Susan Brown

[Susan Brown]

A good video to watch to help you with those two questions is the “loose connection” video at the end of Unit 3.

The scenario is similar – two loads in series, and you want to calculate the heat generated by them. One load happens to be a loose wire connection. So, even though it’s not intentional, because the loose connection has a small resistance, it acts like a load on the circuit. Watch that video and see if it helps.

[Susan Brown]

Okay – I just reset you.

[Susan Brown]

P.S. Do you want me to reset that quiz?

[Susan Brown]

If you look at the written info in that unit about symptoms of clogged orifices, one of the answer choices we give you is NOT on that list.

It’s also kind of intuitive – if the orifice is blocked somehow, what would you NOT expect?

[Susan Brown]

Look back at the Unit 8 quiz (Voltage Drop and Load). Do you know how to calculate the correct answers to #7 and #8? (You also missed those on the Module exam). You’ll need to be able to do this type of calculation on the Midterm exam as well.

[Susan Brown]

Hi Everardo – I just sent you an email.

[Susan Brown]

Hi Darren,

He uses a standard Apple iPad that he got a few years ago.

Here’s the page at the Danfoss website with various downloads for the Refrigerant slider (now called Ref Tools), including a web version you should be able to use on your laptop.

https://www.danfoss.com/en-us/service-and-support/downloads/dcs/ref-tools/#tab-overview

You can download apps onto any devices that you have!

[Susan Brown]

Okay, Everardo. I reset you.

[Susan Brown]

Hi Aaron,

The thing to remember about Module exams is that you’ve seen these questions before in the Unit quizzes. So, it is giving you important information about which areas you still need to get stronger on.

Go back through your Module 3 quizzes. Be honest with yourself – which questions/answers do you understand, and which ones are you still shaky on?

For the ones you are shaky on, review the material in the unit and/or get help from us here in the Forums.

Are you keeping a notebook? Taking notes as you go through the units is really important.

[Susan Brown]

We told you about the book in the “preflight checklist” in the Orientation.

We also told you about a .pdf version that you can use for free in Module 3, unit 1.

See this topic

https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/free-pdf-textbook/

[Susan Brown]

I’m going to help you with #9 tomorrow. You don’t seem to have been following the train of thought with that one.

[Susan Brown]

So that mean, are booster and ignitor in a parallel circuit?

correct. Does any current go through the Main coil?

can you email me #7 answer that you hid it.

You should be working on these questions in your notebook, and I would hope you are understanding the answers well enough to re-create them.

[Susan Brown]

#8: If you become the Booster, why would you not make the same choice as the ignitor and go through the detector switch to get to N? (electrons don’t “see” or feel any bends or turns in wire. They only feel resistance and voltage.)

#9: that’s correct. We know the element has continuity, so there must be an open on one side or the other. How do the measurements in Figure 2 give us more information about the location of the fault?

[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?

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