Susan Brown

[Susan Brown]

I am happier! 🙂

#5 and #7 are now correct. (I hope you are writing all of these down.)

Another way to think about #7, part 4, is that we’ve lost current in one branch, but it is unchanged in the other, so the overall current is lower.

#8
Yes, the ignitor reaches N through the detector switch.
What about the booster? Do the Zen trick on that.

#9
There is still more detail we want to see. Please answer this:
What do the measurements in Figure 1 tell us about current flow? (Think about what creates voltage drop, and the fact that we are measuring 0vac across the element.)

[Susan Brown]

#4 is correct. I am going to hide your answer so we don’t give it away to other students.

#5

#5 = I/ R1+R2+R3
Therefore:
1/ (1/10ohm+1/20ohm+1/40ohm) 1/ (0.1+0.05+0.025)
1/0.175 = 5.71ohm

You’ve got the correct formula for calculating equivalent resistance.
But look at the diagram again for Question 5. Can you see what you did wrong above?

#7: part 4 is incorrect. Watch the last video in Unit 4.

#8: do the “zen trick” on the ignitor. How do you reach N?

#9:

Suppose to be 240 ac V

true

but as it shows L1 & L2 have no power supply both line somewhere disconnected.

Are you saying that both L1 and L2 appear to be disconnected? Then where is the 120vac that we’re measuring coming from in Figure 1?

Look at Figure 1 again. We know that the element has continuity (we say that in the problem statement.) We know there is some voltage present.

Question: Is current flowing in the circuit?

[Susan Brown]

We’re happy to help.

A couple of questions to start with:

1. You had gotten 4 and 5 correct on your first attempt. Do you not remember how you did those?

2. Did you read through the Help Page and follow any suggestions there?

[Susan Brown]

Hi Ted,

Glad to help. Which unit in Module 3 is this about?

[Susan Brown]

Okay, great! It will be worth the effort, and we’re here to help you as much as you need.

God bless you, too!

[Susan Brown]

Hi Everardo,

The questions that you keep missing are important concepts that you need to know for electrical troubleshooting.

Do you keep a notebook and take notes when you watch videos and go through each unit?

Have you gone back over each unit quiz and made sure you understand all of the questions and answers?

[Susan Brown]

Hi Jason – no worries. We’ll grade it and if you don’t get what you need to pass, you’ll have a second attempt.

[Susan Brown]

Hi Everardo,

Yes, I’m on the east coast, so the timing isn’t great for you. Don’t worry – once you get through the Midterm, you will be able to move more quickly. This Basic Electricity information is often the stuff that takes longer for students to get.

I=E*R 240*37=6.5 amps
P=IxI R=6.5×6.5×32=1350 watts.

You are close! The question we are working on is this

Question #38 – You’re working on an appliance with the circuit configuration shown below. R1 = 5 ohms. R2 = 32 ohms. If everything is working correctly, what is the expected heat produced by R1?

R1 is 5 ohms. So to answer #38, your second step should use 5, not 32, in the calculation of P.

I reset the Module exam for you.

[Susan Brown]

I’ve asked Sam to help you with the first part of your question.

Here’s the Fluke product he was talking about in that video

https://www.fluke.com/en-us/product/accessories/probes/fluke-80pk-8

[Susan Brown]

There’s not a practical difference between 18.75 and 20 ohms in terms of the loads and circuits we deal with in appliance repair. Those two numbers can be considered the same.

In fact, you can generally get by in the field with just remembering the rule of thumb, that the equivalent resistance will be something less than the smallest resistance.

But occasionally it will be helpful to know approximately what that actual eq. resistance is.

[Susan Brown]

Thanks! Glad to help 😀

[Susan Brown]

Try the calculation again and let me know what you get.

[Susan Brown]

Hi Ted,

If you are off by just a little bit, it could be from how you round the decimal numbers.

For example, 1/30 is 0.0333333…

If you round that to 0.03 and continue on with the calculation, you’ll get around 20 ohms.

If you maintain a few extra “3s” on your calculator, you’ll end up with 18.75 (ish).

If you aren’t getting close to that number, then watch this video, which should help you:
(the numbers are slightly different, but you’ll see the technique.)

https://www.youtube.com/watch?v=iHB3lxdc68E

[Susan Brown]

Sorry – I had replied to your earlier question, but for some reason it didn’t get saved. Annoying!

But first, no, the bonus info is not on the Final.

The wire harness:
He’s never encountered a problem with a wire harness in this kind of setting, where there isn’t regular flexing of the wiring. You might see that in settings where wires are near a door/door hinge and get a lot of motion or friction.

If it did occur, you could do a wire splice to get the right connector configuration.

He likes the little Klein because it has more than one function and is small enough to fit in the little admin bag he carries, without having to open and dig into his larger tool backpack. It’s not quite as accurate as the gun, but accurate enough for most applications. That being said, he used an IR gun for years before he got this one, so either is fine.

[Susan Brown]

You are getting close.

You found the current that is flowing through the circuit, 6.5 amps.

You used the correct formula for the heat generated, P = I x I x R
(or, I squared times R)

The mistake is that for R, you used the total resistance in the circuit, which is how you would fine the heat generated by both loads combined. (Side note: your math doesn’t work out… 6.49 x 6.49 x 37 is 1558)

We just want to know the heat generated by the 5 ohm load.

[Author]

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