Susan Brown

[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…”

[Susan Brown]

Do you know the rule of thumb for equivalent resistance of loads in parallel? That’s what you need in order to find the correct answer out of the choices we give you.

[Susan Brown]

You are correct about the safety. It’s all by itself in a circuit, so will drop the entire source voltage. (No need to know its resistance, since you didn’t have to actually do a calculation.)

But, there are a couple of things we still need to straighten out.

First of all, the detector is a switch, not a load. (Is there any voltage drop across a closed switch?)

Second, remember that electrons don’t “see” any turns or bends in a wire. Unless there is a shunt, they will take any valid path. So, if the main coil were getting current, then it would be in series with both the igniter and the booster.

But – do the “Zen trick” on the booster. How do you reach out and touch Neutral?

[Susan Brown]

Hi William,

Each load does have a resistance, but we don’t tell you what it is because you actually don’t need to know it in this particular scenario.

The first step is to decide if each load is in series with any other loads. The “Zen trick” on the booster, igniter, and safety is a helpful tool.

Then, if you have that determined correctly, you just use what you know in general about voltage drop and voltage supply to answer the question – no calculations necessary!

What do you think? How are the loads in relation to each other? Do they all have current flowing through them?

[Susan Brown]

The reason I asked about the supply voltage is that the answer you gave – 40 vac and 80 vac – only adds up to 120vac. As we taught you, the sum of the voltage drops across loads in series will equal the source voltage.

But we indeed were asking you to calculate the voltage drop across R1 AND the voltage drop across R2.

And the correct answer is one of the choices given.

[Susan Brown]

Hi Jim,
Thanks for posting a question!

What is the source voltage for this circuit?

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