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Just to make it a lighter half joke, “ clear as mud to this crawdad!”
Like Susan said without a diagram it’s hard to say what is going on. In your example we see a L1 going to a switch on one side the other side of the sw going to a light bulb the other side of light bulb going to L2 or N. now if that “shunt is on the leg of the sw going to the bulb there will be resistance yes but no current flow as it is required to go on thru the bulb. So no lite & didn’t matter what the resistance is the flow stopped when the bulb went.
Danny I can’t speak for Mr Samurai as to your time scheduling. I can assure you Samurai is a reasonable person that will weigh the facts and come to a reasonable conclusion. Do all you can while waiting. And if you want I assume you could post questions here to get help in understanding electrical issues. Or any issues for that matter relating to repair of appliances.
One side to the other on the open switch will result in line voltage showing up at the switch
On a related note. Since the recomended grades on quizzes is either 80% to 90% do i have to lower my score I f i get a 100% grade?
Just a smart hind end question.Lucky for me I read this thread a few weeks ago and forgot all about it. Then the quiz asked close to the same question. Intoo did a quick lookup and used the formula and got the answer correct. Thank you Susan and Sem
This question is not clear.
It asks where eeps are, in my way of original reading. These 2 points can be one end of an eep but by definition given in text they are truly power points not eeps.So the question is or maybe a more leading question would be where are the power points and the eeps on the broil element? Otherwise its unclear as to the true nature of t he question. A note at the end might also clear this up by stating that there are power points that can have eeps on each side of the broil element.
A point or question could be ” concerning EEPs they can be found only on one side of the load on a circut at one time?” True or false? True
Also because the frost – water resulting from defrosting the pan that catches it is under the Evaporator.
Therefor it can be called correctly the evaporator drain pan. But I could be wrong just using logicOk so now I’m better understanding what I’m suppose to look for. I’ll see what comes next on the test.
Thanks for the help.
Here is my explanation
Ed=IxR
R1 = Ed = 10×1.7 = 17.1Vd
R2 = Ed = 20×1.7 = 34.2Vd
R3 = Ed = 40×1.7 = 68.4Vd
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Total is shy of original
VD =119.7Total
When I did this in the car I got 112vd miss added I guess. No I wasn’t driving. People see me coming and get off the sidewalks.Here is my explanation
Ed=IxR
R1 = Ed = 10×1.7 = 17.1Vd
R2 = Ed = 20×1.7 = 34.2Vd
R3 = Ed = 40×1.7 = 68.4Vd
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Total is shy of original
VD =119.7TotalI see your reason and logic. Let me get back to you on this
That’s a good question. One that evidently have to extract by logic from the information given.
Since none of these resistances in the can series circuits can be known some estimate of voltage drop will be less than that of the original voltage the loss being, I reason, to the resistance in watts and thus heat.
Where would I explore to find the formula? In the previous text?
Exactly
Therefore if the points are on the same side of L1 or N no reading on any point on either of these 2 main points will show any potential. Therefore the word point is general in meaning and kinda miss leading, in general.
Clarify where the points are from a supposed potential to the other side of the potential circuit. Otherwise points could mean points on the same side of the load.
That’s what I mean about points not well taken.Voltage drop is going to be 120vac.
Because the voltage source is 120vac and all the loads are in series so the voltage drop across the whole circut has to be 120vac.
I get this one.
Now to calculate the Vd across each load in a series? -
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