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Air distribution system and temperature sensing device
Wow – great job spotting that typo – thanks! We’re going to change that to “start and run winding” (start and main would also be correct, of course).
You’re welcome! Always glad to help.
March 10, 2020 at 1:49 pm in reply to: Module 1, Unit 7, Ref. Air Dist. Sys.: Evaporator Split or Approach #18693Thanks for the input!
The “split” terminology that Scott uses is what he learned in Refrigeration Engineering (something he studied and practiced once upon a time). There are often cases where different disciplines use terms in slightly different ways. As long as we have a functional understanding of the phenomenon being described, we’ll be fine.
Hi Joe,
You are correct on all 3 accounts.
There are two loads in the circuit – the light bulb and the heater.
When the switch is open, electrons only have one path to travel between L1 and N: through both loads. Thus, both the heater and the light bulb with do work.
When that switch is closed, it is a shunt. It might help to think of the term “bypass” instead of shunt. A shunt deliberately bypasses one or more loads in a circuit, but NOT all of them. In this circuit, the heater is bypassed when the switch is closed, but the bulb still gets current.
A circuit will never be designed with a short. A short is a result of some kind of failure or fault condition.
Does that help?
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.
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