values in the hundreds or thousands, a few watts one way or the other in the “ones” place doesn’t matter.

Amps is where you usually will see smaller, more precise numbers, like 3.2 or 6.5.

Hope that helps!

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