Christopher Dehnart

[Christopher Dehnart]

Absolutely, thanks Susan 🙂

[Christopher Dehnart]

Mondays are a busy day for me, but I will try and make it to the next one! So to clarify, when Scott said “I know what your thinking, but hey Scott you were just saying I have the potential voltage of 120 here and its going to drive electrons through each of these regardless of whats going on.”

I believe you said earlier that what happens in one branch of a parallel circuit will not effect the other branches. UNLESS, if I understand correctly, it is a shunt. Which technically is a bypass that controls the operation of certain components.

Meaning a branch off must contain a load in order to be considered a parallel circuit, and if it does not contain a load then it would be considered a shunt.

I think what was hanging me up was a shunt looks much like a parallel circuit on a schematic, and earlier you were saying that what happens in one branch, would not effect the others.

I was thinking that even if you added a shunt, the other two bulbs(loads) that you were bypassing should still come on because the potential voltage of 120 is present at one end, and a direct shot back to Neutral would complete the circuit.

So my misconception has been identified and eliminated (right?)

[Christopher Dehnart]

2880Watts
Which granted is hotter but if I, P, E, and R are all relevant, meaning if you have 2 you can calculate any of the others using Ohms Law Pie Chart

Then if you keep your Current the same at 6amps (which normally would change depending on your resistance) and up the resistance to 80 then a equation of E = I x R would mean that your voltage would be 480 volts. In our example we are working with a 240 volt system with a loose connection which would only raise the resistance, lowering your wattage. Which is what is confusing me when you think about a loose connection burning up.

[Christopher Dehnart]

So for example 1800 Watt heater, 32 ohms, 7.5 amps with 240 volts.
Using the equation.
p = I(squared) x R
1800Watts = 7.5amps(squared) x 32ohms
So if we increase the total resistance (ohms) to say 40ohms we have:
1440 Watts = 6amps(squared) x 40ohms

Your statement “From this equation you can see that increased resistance can increase the heat generated” if I am working this equation right, you are wrong. Increased resistance LOWERS Watts which is your heat.

I am assuming you are right and that I did something wrong, but I am not seeing it.

The Watts, which Scott described as “load performance. Example: light bulb or heat from a heater” goes DOWN along with the Amperage when resistance is increased.

Why then does (the loose connection) increase in heat, when the equation shows that it should be reducing in heat?

You could classify both the loose connection and the heater as a resistor with a measurement of resistance. If one or both increases in resistance for what ever reason then it should lower the heat(watts) of both.

A heating element with a higher Wattage (which if the voltage stays the same, means less resistance) will give off more heat. A element with a lower Wattage (same voltage, greater resistance) should give off less heat.

The greater the resistance the less heat(watts) we should be getting according to the equation.

So a loose connection will not increase in temperature, but decrease.

My assumption before this class was that a loose connection heated up because it was loose and then would get hotter and hotter as it deteriorated until it burnt up completely. According to the equation that is wrong.

But what is really happening (if I now understand this correctly, please inform if I still have it wrong) is that a loose connection at the time of the slightest signs of resistance will be at its hottest temperature and because it is not designed to get hot erodes away getting cooler and cooler until it finally (and this does not make sense) burns apart. If at its worst condition(high resistance) then it should be at its coolest temperature, which means it shouldn’t “Burn up”

I know I am missing something here, but I dont know what. Can you see the error in my thinking?

[Christopher Dehnart]

You know what Scott, I think your right… Ive been working as a tech in the business for close to 9 years now and thought I had a decent grasp on real diagnostic work, or at the very least a good basic understanding of electrical knowledge.

I think what I really have is “good” pattern recognition “skills” Which as I continue to learn through your courses, is not going to cut it. I do use my multi meter on close to every call. But I see now that it is limited greatly by my own understanding of how things work.

I was excited to start learning the real in depth stuff thinking I had enough experience and know how in the basics to understand it all. Obviously I was wrong, humbly I will start at the beginning.
Guess a cup truly can not be filled unless it is first emptied.

Thanks again Scott.

Chris

[Christopher Dehnart]

Haaaaboy…. haha ok I see it now. I had this exact problem occur on one of these maytags last year, and I ended up calling techline (who had me replace the timer) because I missed that contact on the centrifugal switch then, just as I have now… I was thinking that was a open ended contact much like the second switch/contact that opens for the start windings. Embarrassingly, I started at this schematic for over a hour trying to see what I had missed thinking maybe I was not understanding the time chart correctly!

Thank you for clarifying this for me. Kills me that I missed that! I consider myself to be a pretty observant person… Which makes this a humbling moment hehe.

Thanks again Sensi Master Samurai Repairman!
*bows in respect*

[Christopher Dehnart]

I apologize Scott, the Time Chart I posted does have your markups on it in addition to some of my own and I did use a screen shot to capture the image so it has your logos, mouse cursor and youtube references. But the Schematic is MY OWN NEW markups. It looks close to yours as I used the same colors but the difference is my markings of Cams 2 and 4. I dont believe you addressed Cams 2 and 4 in your video.

I did however re-read your post carefully and using your links marked up a second and third schematic to show you my thoughts. I know your time is valuable and that trying to explain these concepts to those taking your class can be aggravating. So I apologize for my inconsistencies and mistakes.

Everything makes perfect sense to me on the regular cycle.
https://drive.google.com/file/d/0BymWvdFqq7ofd19xc3Z2T2twX28/view?usp=sharing

However mapping it out with my markups for the delicate cycle confuses me as I do not see a pathway to complete the circuit, providing a Neutral to the Start windings on the motor.

https://drive.google.com/file/d/0BymWvdFqq7ofYVJUTGpTQk54aTQ/view?usp=sharing

[Christopher Dehnart]

Sorry Scott, did the pictures I added to my post/question not show up? I know your forum shrunk them down a bit but here are the links to the images themselves with a different web host. The images I posted were blanks that I marked up to show my thoughts.

Here are the links
https://drive.google.com/file/d/0BymWvdFqq7ofY3lNZEhqU3NaSkk/view?usp=sharing
https://drive.google.com/file/d/0BymWvdFqq7ofb2hpamJMVEVtajg/view?usp=sharing
Thank you Scott.

Chris

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