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For the element to get hot, it needs both L1 and L2. If the L1 to H1 contacts are stuck closed, does that mean that the element would also get L2?
Good question!
The electrons will move IF they are in a conductive material (AND have a complete circuit from a higher charge to a lower one).
Copper wire (like most metals) is very conductive. The electrons move pretty readily in it. Other materials are not at all conductive, such as rubber.
If you only touch the insulated part of the wire, that protects you. That insulation prevents electron flow. If you touch a bit of bare wire that is live, however, you can get shocked.
Hi Steve,
Good job working to understand all of the questions and answers, even if you passed the exam!
You’ve got most of the correct idea – the condenser does remove enough heat to condense the refrigerant into a liquid. However, it remains a hot liquid, not cool.
Hi Shaya,
This video may help you see how to calculate equivalent resistance in parallel circuits:
https://www.youtube.com/embed/iHB3lxdc68E
No worries!
Hi Kris,
14,400 watts should definitely sound wrong to you – think of how hot a 100 watt lightbulb feels!
Here’s why your calculation didn’t work: there is another load in that circuit – the element.
In the P = E^2/R formula, E is the voltage dropped across the load. It is only equal to the supply voltage if there is just one load in the circuit. In the loose connection scenario, you have two resistances in series.
So, if you want to use this formula for P, you have to calculate the voltage drop across the loose connection first.
Alternatively, and this is what we show in the loose connection video in Unit 3, you can use P = I^2 x R. This means you have to calculate the circuit current first, but that’s pretty straightforward (I = E/R, where E is source voltage and R is total resistance in circuit).
February 28, 2020 at 9:28 am in reply to: what should i study for the core exam final part one what should i focus on most #18595The Final Exam covers material from the entire course. It’s a good time to do a quick review of each unit to refresh your memory, and skim through your quizzes. If there are any lingering areas that you don’t feel strong on, ask us for help!
Hi Ed,
I just emailed you about this…A short or an open? These terms have precise meanings.
From Mod 3, unit 1:
Short: A condition that creates a circuit that offers no resistance to the current flowing through it. A direct short will cause a fuse to blow, or possibly start an electrical fire.
Whereas an open is just that – some break in the circuit (open switch, broken wire, for example) that prevents current.
If the fault is on the L1 side, how do you explain this measurement?
You use your meter to test for 120 V AC from one of the terminals of the light to neutral and you read 120 V AC on your meter
Hi Darren,
The Final Exam covers material from the entire course.
🙂Yes, that’s correct!
You haven’t been asked this exact question, but you just have to think about what the measurements are showing you.
It’s a safe assumption that we’re talking about a 120vac circuit.
We’re telling you the light bulb and receptacle that the light bulb is in are good.
There will be two wires going to the bulb, correct?
One is Line, what is the other?
Then, think about these two voltage measurements, and what you can conclude:
1. one side of the bulb has 120vac wrt N
2. when you read across the bulb, you get zero.Let me know what you think…
What
Well, much better to over-think than to under-think! In my opinion, you were just thinking it through. We’re always happy to help.
Correct, 1 and 2 would not be able to get to N if a break happened there.
This does not change their status as being in parallel with each other. The electrons that go through either 1, 2, or 3 (when there is no open anywhere) do not have to go through one of the other loads to get to L1 or N.
Both of those diagrams show 3 parallel circuits. If they are functioning, they are electrically equivalent. (Assuming the loads are the same.)
The slight difference in configuration only makes a difference in terms of how a break in the wire at certain points would affect the other branches, which seems to be what you’re getting at.
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