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That’s exactly right. You’re not going to mess around with calculating ohms. You WILL use the wattage spec of an AC load with its voltage supply spec to calculate the expected amp reading on your meter to see if the load is operating in spec electrically.
In a circuit with a single load, such as a dryer or dishwasher heating element, all of the supply voltage will be dropped across that load. In this case, voltage supply and voltage drop will be the same.
I have a good webinar recording that goes into this in more detail that you should watch: https://appliantology.org/topic/72423-voltage-voltage-drop-loads-switches-jumpers-cheaters/
Let me know if you have any other questions.
It would behave the same way. Functionally, the only difference between a 3-wire and 4-wire system is the Neutral-Ground bonding point. In the 3-wire system, Neutral and Ground are bonded at the appliance. If the 4-wire system, Neutral and Ground are bonded at the circuit breaker box. Electrically, either system would behave the same way in the failure scenarios discussed in the video.
This webinar recording at Appliantology discusses this in some detail: https://appliantology.org/topic/82369-dryer-cord-wiring-weirdness/
July 26, 2022 at 11:23 am in reply to: Circuit Components and Series Circuits (calculations series circuit) #24029You usually won’t do these types of calculations in the field but they are in the course to help you understand how electricity actually works. Since we can’t see electricity, we can only understand it by using the ohms law equations, which are essentially a few ratio relationships between volts (E), amps (I), ohms (R), and watts (P).
The other equation you mentioned is actually P=E^2/R– E X E DIVIDED BY R. You would still need to use the voltage drop across the load in question (E) and the resistance of that load (R). It’s useful when you know E and R but don’t know (or don’t need to know) amps.
It’s more correct to say that copper has very *little resistance* compared to loads. Copper actually does have some resistance. Power companies deal with this when distributing high amps through power lines. It’s call “I-squared R losses” because it’s based on the ohms law equation, P= I^2 * R. Where I is the amps moving through the lines and R is the copper resistance. This is the loss of power in heat just from moving high amps through power lines. The resistance of the copper is what it is so nothing you can do about that. But the delivery amps you can do something about. If you jack up the voltage, then you need to move fewer amps. This is from P= I*E where P is the deliverable power, I is the line amps, and E is the line voltage. That’s why power companies deliver power at high volts, say around 14,400 volts and then step it down at transformers on the pole near the house to your 120/240 supply.
With windings, you’re dealing with lots of copper coiled together. So you have some copper resistance there. But bigger than that, you have inductive reactance, which is another opposition to electron movement (amps). Your meter on ohms only measures resistance, not the reactance. Reactance only appears when electrons are moving back and forth in AC power. In motor windings, the reactance will be much higher than the copper resistance and is the main opposition to electron movement. That’s another reason you want to rely on amps (as a proxy for watts) when diagnosing AC loads like motors.
Measuring for resistance in this case is iffy because the resistance you’re looking for is so low– just a few ohms. Better is to work on the live circuit and measure voltage drop across the various connections. In a 240 VAC circuit, even a resistance of just a few ohms will show up as a big voltage drop.
Is the UI connected to the main board by a ribbon connector? Those are famous for getting some corrosion or oxidation at the ribbon connector pads. Sometimes, just cleaning the pads with a pencil eraser is all that’s needed. Also, what kind of keypad is it: membrane or capacitive touch?
These kinds of specific troubleshooting questions are best asked at Appliantology where we can link you to resources and other techs can chime in. That’s why we give free Appliantology membership to MST students. The MST forums are for questions about the course material. I’m happy to help you with this question but please keep this in mind for future questions on specific service calls.
I presume you were on the call in the first place for a warm compartment complaint. On these calls, your first troubleshooting move is to half-split between the sealed system and everything else. There are three common ways of doing this explained in this article at Appliantology: https://appliantology.org/blogs/entry/1227-3-ways-to-half-split-warm-refrigerator-problems/
From your description, it sounds like you’re dealing with a split phase compressor as opposed to a variable speed unit. If this is the case, it’s very likely that the sound you’re hearing is the famous hum-click cha-cha. This is usually caused by a failed compressor start device that is no longer powering the start winding in the compressor. Without the start winding supplying that temporary second phase, the compressor main winding is experiencing high amps and you’re hearing the 60 Hz hum because the compressor is unable to get the rotor spinning from a dead stop. Those high amps create heat. If this goes on very long, the compressor windings will get damaged from the heat and the compressor will need to be replaced.
An easy test for the start device is to remove it and shake it. If it rattles, it is fractured. I have a webinar on split phase compressors and PTC start devices at Appliantology: https://appliantology.org/topic/64098-mst-office-hours-8212017-split-phase-compressors-and-ptc-start-devices/
Another half split on the sealed system is whether or not the freezer is at 0F. This can still be true regardless of frosting, depending on how big the opening is. If the freezer is at 0F, this eliminates the sealed system right off the bat. Also, frosting in the freezer (or condensation in the beer compartment) is almost always an air leak problem, not a sealed system problem. It’s the warm air infiltration that brings in the humidity that then freezes (or condenses ton liquid) and forms the frost.
Also keep in mind that a normal condenser split presumes that you have normal heat transfer at the evaporator coil. If that is not true, then the condenser splits will be low. For example, if the evaporator coil is caked in with rime ice, you will not have normal heat transfer occurring. Also, you need to ensure that the evaporator fan is running with the cover in place to direct air flow across the coils. And if this is a variable speed system, you need to make sure the compressor is running at full speed. Without all these conditions true and in place, condenser splits will be low and could lead you to incorrect diagnostic conclusion.
There’s a post at Appliantology that goes in to more details about the condenser half-splitting along with a couple other methods for half-splitting between the sealed system and everything else: https://appliantology.org/blogs/entry/1227-3-ways-to-half-split-warm-refrigerator-problems/
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