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For #7:
You didn’t show your calculation of how you got 1800 watts, so it’s hard for me to see what you did wrong.
P = I^2 x R is the best way to calculate this. We give you R. But first you need to calculate the circuit current, I.
What did you use for I?
There’s a lot of complexity to the relationships between voltage, current, resistance, and power, but we’re trying to present it as simply as possible – just what you need to know to help you be a strong electrical troubleshooter.
Which Ohm’s Law formula you use depends on what you are dealing with.
For example, in a circuit, the current is I = E/R.
Higher resistance will mean lower current. Inversely proportional.
Higher voltage will mean higher current. Directly proportional.The way the formula is written shows you that.
If you want to calculate the voltage drop across a load, and you know the circuit current and resistance of the load, then you’d use E = I x R.
When you are talking about power (watts), then you might be using P = I x E. For example, if you know that a single load in a 120v circuit is supposed to generate 120 watts of heat, then you would calculate that there should be 1 amp of current in that circuit.
This module takes more effort than most to master, but it will really help you down the road.
Yes, if loads are in series, the total resistance of that circuit is just the sum of the resistances of the loads.
That calculation is correct.
For Questions 7 and 8, review the video at the end of Unit 3 about the heat generated by the loose connection. Very similar scenario.
stumped?
Nope! We get asked this question periodically. You happened to ask this on a Saturday night… we work a lot, but we do take breaks here and there!
The power company is supplying power.
P = I x E.
So, if your goal is to deliver a certain amount of wattage, then you can see that a higher voltage will mean a lower current.
If you want to have 1000 watts, for example, you could have 100 volts and 10 amps, or you could have 200 volts and 5 amps, etc.
When you are looking at the voltage drop in a circuit, that’s when we use E = I x R, and current is directly proportional to voltage.
Yes, surface control. The L2-H2 contacts have to be closed as well for current to go through the element.
1. We don’t base Certification on the overall course grade, which is an average.
2. Certification is based on EACH individual unit quiz being 80% or higher. Yours are all 90% or higher at this point, so you are fine so far.
3. The exams in the course (Module, Midterm, Final) have to be 90% or higher on EACH one.So far, so good!
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… -
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