Susan Brown

[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.

[Susan Brown]

Can you do the calculation for #38? Let me know what answer you get.

[Susan Brown]

Here’s what I wrote to you:

Before I reset you, I’d like to work with you a little bit to help you get a better understanding of the material.

Please look through the results of the exam, at the questions you missed. Are there any of those that you don’t understand what the correct answer should be and why?

I would like to help you!

Also, do you have a notebook where you make notes on the course material?

In other words, please let me know a few of the questions that you don’t fully understand so we can talk through them.

[Susan Brown]

Sounds like you’ve got it.

If 120vac is present at the load, this means there is no break on the L1 side of the load. Thus, the open is on the neutral side.

If the fault were on the line side of the load, you would not measure any voltage at the load.

Make sense?

[Susan Brown]

Yes, but if there’s an open on the L1 wire somewhere between the outlet and the load, how would that voltage be present at the load?

[Susan Brown]

I’m not sure what you mean about measuring 120vac “through” neutral. We measure it “with respect to” neutral. Meaning, one probe of your meter is at the terminal on one side of the bulb, and the other probe is on a known neutral point.

On the other hand, when we place our probes on either side of the bulb, we get no voltage difference – they are at the same potential. Thus, we know that there’s no current flowing, and the bulb is just acting like a wire.

If the open in a circuit is on the “hot” or “line” side of a load, where would you be getting the 120vac from?

[Susan Brown]

Correct – there is an open in the circuit.

That’s part of the answer. The other part is which side of the circuit has the open.

It’s a 120vac circuit, so line is on one side of the load, and neutral is on the other.

We measured 120vac at one side of the bulb. What does this tell us about the location of the open – which side of the circuit is it on?

[Susan Brown]

I’ve been replying to your reset requests, asking you to work with me on some of the questions you missed. I haven’t heard back from you. Have you received those emails from me? (if not, check your spam folder.)

[Author]

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