Susan Brown

[Susan Brown]

Hi Joe,

Thanks for asking a question!

Think about the two basic components of electrical energy: current and voltage.

Which one gives the potential that the other one will happen?

(You can also revisit Unit 1 for help.)

[Susan Brown]

You’re welcome! Just so you know, I’m going to hide parts of your answers so we don’t give too much away to others.

[Susan Brown]

Glad to hear it!

[Susan Brown]

Yes! As it would also be for the booster. Does this clear up this scenario, or so you have any other questions?

[Susan Brown]

yep!

[Susan Brown]

That’s correct! I’m glad you were a little wary of saying “sealed system”, because of the whole “shot of freon” thing, but in the situation where both compartments are warm, that’s the first area you want to check and either confirm (and often tell them to buy a new one, if it’s an older fridge) or rule out and move on to the controls.

[Susan Brown]

Okay – so if the ignitor [answer hidden] what is its voltage supply? What is its voltage drop?

[Susan Brown]

Here’s how I get that:

1/30 + 1/50 = 0.0333 + 0.02 = 0.0533

Then 1/0.053 = 18.75

[Susan Brown]

Hi Kyle,

Check out these two previous forum topics and see if they help you. Feel free to ask more questions after looking over these:

https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/calculating-equivalent-resistance/

https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/equivalent-resistance-in-parallel-circuits/

[Susan Brown]

Hi Kaleb,

You will be encountering similar calculations as you move along, so it is awesome that you are asking for help understanding this, even though you technically passed the quiz!

First of all, we step you through a similar calculation in the last video of this unit. That’s helpful to review. And read the caution under that video about using the P = E(squared) / R equation. That only works if you calculated the voltage drop across the loose connection first – you can’t use the entire source voltage. Since we haven’t yet covered voltage drop in detail at this point in the course, there’s a better equation to use for P that involves I and R.

Rewatch that video and see if that helps you. If there’s something you can’t quite follow, let me know!

[Susan Brown]

correct – if a load is by itself in a circuit, then we know it will drop the source voltage across it.

Okay – now onto the other loads in the gas valve circuit that you are having trouble with. The next step is to make sure you know which loads are getting current, and what their relationship is with each other (series or parallel).

1. Which loads are getting current? Which are not?
2. Do the “Zen trick” on the loads that are getting current. Which ones (if any) are in series?

[Susan Brown]

Correct!

So, if you have only one load in a particular circuit, do you need to know its resistance in order to know what its voltage drop is?

[Susan Brown]

Jordan – where’d you go? 🙂

[Susan Brown]

Correct!

One way you can think of it is that “voltage” is a cause – it causes electrons to flow. “Voltage drop” is an effect – the result of electrons flowing through a load.

And you are correct that if we have loads in a series, the voltage drop across each load is based on the resistance. We can calculate that using E = I x R.

What is true about the sum of all of those voltage drops (if you have two or more loads in series)?

[Susan Brown]

Hi Jordan,

Thanks for posting a topic! I’m going to step you through with some questions to help you with this, so we’ll go back and forth a bit. Be sure to keep an eye out for my replies!

Let’s first just make sure you understand voltage drop. The general term “voltage” means a difference in electrical charge between two points. Voltage drop is a more specific type of voltage – can you describe in just a few words what voltage drop is?

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