Susan Brown

[Susan Brown]

I reset you

[Susan Brown]

Sure! That will be a good review for you. These units are very common ones to need a little more help on (and often resets). Be sure to ask questions if you need to.

One other thing about your original question. On my dashboard, I can only see the score from the most recent attempt. If you ever have a situation where your first score was higher (and qualified for Certification), you would have to send me the email you received with that score. Otherwise, I’d have no way to know.

[Susan Brown]

Hi Ethan,
First of all – you indicated in your student info that you want to earn Certification for the course. Do you recall the requirements for that?

You must earn 80% or higher on EACH unit quiz, and
90% or higher on EACH exam in the course.

So, not only would you need to retake the exam anyway, but there are two unit quizzes that are below 80%. In order to retake those, you would need to get set back to Mod 4, unit 5.

Let me know!

[Susan Brown]

Hi Michael – I reset it so you can start fresh and answer all the questions.

[Susan Brown]

You got it!!!

(FYI, I’ll hide this answer so we don’t just give it away to other students. They need to experience their own “a-ha!” moment 🙂

[Susan Brown]

Hi Victor – I reset that quiz for you.

[Susan Brown]

Hi Kenneth,
Right, because if you only have one load in a circuit, the voltage drop will equal the source voltage.

The key here is the impact of the closed detector switch on the circuits with the Ignitor, Booster, and Main.

If you do the “Zen trick” on the Ignitor or the Booster, how do you “reach” N? Through the closed switch, through the Main, or both?

[Susan Brown]

Hi Chris,

“1 over” is the same as “1 divided by”. So, for example, 1/10 is “1 divided by 10”. When you do that on your calculator, you should get 0.1 (“one tenth”)

1/2 should result in 0.5 (“five tenths”, which is the same as one half)

So, let’s say you have two 10-ohm resistances in parallel.

The calculation would be
1/(1/10 + 1/10) = 1/(0.1 + 0.1) = 1/(0.2) = 1 divided by 0.2 on your calculator = 5 ohms.

Do you get that when you do it on your calculator?

What do you get if you try it again but with two 20-ohm resistances in parallel? Let me know!

[Susan Brown]

It would be similar to a refrigerator – the point is that it’s a cold, damp environment. So the choice of connector and other precaution are important. You had a quiz question about this in that unit.

[Susan Brown]

Hi Raja,

This is a little Basic Electricity practice.

In order to get current to flow through a load, you need a voltage and a complete circuit.

And by voltage, we mean a voltage difference. All voltage is expressed as the difference in charge between two points. A classic reading of voltage potential is measuring from L1 with respect to Neutral (where you use a known-good neutral point as reference). Reading voltage drop is when you are measuring across a load (a component with resistance that does work – element, pump, bulb, etc.). Current flowing through a load produces voltage drop.

So – if you have L1 on both sides of a load, there is no difference in voltage that will drive current through that circuit. So, no current.

[Susan Brown]

If there is only one load in a circuit, do you know its voltage drop? (think about what Kirchhoff’s law teaches)

[Susan Brown]

Hi Troy,

No calculations are needed to answer Question 8, and you definitely don’t need to make something up. The key is seeing the impact that the closed detector switch has on the circuits.

Let’s back up a bit – if you look just at the Safety, do you know what its voltage drop is?

[Susan Brown]

Hi Brian,
I reset you. FYI – it’s best to use the Quiz & Exam Reset Request form when needed (in the “Campus Support” menu).

~ Susan

[Susan Brown]

That’s correct

[Susan Brown]

Also – go to the Core course, Mod 4, unit 4, and watch the second video. I start with a circuit with a single 60 ohm load, then add a 40 ohm load in series. The current in the circuit decreases, and the voltage drop across the 60-ohm load decreases with the addition of the second load. But obviously the 60-ohm load still has 60 ohms.

The main point of the question you are writing about is to make sure people understand that resistance is an inherent property of a load and in the types of Ohm’s Law calculations we are likely to do as appliance servicers, if you are given a value for resistance, that value is fixed. It won’t change in response to a change in current, for example.

(Note – one exception to this discussion is that some material’s resistance will change some based on temperature. Some types of sensors are an example of this.)

• This reply was modified 1 year, 2 months ago by Susan Brown.

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