Susan Brown

[Susan Brown]

Hi Jared,

Thanks for posting a question!

Question #2 asks about DC current. The part you copied from the lesson is talking about DC voltage.

Very different things!

From Unit 7:

The three most common electrical measurements you’ll be doing as a professional appliantologist are:

– Voltage, either DC or AC
– Resistance and its daughter test, Continuity
– AC current, which is always done with a clamp-on amp meter (or ammeter). (Note: You will never need to measure DC current in appliance repair.)

[Susan Brown]

Okay – it sounds like you’ve got it figured out. I’ll reset that quiz for you. Feel free to ask more questions in these Forums as needed!

[Susan Brown]

Hi Josh,

Thanks for posting a question!

A great video to review is the one at the end of Unit 3 in Basic Electricity. The one about the heat produced by a loose connection. Watch that and see if you can duplicate the calculation on your own. If there are any steps that you don’t understand, let me know here and we’ll discuss it further.

Let me know if that helps.

[Susan Brown]

Hi Mike,

Fundamentals has taught you the basics of reading schematics. But that’s like learning to read music. You will need to practice in order to get more “fluent”.

Oven & Range Repair and Advanced Troubleshooting have the most practice on schematics compared to the Refrigerator course.

Then, once you’re done with the courses, many of our webinar recordings over at Appliantology give you more exposure.

[Susan Brown]

Hi Josiah,

Did you see the explanation that showed up after you took the quiz?

Although it is true that a heating element is composed of high-resistance wire, this is not what makes it a load. For example, a heating element in a box sitting on the shelf is also made of high resistance wire but it is not a load because it isn’t in a circuit with current flowing through it.

A load in the context of appliance repair is:
– technically speaking, a component that does work – produces motion, heat, light, etc. – when it gets power
– electrically speaking, a component that produces a voltage drop when current flows through it.

[Susan Brown]

It’s just a humorous one-question quiz. 🙂

[Susan Brown]

Hi Mike,

It’s not so much of what type of learner you are, it’s more that you just need to practice. That’s true for everyone who takes our courses. You have to practice to really get everything to click and to gain proficiency.

We suggest starting with your own appliances. Get the tech sheets and compare them with what you see. Look for free or cheap appliances that are banged up or even broken (Craigslist, town dump, etc.). Use Appliantology to get the tech documents and start playing around.

The only way to post an image directly in these Forums is if the image is on a website and you can give the link. You can also just email the photo to us and we can upload it here.

[Susan Brown]

Just did!

[Susan Brown]

Hi Mike,

Good question! If you re-listen to that part of the video, Samurai mentions that the power company is trying to deliver a certain amount of power along the lines. And we know that Power = Current x Voltage (P=IxE). So if you have a certain goal for the amount of power you want to deliver, the higher your voltage, the lower the current needs to be.

But you are correct that if we are looking at a circuit, and it’s got a fixed resistance, if you changed the voltage supply from a 120vac to a 240vac, then the current would increase according to I = E/R.

But, say you need to generate 1000 watts of power. Play around with the numbers, using P=IxE. What would the current need to be if you used a 120 vac supply and then a 240vac supply? You’d need half as much current if you use 240vac.

Make sense? There are various “moving parts” to creating the electrical situation that an engineer wants in a circuit: power, voltage, current, and resistance. And they all act according to Ohm’s Law equations!

[Susan Brown]

That will work ONLY if you have 2 resistances in parallel.

If you take 1/(1/R1 + 1/R2) you can rearrange it to equal (R1xR2)/(R1+R2)

This does NOT work for 3 or more loads in parallel.

[Susan Brown]

Thanks, Gaspare! I was just getting on here to link you guys to the previous threads on Equivalent Resistance.

Jonas – be sure to click and read that thread above. And in that thread, you’ll see a link to another one where we break down the calculation for you.

Let me know if you have any other questions.

[Susan Brown]

Hi Raymond,

Those two statements are actually saying the same things. We just changed the wording slightly for the sake of being a little challenging.

The main point is the direction of electron movement. Both cases indicate moving from the cathode end to the anode end and not vice versa. This makes sense, given that the voltage is more positive at the anode end than the cathode end, and electrons move towards more positive charge.

Hope that helps!

[Susan Brown]

“Shunt” is correct in that scenario.

Or does shunt lead to another load, and that’s what makes a shunt (vs. a short)?

That’s more like it. A shunt will be deliberately used to bypass one or more loads, but there will still be a load in the circuit somewhere.

A short creates a condition where there is no longer a load in a path from line to neutral/ground, so there’s nothing to hinder the flow of electrons, leading to a failure of some type (blowing a fuse, burnt wire, etc.).

[Susan Brown]

Hi Jonas,

Thanks for posting a question!

Yes, the * symbol is the same thing as x . Both are “times”, or multiplication.

You might have just made this up as an example, but I want to make sure you know that V=W*R is not a valid Ohm’s Law equation! It would be V= I * R (voltage equals current times resistance)

[Susan Brown]

Hi Abbey,

Thanks for posting a question! We show this in several examples in the Basic Electricity module, but just to be clear: when you have loads in series in a circuit, the total circuit resistance is simply the sum of the resistances of the loads.

Again, this is a series circuit only. Loads in parallel are different. We can calculate an equivalent resistance with those, which we describe more in Unit 5, and also in this Forum thread:
https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/equivalent-resistance-in-parallel-circuits/

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