Forum Replies Created
-
AuthorPosts
-
If you use the normal VAC function on your multimeter, you would read 120V across that push to start switch — even though you have no connection to neutral because of the open door switch.
A quick correction to what I said here — because the door switch is open and therefore not providing a path to neutral, you would most likely not read a full 120 VAC across the push to start switch with your standard VAC function. This is because, with that wire not being connected to neutral, it’s just floating there with some amount of ghost voltage sitting on it. So when you make your non-loading measurement, you would get a weird, indeterminate voltage reading — less than 120, but more than 0, probably with some fluctuation.
Here are a couple videos by the Samurai that illustrate this concept very well:
Hi George,
We’re dealing with a pretty simple circuit here: just the two switches in series with each other. Let’s say that you’re measuring across the push to start switch with it not actuated (and therefore open), and the door switch has failed, going permanently open.
In this case, the reading you would get depends on the meter you use. If you use the normal VAC function on your multimeter, you would read 120V across that push to start switch — even though you have no connection to neutral because of the open door switch. However, if you make the measurement with a loading meter, you would read 0V across the push to start switch, since even with the 120V present at one side of the switch, there is no valid path to neutral. That’s a much more useful measurement, since it shows you that something in the circuit is interrupting your power supply.
This is one of many cases where using a loading meter instead of a normal voltage meter is essential. The standard VAC function would lead you to believe that everything about the circuit is fine. But the loading meter clearly shows that there is something wrong with the circuit.
June 14, 2017 at 11:07 am in reply to: 1.3 Fundamentals -Basic Electricity end of Module exam #12147Yes, that’s the correct answer. Good job!
When someone says “there are 0 volts from L1 to neutral”, that’s just a shorthand way of saying “when I take a voltage measurement with one lead on L1 and the other on neutral, I read 0 volts.” You could also say “there are 0 volts between L1 and neutral.” It all means the same thing — that is, that you’re measuring L1 with respect to neutral. And yes, even though neutral isn’t involved in this 240V circuit, you still use it as a reference point for measuring the presence (or absence) of voltage.
As for question #6, your answer is a bit off. Could you show me your calculations?
Sure thing — I’ve sent them to you.
Yes, I am the Samurai’s son! I doubt I can give him a run for his money yet, since everything I know I learned from him, but maybe someday…
And yes, the capacitor would be charging and discharging at 60hz, in time with the voltage’s reversing polarity. That’s how capacitors behave in any AC circuit.
Yes, exactly right! Your answers were so good, in fact, that I’ve hidden them so that no one steals them. 😉 Great work!
Hi Josh,
Let me answer each of your questions one at a time.
So, I’ve heard some people call capacitors, “batteries”.
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 doesn’t 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. 🙂
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!
-
AuthorPosts
