Susan Brown

[Susan Brown]

Hi Mohammed,

You kind of skimmed through Modules 5 and 6 in the Core course. It would help you a lot of go over that material more attentively.

For example, Mod. 6, unit 4 is “Using Schematics to Troubleshoot Appliances, Part 1” and includes a video discussing using jumpers. (“Voltage & Voltage Drop, Loads & Switches, Jumpers & Cheaters”)

[Susan Brown]

1/80 is 1 divided by 80 is 0.0125
1/480 is 1 divided by 480 is 0.002

1/(0.0125+0.002) = 1/0.0145 = 1 divided by 0.0145 = 68.6 ohms

Does that make sense?

[Susan Brown]

Yes, you can do that with pretty much any powered load in any type of appliance. We teach the basics of doing this in the Core course.

[Susan Brown]

Yeah, that is odd. We often see k-ohms (1000 ohms) in appliance repair.
The Kleinert book has a lot of helpful reference stuff, but it has parts that are aggravating.

[Susan Brown]

They probably mean Megaohm (MΩ)

[Susan Brown]

I think 7-13 says “another” in reference to 7-12. I agree that 7-11 and 7-13 appear to be the same.

[Susan Brown]

Hi Rudi,

The screw driver method is fine. The concern over damage to the capacitor is from arcing that may deform the capacitor connection terminals. But we’ve never had a problem doing this. I think a lot of these cautionary instructions are there for liability reasons.

[Susan Brown]

We want you to focus on the circuit analysis for this question on the exam, and what the measurements tell you. Since this is at the end of the Basic Electricity module, that is the focus. “Missing” is not a very precise term, so think instead about there being an “open” somewhere in the circuit. You are correct that we are not measuring any voltage drop, so there is no current. This lets us know that there is an open circuit that we need to hunt down. That is actually the first part of the answer to this question.

[Susan Brown]

Hi Chris – see the reply I just sent to your email.

[Susan Brown]

The difference is coming from the number of decimal points you are preserving.

1/80 = 0.0125
1/480 = 0.002

1/(0.0125+0.002) = 1/(0.0145) = 69

The most important thing is to know the rule of thumb – that the Equivalent Resistance will be less than the smallest of the resistances.

You can estimate it by knowing two things.

1. The closer the resistances are to each other, the Req will be smaller – but no less than half of the smallest resistance. Test this by calculating Req when both of the parallel loads are 10 ohms. You will get Req of 5 ohms.

2. The further apart the resistances are to each other, the Req will be close to the smallest one. Test this by calculating Req when one load is 10 ohms and the other is 1000 ohms. You get an Req very close to 10.

[Susan Brown]

Hi Michael,

#7: your first two are correct, but not the signal. You’re looking for another “power supply” answer.

#10: your inputs to the board are going to be signals/data. Thermistor temperature is correct. And the other one would be the motor speed.

Did you see this topic?
https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/module-5-unit-5-2/

[Susan Brown]

1/10 = 0.1
1/1000 = 0.001

Req = 1/(0.1 + 0.001) = 1/0.101 = 9.9 ohms

What are you getting?

[Susan Brown]

Hi Andrew,

We did a big upgrade of the Advanced Refrigerator course at the end of November. We sent out a newsletter to announce it and put notices at the beginning of each module and on the Contact page explaining what happened.

We improved the Sealed System units and rearranged a few others so that the flow of the course is more logical. I’m sorry for any confusion it caused you, but you do have some great new material to go over! All you have to do is look in the sidebar navigation to find units that you will have to go back and do first.

If you need any further assistance with this, please use our Contact form so we can email privately.

[Susan Brown]

the lights bi metal switch and the infinite switch l1-H1 Bk(j).

Those are correct.

[Susan Brown]

Current is directly proportional to voltage in a circuit in the context of voltage drop. I = E/R; or E = I x R

You might be thinking in terms of power (P = I x E), in which case for a fixed value of P, I and E are inversely related.

I’ll clarify that on the quiz. Thanks for the question!

[Author]

Posts