Susan Brown

[Susan Brown]

The question is, “The LP gas pipeline and distribution system is very similar to that for natural gas.”

The answer is “false” because LP is not distributed via pipelines, but is always in tanks on the property.

[Susan Brown]

Hi Chad,

The disconnection point is on the centrifugal switch side (L2) of the element – we are disconnecting the wire at that point. The controls side of the circuit (L1) is still connected to the heating element. So the point is that after doing that half-split, we measure 120wrt N on the “disconnected” wire (L2) and 0v wrt N on the connected side (L1). So, we know that there is an open somewhere on the L1 side. This the side with the controls, so the correct answer was “There’s a problem somewhere in the control circuit (thermostat, hi-limit, timer, etc.)”.

It’s not important to the question whether the motor is running – the 0v wrt N reading is abnormal and indicates a problem. That’s the point of this question, to understand the basics of half-splitting in and L1-L2 circuit.

[Susan Brown]

Sure, I understand your point. This isn’t a question we see many people stumble on, however. I think sometimes someone with more of a background in this material can “overthink” it, if you know what I mean 🙂

[Susan Brown]

Hi Chad,

The answer we are looking for here is “identical”, because that was one of the main points of the video – that identical circuits can be represented in ways that look different at first glance. For example, drawing a circuit as a loop vs. a straight line. Or some parallel configurations can look different, but when you look at the relationship of the loads more closely, you realize they are identical to an electron.

Note – we don’t mean that all the circuits in the entire video were identical to each other. Just that he would show two on the same slide that were.

I hope that clears it up for you!

[Susan Brown]

haha, we’ll look the other way whenever you exceed expectations 🙂

[Susan Brown]

Great!

[Susan Brown]

P.S. When I send emails to you, they come back saying that they were blocked. Are you not getting them?

[Susan Brown]

Hi Dave – same Module (Basic Electricity)

[Susan Brown]

Here are the sections from the unit that answer these questions. Let me know if you need any further help.

Question 4
If the motor is running, but the machine is not agitating, then you are most likely facing one of two issues, depending on the method your model uses to drive the tub. On belt-drive models, a worn or broken drive belt causes this issue.

Alternatively, on direct-drive models where the motor directly drives the transmission, the most likely cause is a broken drive coupler. This is a small plastic component that connects the drive motor to the transmission.

Question 5
In order to properly drain the water from the appliance, the drain hose must be looped higher than the top of the tub and not pushed too far into the drain pipe (if applicable). If the hose is not installed properly water entering the tub may drain out simultaneously, preventing the washer from completing fill.

[Susan Brown]

Hi Carl,

Have you submitted a Contact form at Appliantology? That’s the fastest way to get an answer.

All I can see from my end is that you signed up for your free 6-month student membership in July, so it would have expired this month. Did you ever submit your application for the Alumni membership? We would have sent that link to you in August, when you passed the Core course.

If not, then perhaps that is your next step:
https://appliantology.org/blogs/entry/1005-the-master-samurai-tech-alumni-program/

[Susan Brown]

Yes – that’s it!

[Susan Brown]

Hi Sem,

I can’t remember off the top of my head if we have that conversion somewhere in Core, but a quick internet search told me that

1 BTUH = 0.0004 hp
1 hp = about 2500 BTUH (2544 to be exact)

[Susan Brown]

Here’s what we teach in the unit:

“In general, larger diameter conductors can carry more current. But, as you can see above, there are other important factors to consider such as the metal of the conductor (eg., copper vs. aluminum) and the temperature rating of the insulation.

“The diameter of a conductor is given as a standardized size system called American wire gauge (AWG). The thing to keep in mind with AWG, often just called the “wire gauge,” is the wire diameter increases as the AWG number decreases. You can see this in the image to the right.”

So the higher the “wire gauge”, the smaller the diameter, thus the smaller the amount of current it can handle. Which means the correct answer to the queston you keep missing is “decreases.”

It’s annoying that someone set it up that way, but it is what it is!

I reset you.

[Susan Brown]

That is an excellent question – it shows you are thinking things through.

It all comes down to *power*.

P = I x E

Distribution lines are delivering a certain amount of Power (watts). So, looking at P = I x E you can see that if E is higher, I will be lower.

Power also has to be in the top of your mind when it comes to appliance circuits.

When we analyze a circuit, we have to keep in mind causes and effects, and what the primary job of the power supply is.

And that is: to deliver a certain amount of power (watts) to a load. Engineers figure out the optimal wattage for a load to do the work they want it to do (heat, motion, light, etc.), then figure out the best way to deliver that.

In the case of an electric oven or dryer, the wattage needed for the desired performance is high. If the regular 120v household supply was used, the required amperage (to achieve the desired wattage) would be so high as to require higher gauge wiring and connections – which is expensive. But if 240v is used instead, the amperage is at a lower level so that the regular wiring is fine.

In other words, the power required and the voltage used to deliver that power ultimately decides the current in the circuit.

We can use I = E/R to determine the current in a circuit if that is all we are given. E is actually “voltage drop” in this equation, which will equal the supply voltage if R is the total resistance in the circuit. This equation is actually just showing us the result of design decisions that were made to achieve a certain wattage.

I know this stuff makes my brain hurt a little bit, but keep thinking about it, and remember that the desired power/watts is always the overriding factor that determines the particulars of a circuit.

I hope this helps!

[Susan Brown]

Yep!

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