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The quiz questions come directly from the content in the course, so it’s best just to review the lessons. Rewatch the videos or read the transcripts, and go through each quiz. Let us know if there are any questions or topics you don’t understand.
I know the exam takes time, but you can have two more attempts after this one if you don’t pass it, if that helps you to feel better.
We have a circuit with two loads in series.
R1 = 5 ohms
R2 = 5k ohms (5000 ohms)The question is: “how much work is R1 doing compared to R2?”
The work done by a load is directly proportional to resistance. (Voltage drop has the same relationship, FYI.)
So, higher resistance = higher voltage drop and higher wattage.
Since it is directly proportional, you don’t actually have to do the calculations to answer. R1 is 1000 times less than R2, so P1 is 1000 times less than P2.
But as you discovered, if you aren’t sure, you can always run the calculations to figure out the answer.
I got this question wrong because I did not run the numbers correctly. I assumed R1 would do more work(Watts). R1, 120v/5ohms=24watts. R2, 120v/5000ohms=.024watts. 24 is 1000 times MORE than .024. Did I do this calculation for a parallel circuit?
Yes, you assumed that R1 and R2 would both have a 120V drop across them, which would only be the case if they were in parallel with each other, not in series.
Your second set of calculations treated them correctly.
Sure!
im (or IM or I/M) is ice maker
FF is fresh food compartment (in contrast to the freezer compartment)
Our printer just produced the next batch of Certificates, which includes yours. I am out of town right now, but will be able to get them in the mail later this week!
Hi Sentayehu,
Please see the email I sent you. You do have another attempt available on the Module 2 exam (Module 2, Unit 7). Please retake that and let me know, and then we will move forward from there.Okay, let us know if there are any you need more help with.
Yes, that will help. But remember that on the Module Exams the mix of questions changes each time you take it. Some will be the same. Have you gone back over the Unit Quizzes and figured out the answers to the questions you missed?
Some of the questions you got wrong I’m sure you can confidently find the correct answers by going back over the material. Searching the video transcripts is a great way to look for the answers.
Please just send us the questions where you tried to find the answer and really couldn’t, then we can help you.
Yes! I got your Reset request by email and just reset you. Let us know if you need any help.
Yes, heat produced is power. The units are watts.
When you have two loads in series, it makes things a little more complicated because the voltage drop is divided up between the two loads.
When you use P = E^2/R for one of the loads, E must be the voltage dropped across the load in question, not the entire circuit.
You can calculate the voltage drop using E = I x R, meaning you would still need to calculate current (I) first.
Once you know I, you may as well then just use P = I^2 * R. It’s a little more direct.
I is easy to calculate. I = E/R, where E is the total voltage drop in the circuit (which is equal to the source voltage) and R is the total resistance in the circuit.
Does that help?
Hi Matthew,
The first answer is “A low side leak”. This is from the 3rd video.
The answer to #3 is 250 pounds of pressure. See the first video, just before the 1 minute mark.
Hi Matthew,
You would use the same vacuum pump as for R134a.There are some differences with R600a but they have to do with pumpout times.
Here’s a discussion you might find interesting:
https://appliantology.org/topic/85933-25-year-to-date-lg-refrigerator-sealed-systems/page/10/?&_rid=4#findComment-580514This ignitor is wired in series with the gas valve, meaning that current is flowing through both whenever the circuit is powered.
By saying “the circuit is powered”, we mean the oven is on and calling for heat by supplying voltage to the ignition circuit. Because this circuit has both the ignitor and the gas valve, it will remain powered the whole time heat (a flame) is desired, not just when it’s igniting after being off.
Does that help?
Here’s a video at Appliantology that goes into more detail, FYI:
https://appliantology.org/topic/62940-mst-office-hours-582017-gas-oven-service-call-after-a-parts-changing-monkey/You are allowed to post any question you want to! If the question that you typed in didn’t post, then there must have been some kind of connectivity glitch.
Hi Lukas,
Thanks for the comment.Since we only mention COM and NO as the contacts of the switch, we are implying that it is a 2-contact switch.
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