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Steve, I see that you replied to the notification email telling you that I had replied here. Please post your followups here in the forum instead of replying to the notification emails.
The question was not referring to how a GFCI works.
You are correct- the question is not asking how GFCIs work:
Question #12: A ground fault is
Am I missing something in your question?
Hi Joe,
In theory, if there was not a near short or low impedance fault in the motor, this would work with better meters. Some meters won’t like that you’re running AC through what is meant to be a DC function and you could get confusing results and, instead of clarifying, adds another layer of uncertainty, “Is my meter playing games with me or what?”
In general, I want to observe and use my meter in the manner it was intended. For example, not use a DC current function to complete the circuit for an AC load and power supply.
Most meters will have a fuse in the DC current circuit but you would need to verify this with your meter.
Bottom line is that I don’t see a compelling reason to risk the meter or introduce uncertainty (due to unknowns about the meter behavior) when a simple cheater cord or jumper wire eliminates all that.
Just curious where you heard about using your meter this way?
Ahh, grasshoppah, you have snatched the pebble from my hand!
I did that deliberately to see if students would catch it. You are one of the few. Congratulations!
(And the answer to your questions is YES– the meter should be set on LoZ or using a wiggy)
Hi Joe,
Sorry for the delay in replying– this one slipped by me.
The presentation dealt with the most common type of inverter-driven compressor motor: the three-phase BLDC motor. These units have been very reliable, both the motor and the vapor pump (compressor proper).
The other type of inverter-drive compressor out there is the linear compressor. So far, only LG is using these. The linear motor in these units are very reliable. But the vapor pump, not so much– lots of problems with suction reed valves breaking apart. I suspect this is the type you’re seeing the prevalence of compressor failures on.
There’s a webinar recording on linear motors and compressors that you may want to watch for more detailed information. https://mastersamuraitech.com/webinar-recording-linear-motors-linear-compressors/
I am enjoying the course.
Glad to hear it! ?
And good job asking questions here in the forums. We love helping students!
Do some GE refrigerators require seperate components to do diagnostics or tests such as forced defrost? If so do you always carry these?
On some of the older Arctica platform models, you needed a separate cable and touchpad to enter diagnostics. Although I have had them for years, I have never used them once because I can diagnose everything I need from the motherboard.
Will a Triac turn on with a steady DC voltage on the gate or must it be pulsed?
Triacs will turn on and stay on with steady DC but it is usually pulsed.
Triacs are current controlled devices which means that in order for them to “turn on”, there has to be current flowing through the gate and through the anodes. Since the triac is handling AC voltage through its anodes, the voltage goes to zero 120 times a second– each time the it changes polarity. When voltage goes to zero, current goes to zero because voltage drives current. When current through the anodes go to zero, the triac “turns off” and must be “re-gated” to start conducting again. This can be done by either supplying a pulse to the gate at the correct time or just keeping steady DC on the gate.
I wrote a good, detailed article on triacs for appliance techs which you can read here: https://appliantology.org/blogs/entry/953-triac-operation-for-appliance-techs/
To get 240 VAC at the shop, you would need to bring in the Line that isn’t already there, either L1 or L2. You would need to identify this at the circuit breaker.
If you ran the same Line out there that already exists (example: L1 already present and you run another L1), this will not give you 240 VAC, just another 120 VAC circuit on a different breaker. So to answer your last question:
Wouldnt i just have 2 120v wires from the same circuit, which would be in phase with one another, therefore incapable of producing 240vac
Yes, Just 2-120 VAC circuits. No 240 VAC in this case.
Ideally, the L1 and L2 run out to the shop would originate from a ganged breaker where both L1 and L2 trip together. This may be required by the electrical code, I don’t know.
Putting aside code issues, running the missing Line out to the shop from a separate breaker would technically work. You would want the breaker to be the same amp rating as the existing breaker for the Line that’s already there. You would then pick off your 240 VAC from each separate line. How that would look like physically depends on how you install it in the shop.
If there’s already a Neutral run out there (which there must be if you currently have 120 VAC… unless you’re using Ground as the return path, a big no-no), then each Line could use the same Neutral for return.
As I said, I’m sure this would be in all kinds of code violation but, from a purely technical standpoint, it would work.
Does that help?
January 13, 2018 at 11:52 am in reply to: unit 3 / quiz / question 11…" loose connection, increased resistance in circui #13769Hi James,
Glad to hear you’re enjoying the course! You’re doing great, keep up the great work and it will payoff bigtime!
The first thing to recognize is that the current in a series circuit is the same throughout the circuit and through every component in that circuit. So the current through the loose connection will be the same as the current through the heating element.
The other important thing to know is that in a series circuit, the sum of the voltage drops across all the components will add up to the source voltage. So the voltage drop across the loose connection plus the voltage drop across the heating element will add up to the supply voltage.
Knowing this, we don’t need to calculate the voltage drop across the heating element because this does not help us in calculating the wattage or voltage drop produced across the loose connection.
But total circuit current does.
If we know the circuit current, we can use P=I^2 x R to directly calculate the wattage (heat) produced by the loose connection. We could also use E=I x R to directly calculate the voltage drop across the loose connection connection.
So how do we calculate the total circuit current?
Hi Kyle,
Just to add something about 240 VAC elements that’s true regardless of whether it’s a 3-wire or 4-wire configuration or whether it’s a dryer or oven element.
In a normally functioning 240 VAC element, there should be no current path between either of the power terminals and ground/chassis/Neutral. This means that your ohm meter should read open (usually “OL” on a digital meter) when you have one probe on the power terminal and one probe on the housing or chassis. Of course, power off and at least one wire removed from the element so you’re not reading back through a parallel circuit!
Let me know if you have any further questions on this.
Scott
Hi Sam,
Good question! The difference has to do with the operating temperature of each coil. Since the condenser operates at a higher temperature, the heat transfer rate — the speed at which heat leaves the condenser coil to the outside air— will be much faster than in the evaporator. So the condenser coil diameter doesn’t need to be a large (less surface area).
The other reason the evaporator needs larger diameter coils (and hence more volume) is because it operates at a much lower pressure than the condenser. The diameter is chosen to match the compressor capacity and desired low side pressure.
Remember also what’s happening to the refrigerant in the evaporator vs. the condenser. In the evaporator, the liquid refrigerant is boiling. The tubing has to have enough volume to accommodate both liquid and vapor to allow this boiling to take place. In the condenser, the superheated vapor from the compressor is quickly cooled and condensed or “squeezed” back into a liquid so it doesn’t have to have the space to accomdate boiling.
Finally, I should point out, that the question is inelegantly worded. Just because the evaporator can hold more refrigerant than the condenser doesn’t mean it does. It really should ask, “The condenser has a ____ volume than the evaporator.” Choices: greater, lower. Correct answer: lower. We will edit the question accordingly, thank you for bringing it to our attention.
I hope I’ve answered your question. Please let me know if you have any other questions.
Scott
Hi Troy,
attempts to reignite
Meaning if fuel is present in the correct amount and location, then ignition occurs. As far as the spark module is concerned, it is attempting to reignite by sending high voltage pulses that produce sparks. If it senses the same amount of current in both voltage polarities, then it knows flame has not been established so it keeps trying.
It think that’s what you’re asking. If not, please let me know!
I’ll answer your question about Module 7, Unit 4 in a separate post.
October 11, 2017 at 12:00 pm in reply to: Module 3, Unit 3 Door lock motor assemblies, how to open locked door #13326Hi Troy,
Great question! The short answer is that it depends on the configuration you’re dealing with.
For models with the door lock motor in the front, right at the door, you can usually make a hook out of a coat hangar, for example, fish it in the gap between the top of the door and control plenum, hook onto the latch and pull it. It usually doens’t have to move much to allow the door to open.
For models where the door lock motor is in the back and connected to the door latch assembly via a mechanical linkage, you’ll need to access the back of the range and disconnect the linkage from the motor assembly. This video shows an example of this situation:
https://www.youtube.com/embed/lqWCnQlw6YI
Let me know if you have any other questions!
October 9, 2017 at 1:27 pm in reply to: An outlet can test good on a meter even if the neutral is open #13317Hi Richard, Welcome to the Master Samurai Tech Academy!
The phenomenon here is called ghost voltage. This is voltage capacitively coupled into an open conductor. It is essentially like static electricity. The amount of ghost voltage will be a function of the physical proximity of the open conductor to the live conductor, the length of the open conductor, and magnitude of the supply voltage.
While it is true that 70 VAC is a common presentation of ghost voltage when measured with the VAC function on a DMM, it need not be. The “approximately 70 VAC” reading is just a rule of thumb. Depending on the physical arrangements mentioned above, this voltage can be much more or less that.
It is also true that the meter reads the voltage based on current draw. But it’s all about relative magnitudes. The VAC function on a good quality DMM will draw less than 50 micro-amps of current. At this tiny current draw, it would take a much longer duration of time relative to the duration of the actual measurement to see a decrease in the measured voltage.
By contrast, the current draw on a loading meter, such as the LoZ function or a wiggy (wigington tester) for a 120 VAC circuit is about 40 miliamps— an order of magnitude higher than the current draw for the VAC function on a DMM. At this higher current draw, the ghost voltage is “bled down” almost instantly and the loading meter would correctly and instantly show 0 effective volts available to do work in a live circuit.
After you’ve studied the supplementary information below, you should see that the answer is, in fact, quite correct. However, if you have any other questions, please let me know.
A short demonstration of ghost voltage in a 240 VAC circuit:
https://www.youtube.com/embed/yQb2JgDh-10
A longer webinar presentation on voltage measurements and loading meters:
https://www.youtube.com/embed/VvzvL9y4ev4
An article from Fluke Corporation on ghost voltage and dual input impedance meters: http://support.fluke.com/find-sales/Download/Asset/2718074_6116_ENG_A_W.PDF
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