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Hi Darren,
They are similar in the basic function – to prove that there is a flame/ignition source before opening the valve to let the gas in. But the designs are very different.
A standing pilot system is very old technology and you’ll rarely come across it anymore except in off-grid applications or very old homes. No electricity is required for these.
We cover the glow-bar ignition systems and valves in much more detail in the Oven & Range course, so you’ll learn even more there!
Just reset you!
In the future, please post separate questions as separate topics, and be sure to put them in the appropriate forum. It is difficult to have two conversations in the same thread.
#7: the circuit current is calculated with I = E/R where E is the source voltage and R is the TOTAL resistance in the circuit. Remember, in a series circuit, the current is the SAME at every point in the circuit.
Once you have the correct circuit current, then you’ll be able to get the correct answer with P = I^2 x 32 ohms. (Note that here we just use the resistance of the load in question, not the total resistance in the whole circuit.)
#9:
Explanation: Remember that in parallel circuits, the voltage SUPPLY to each branch will be the same as the source voltage. Each branch is connected independently to to the source; in effect, parallel branches are independent series circuits. And in any series circuit, the sum of the voltage drops must add up to the source voltage. So with only one load in a series circuit, the voltage drop must equal the source voltage.
R2 is 5 times greater than R1. L1 voltage is 120. The correct answer is both are 120v? because 5 times R1 is 600 and thats wrong! why. is the 5 times thing a lie. Ive failed these two questins 4 times. theyre gonna kick me out. please help
Did you read the explanation that we gave in the quiz results? The fact that these loads are in parallel, not series, is key.
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
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