Susan Brown

[Susan Brown]

E or V is fine for volts. Our chart in Unit 3 uses E. Did you see my reply above?

[Susan Brown]

Actually, the chart in Unit 3 uses “E” for voltage, but some other charts do use “V”. As long as you know either of those letters is “voltage”.

So, you are given “P” and “E”, and we ask you to find R.

This means you need to find an “R = …” equation in the pie chart that uses E and P. Can you see which one?

[Susan Brown]

Hi Rees,
Thanks for posting a question here! Yes, it can take a little time getting used to working with equations and such if you haven’t had much experience with it. But we are happy to help.

First of all, the Basic Electricity module is one of the toughest ones for many of our students. I tell you that just so you won’t get discouraged. Some guys mistakenly think if the 3rd module is difficult, they won’t be able to handle the rest of the course. Not so. So, hang in there!

Secondly, some of these units do require going through a few times. Are you pausing the videos as you watch them to take notes? Do you try to recreate the calculations that we show you in your notebook? These are the types of study habits that will help you.

The other thing that would be helpful is to identify more specific points in the unit where you don’t follow what we are showing you, so we can give you more help here in the Forums.

Let’s start with Question 7. “A heating element has a rated wattage of 5600 watts and is connected to a 240 V AC source. What is the resistance of the heating element?”

The first way to organize your thoughts is to identify what information we are giving you, and what we are asking you to find.

Given:
heating element – 5600 watts
power source – 240vac
Find:
resistance of element

Can you tell me what the symbols (letters) are for those 3 items? (the ones we used in the Ohm’s Law chart)

[Susan Brown]

Great! Don’t hesitate to post in the future if there’s anything else you’d like more help with.

[Susan Brown]

Yes! I know it might seem like a minor point, but for a load (or loads) to do work, the circuit needs both voltage and current. (P = I x E) A lot of techs think in terms of ohms (which is very weak) or perhaps voltage or current, but rarely in watts. We’re wanting our students to keep power in the forefront of their thinking when it comes to loads doing work.

Older appliance tech sheets nearly always gave specs in “watts” for loads. Now we notice that they don’t do that as much, because too many techs didn’t know what to do with watts, so they often just give the resistance (ohms). Problem is, ohms is not always a reliable test. Watts is the stronger one.

Thanks for posting this question! Let me know if you have any other questions about that.

[Susan Brown]

You’re getting warmer! Voltage drop is actually an effect of current going through a load. (On the other hand, “regular” voltage is the difference in charge that causes current to flow.)

Let me try this a different way. The answer is one of the terms in the title of Unit 3!

[Susan Brown]

Hi Phillip – our downloads section is over at http://www.Appliantology.org

[Susan Brown]

Hi Mohamed,

The best review to help you understand the calculations needed for Questions 7 and 8 is the video at the end of Unit 3, about the heat produced by the loose connection. The circuits are very similar – two loads in series, L1-L2 circuit. Rewatch that and see if it helps. If you don’t quite follow the calculations, let me know.

Any reading assignments are listed in the unit in italics, and look like this:
Reading assignment in the Kleinert text:
– page 84

[Susan Brown]

Sorry – Module 3, unit 3

[Susan Brown]

Hi William,

We’re trying to get very specific with the answer here, as it can really help when you are on the job doing electrical troubleshooting.

We talked a few times, in a few places, about what it takes to produce work in a circuit. Current is just part of it. We talk about the measure of work in the first video in Unit 3. Do you see the term we are looking for?

[Susan Brown]

Hi Richard,

For Question 8, you did get the voltage drop across the safety correct – it’s 120vac. How did you determine that? Did you do the “Zen trick” to see if it’s got direct access to L1 and N?

Try that same process on the booster and the ignitor – what do you find?

[Susan Brown]

Hi Sean,

That’s awesome that you are taking the time to understand this material better. Unit 8 is challenging! The quizzes are an integral part of the learning experience. Some of the questions deliberately push you to grapple with the material, and not just be a quick check on what you learned. These questions pull together what you’ve learned in several units in this module.

Questions 6, 7, and 8 are all based on the same circuit. There are two loads in series, and it’s an L1-L2 circuit.

Question 6 asks for the current you will measure at a certain point. So, you have to think about the characteristics of current in a series circuit, what the voltage supply is, and which Ohm’s Law equation will be appropriate.

Questions 7 and 8 ask for the heat produced by each load. If you’ve answered Question 6 correctly, you’ve got one of the things you need for this calculation. Then it’s a matter of selecting the correct Ohm’s Law equation.

One of the helpful videos to review, because it’s showing a similar scenario, is the “loose connection” video in Unit 3. Rewatch that, and see if you follow the calculation, and if it helps you to understand these questions better. If not, let me know and I’ll help you further.

[Susan Brown]

Hi Terry,

You have a good point. We were obviously thinking of a load when we wrote the question, but the word “something” could be interpreted the way you have. I’m going to run it by the rest of the team, but we’ll probably update the wording to be more clear. But for your reference, we are talking about a load (which, naturally, has resistance as an inherent property).

The main point we are trying to drive home is that you must have current flowing through a load to get voltage drop. Many techs forget that it takes voltage AND current to power a load, and the concept of understanding voltage vs. voltage drop can help clarify troubleshooting.

Thanks for putting so much thought into this and for posting your comment!

[Susan Brown]

Hi Michael,

You are correct about the safety and also the main.

Electrons will flow through any valid, non-shunted path between Line and Neutral if they have voltage “pushing” them.

So now you just need to figure out what the relationship of the booster and ignitor are. Are they in series or in parallel? Do the Zen Trick on those two loads. What do you find?

[Susan Brown]

Hi Phillip,
Yes – that is correct! You also correctly stated the rule of thumb. So, for Question 18, the correct answer choice is “something less than 58 ohms…”

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