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HI Darron,
What is the power source for the light – L1-N, or L1-L2?
Hi Broc,
The most common mistake folks make on this one is to look at the wrong indicator light on the schematic. There is one “surf ind light” and 4 “hot ind lights”. Which one are you looking at?
Also, I can tell you that there are 2 correct answers, and your last attempt you chose 3 answers. One of the ones you chose is incorrect.
Lastly, did you read the explanation that shows up on the quiz results for that question?
Hi Joe,
Check out these two previous forum topics and see if they help you. Feel free to ask more questions after looking over these:
Hi Kevin,
Team Samurai just had a pow-wow about the 4 main systems and the condenser fan, and where it belongs. After all, even though the condenser fan is not physically in the sealed system, it supports the functioning of the sealed system.
We’re probably going to change the wording of this particular question, and tweak the description of the 4 main systems just to clarify. We’ll dial that in this afternoon and get back to you.
Okay – no problem. Thanks for letting us know!
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.)
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.
Glad to hear it!
Yes! As it would also be for the booster. Does this clear up this scenario, or so you have any other questions?
yep!
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.
Okay – so if the ignitor [answer hidden] what is its voltage supply? What is its voltage drop?
Here’s how I get that:
1/30 + 1/50 = 0.0333 + 0.02 = 0.0533
Then 1/0.053 = 18.75
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:
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!
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