Susan Brown

[Susan Brown]

Hi Kevin,

We don’t hear this kind of thing often, but I think there are occasional internet/connectivity glitches that can mess with the timed tests somehow. It’s not a function of the programming, but of internet weirdness.

No worries – I’ll grade your attempt shortly and you can always retake it if need be. I’ll email you with feedback.

As for comparing diagrams and schematics – you will get better with practice. It does take some time, but it’s super helpful to be able to see things from an electron’s perspective (the schematic diagram) than from just the wiring diagram. Stick with it!

[Susan Brown]

We always appreciate suggestions!

[Susan Brown]

That seems more complicated to me – I prefer one calculation – but it does work out mathematically.

[Susan Brown]

Hi Tyler,

I just looked over your quiz for that unit (it’s Unit 3 that I assume you’re talking about, since your quiz score is low).

Many of the questions you missed are on definitions or descriptions that we gave in those first 3 units of the course. For example, the answer to Question 1 is found in one of the definitions in Unit 1.

When you went through each unit, did you:
1. Print out and fill out the Study Sheets?
2. Take extra notes?
3. Rewatch any videos, or sections of videos?

These items are generally necessary when you are learning this type of material for the first time.

Please let me know of some specific items that you are unable to figure out by going back over the first 3 units of Basic electricity and I’ll help you further.

[Susan Brown]

Hi Carl,

One correction and one comment.

First of all, when we’re talking about parallel loads/circuits, we use the term “equivalent” resistance. Total resistance is when we add together loads that are in series.

That being said, your equation is fine as long as you only have 2 loads. If you had 3 or more, it wouldn’t work and you’d have to use ours.

Your equation is just a mathematical rearrangement of 1/[1/R1 + 1/R2], FYI.

Cheers!

[Susan Brown]

Hi Jim,

#4/#5: in the problem statement, it says, “After verifying that the washer is plugged in to a good power supply, you see that the line cord is connected to a square box component.”

So, we’ve verified that the component (which is a noise filter) has input. The next step is to check its output.

(Did you see the Explanation that shows up in the quiz results?)

#10:
From Mod. 5 unit 3:

1. Triacs are used to control AC power supplies

2. You can think of them as solid state relays

3. Triacs are current controlled devices. This means that you need electrons bustin’ down the Gate to turn it on AND you need load current flowing through them in order to stay on.

So, they need AC supply as well as the DC gate voltage, and their output is AC.

#17:

Yes, the centrifugal switch is at the motor (picture of a typical one at the link below). The schematic shows you that there is a direct wire from the switch to the element, so it makes for a great wire to disconnect for half-splitting. That’s the beauty of schematics!

https://appliantology.org/gallery/image/332-dryer-motor-centrifugal-switch/

Let me know if you have any other questions.

[Susan Brown]

Hi Travis,

From Unit 1:

Voltage: The amount of potential difference between two points in an electrical circuit, and the driving force behind actual current (flow). Voltage is also referred to as “potential energy” or “Electromotive Force” (EMF), the force that causes electrons to move from negative to positive. Measured in “volts,” and usually referred to by “V” or “E” in equations.

[Susan Brown]

Did you receive my email with feedback and a link to our Midterm Help Page?

Please read through those and then come back with any followup questions.

Enlightenment is near!

[Susan Brown]

Hi AJ – a couple of clarifications for you:

1. The criteria for certification are 80% or higher on unit quizzes; 90% or higher on exams (module, midterm, final)

2. Please use the Quiz and Exam Reset Request form (it’s in the Campus Support menu) if you should need a reset in the future. Thanks!

[Susan Brown]

Yes – I reset that quiz for you.

[Susan Brown]

Good! Let us know if you have any other questions.

[Susan Brown]

It looks like you figured it out…

[Susan Brown]

The easiest way to do this is in 2 steps.

1. Find the circuit current, then
2. Use the circuit current and the resistance of the load of interest to calculate heat (P).

We use these two steps in the video at the end of Unit 3 about the heat generated by a loose connection. It’s a similar scenario.

Watch that video and then let me know if you can follow those steps or if you have follow up questions.

[Susan Brown]

From what I said above, with emphasis added:

“Note – I = E/R, so it should have been 120/5005 for the current.”

Voltage drop is E = I x R

I’m sorry I wasn’t clear – that part of your answer was correct. You had not done the calculation of current (I) correctly.

But – you didn’t even need to do those calculations to answer the two questions. You just had to look at the relative amounts of the resistances.

[Susan Brown]

I just wanted you to tell me what Ohm’s Law formula you would use to find voltage drop.

I looks like you used E = I x R in your second step above.

Note – I = E/R, so it should have been 120/5005 for the current.

But the point of the questions you missed was
1. which resistor would drop more voltage, and
2. how much work is R1 doing compared to R2

You can answer both of those by just looking at and comparing the values of the two resistors, since you know that voltage drop and work are proportional to resistance.

[Author]

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