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Yes, exactly right! Your answers were so good, in fact, that I’ve hidden them so that no one steals them. š Great work!
Hi Josh,
Let me answer each of your questions one at a time.
So, Iāve heard some people call capacitors, ābatteriesā.
So does the capacitor hold a charge, and then when the motor is called to start, the capacitor releases that charge to the start winding? And thats what creates the phase angle?Yes, capacitors and batteries both hold charge, but the similarities between them basically end there. Capacitors aren’t used as power supplies. A motor start capacitor (and indeed, any capacitor in an AC circuit) is constantly charging and discharging as the polarity reverses from positive to negative and back again.
Also, the start capacitor does not create the phase angle between the start winding and the main winding. The phase angle already exists, simply due to how those two windings are physically constructed. But the start capacitor does increase the size of that phase angle by phase shifting the current that passes through it — something that’s simply a property of capacitors.
Is that phase angle a literal angle? Like 60 degrees? Or is that just how it is described?
A phase angle is a mathematical abstraction used to talk about reactance in AC circuits. If you want to learn more about this, I recommend you click here to check out a webinar recording that talks all about phase angles and motors.
With Run capacitors, the motor doesnāt need help starting?
How does it start if its just alternating 60hz?As their name says, run capacitors don’t have anything to do with starting the motor. A split-phase motor with a run capacitor will most certainly still have a start device of some kind. I see you’ve already been through our Refrigerators course, but we have a lot in there about motors and start devices, which might be helpful for you to review.
On micro farads.
1. Does the higher the mfd, create a greater phase angle?
2. Does it hold a higher charge?Yes, more microfarads means that the capacitor holds a greater charge. And it does also mean a greater phase angle, but only to a certain extent. There’s an optimal range of microfarads for a start capacitor to be in, and if it goes above that, it begins to be more wasteful than helpful.
Hi Matthew,
Remember that when we’re dealing with a 240 volt circuit, neutral isn’t actually a part of the circuit. In a 120V circuit, neutral is the return path for line voltage. However, in a 240V circuit, L1 is the return path for L2, and vice-versa. Neutral isn’t involved. This may seem odd, but it’s just how 240V circuits work. Each line is providing 120V while at the same time serving as the return path for the other.
The reason why we bring up neutral at all when talking about this 240V circuit is because we need to use it as a reference point for measuring voltage, sine it has an electrical potential of 0V. (Remember, voltage can only ever be measured with reference to another point.) That way, by measuring the 240V circuit with respect to neutral (those green boxes in the figures are supposed to be meters) we can tell how much voltage is present at any given point.
If that all makes sense to you, try revising your answers and give me a reply. I’ll let you know if you’re moving in the right direction and explain anything that’s still unclear.
Hi Matthew,
Yes, when you’re dealing with an electric dryer element that’s not receiving its full 240 volts, half-splitting is exactly the technique you would employ. That way, you can identify which side isn’t providing its 120 volts, and from there you can move on to determining which component on that side isn’t doing its job.
Hi Matthew,
The difference between your answers comes from the fact that the Samurai didn’t round up as much as you did. When you divide 1 by 30, you get 0.03 repeating. The Samurai rounded to about the 7th of those 3s, while you rounded to just 0.03.
Your way of doing it is just fine — a difference of 1.25 ohms really doesn’t matter when you’re calculating equivalent resistance. As technicians, we rarely need to worry about getting exact precision with our calculations.
In short, both answers are correct. š
In module 1, unit 2 of the course, we have a link to the Amazon page for the textbook. You should be able to order it from there.
Sorry for the delay in getting back to you. But it looks like you figured everything out! That is the correct answer for #23, and for #54, that process you described will indeed get you the correct answer. Great work!
I’ll go through each of these questions and point you in the right direction if you need it.
#14: From what you said about this one, it sounds like you’re on the right track already!
#23 and #7: Pay careful attention to the wording of these questions. An electron is exhibiting different behavior depending on whether it’s trying to get away from a negative voltage or get to a positive voltage.
#35: Good job! You already figured out your mistake on this one.
#54: Two things to point out here: first, remember that “I” in equations stands for current, not voltage. So you need to use a different equation to figure this out. Second thing is that, since the loose connection is in series with the bake element, it’s only going to drop a fraction of the total circuit voltage, proportional to its resistance.
#55: You already figured this one out by yourself too!
If you’re still unclear on how to get the correct answer on any of these, just let me know and we can go into more detail.
I think the problem here is that you’re misunderstanding the question. The wording is a little tricky, so you’ve got to pay very close attention.
The question is: Which of the following is not an input to the main PCB?
Write me back when you figure out the correct answer!
When you want to test a specific load, you have two options: take a resistance measurement across it, or measure its voltage drop. Measuring the voltage drop is almost always preferable because, while resistance measurements always need to be taken on a dead circuit, voltage drop measurements are taken on a live circuit.
This is important since there are times when loads or switches will fail under load — in other words, they appear fine when you take an ohms measurement, but once they are in a live circuit, they fail. This is something you can only catch by taking a voltage drop measurement, and this is why you should always take a voltage drop measurement instead of a resistance measurement whenever possible.
By “inputs”, we meant electrical inputs to the board, like those generated by sensors. That’s the standard meaning of that word in this context.
For the heater relay, you have to do a little deduction to figure out what it’s switching. First, you look at the picture of the heater PCB, and you can see that the relay has just two terminals, and that all the other connectors on the board have three or more. If you then look at the schematic, even though it doesn’t specifically call out the heater relay, you can tell which connector is the relay from the fact that it only has two terminals — and of course from the fact that the heater is connected to it. Then, to figure out what the relay is switching, you just have to follow the line that does not connect to the heater and see where it’s going.
As for the voltage going to the sub PCB, you don’t need the manual to tell you whether those are AC or DC to know which they are. You will never encounter AC voltages that are in the 5V or 12V range in appliance repair. In addition to that, you also know that control boards run on DC power, so it wouldn’t make sense for these supplies to be AC.
You were right the first time, actually. When the refrigerant arrives at the compressor, it’s a low-pressure, room temperature gas. As it gets compressed by the compressor, the increased pressure causes it to heat up, becoming a high-pressure, hot gas. As it passes through the condenser coils, it cools off, changing states and becomes a high-pressure, warm liquid.
January 12, 2017 at 1:18 pm in reply to: Module 3, Unit 4: DW drain pump circuit is melting my brain #11449I think the big thing that’s throwing you off here is a misunderstanding of how the float switch works. The float switch exists as a safety device in case of overfilling or leaks. It’s called a float switch because it literally floats on top of the water, and if the water level gets high enough to lift that switch, it makes contacts 11 and 14 which, as you correctly said, would run the drain pump for as long as those contacts continue to be made. This is so that it drains all of the excess water, preventing a flood. However, during normal operation, the float switch is in the position shown on the schematic — making contacts 11 and 12.
Another thing to point out is that A9 and A12 are the timer contacts that supply line to almost all of the loads in the machine, including the timer motor itself, which is why those contacts are closed for most of the cycle. All these loads have line present at them for as long as A9 to A12 is closed, but the timer only closes their paths to neutral when it’s time to run them.
And the reason that A1 to A3 close during the drain function is because they’re part of the drain pump’s path to neutral, not the timer’s. Knowing this, try to figure out exactly how the drain pump gets neutral. Let me know if you figure it out or get stuck!
Yes, you’re correct that a reed or centrifugal switch usually won’t have terminals marked as COM or NO. This question is specifically referring to a micro switch, and I’ve clarified its wording to reflect this. The main thrust of this question is just to see if you can determine what the voltage reading would be across the switch in its current state.
It varies depending on model, but most clogged defrost drains are a result of some kind of manufacturing defect that causes the drain to stick closed, such as the duck-bill grommet you’ll find in some machines. To identify this kind of problem, you need to familiarize yourself with the model you’re working on and see what kind of problems it’s prone to develop.
As for the path the water takes, in every case that I’ve seen, the water is simply flowing across the floor of the freezer and seeping out through the bottom of the door gasket. And some of it does freeze, forming the solid ice that you saw in the bottom of the freezer in the video (as opposed to the rime ice formed by water vapor freezing). But once it gets flowing well enough, it’t moving too quickly to freeze, and some escapes from the compartment.
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