Susan Brown

[Susan Brown]

Let’s start with this one: I = E/R

The calculation you are referring to is shown in the “Equivalent Resistance…” video in Unit 5.

It is for a series circuit, with 3 loads. In a series circuit, the current is the same throughout the circuit, and the way you calculate it is to add the 3 loads to find the total resistance in the circuit, then use that in the equation above.

Again – these are loads in series. Loads in parallel are treated differently.

So the equation becomes I (current) = E (voltage) divided by R (total circuit resistance)

You need to add the resistances together first, then put that number in the equation. (In the video, he puts the 3 resistance numbers in parentheses. When doing math calculations, you always do the operation shown in parentheses first, then move on to the others.)

What do you come up with?

[Susan Brown]

Hi Matt,

We helped someone else awhile back with calculating equivalent resistance in parallel circuits. Check out this topic and see if that helps you with your first question.
https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/calculating-equivalent-resistance/

For your second question, I’m not sure if I see what you are showing there. Are you trying to calculate the current (I) when E is 120vac and the total resistances are 58.75 ohms? (those 3 added together). If so, then look at the Ohm’s Law pie chart in Unit 3. If you know what E and R are, and you are trying to find I, what is the formula? (You can just write it in words rather than try to format it)

If I’m misunderstanding your question, please clarify. Thanks!

[Susan Brown]

Hi Matthew,

We like to see students asking questions here, so you didn’t come across in a negative way at all. Feel free to ask questions here about anything you’d like to get help or feedback on.

Yes, your description of current here is exactly what we teach in numerous instructional videos throughout the course, which you’ll see as you progress.

Voltage is what makes the negatively-charged electrons want to move to move towards a relatively more positive charge, and current is the movement of those electrons in the wire.

[Susan Brown]

Hi Matthew,

Sharp eye! Good job looking at the schematic.

That lid switch drawing is very vague and difficult to interpret, and Maytag gives no assistance on the diagram to help us. The symbol that looks like a capital I with bubbles is particularly strange. We have to use our common sense to try to understand what the drawing is showing, but bottom line is we know that the machine is not designed to short out the way it appears to be drawn. In fact, this lid switch assembly has a fuse in it in case a stuck contact creates a short.

Here’s another drawing I found of this lid switch. You can see a few differences, including showing more of a dashed vertical line in that thing I mentioned that looks like an “I”. So that is not a path for current. Hopefully this helps!

[Susan Brown]

Hi Julio,

Thanks for using the Forums to ask a question! We like helping students here.

The important thing is to recognize the shunt, which you did. (Good job!) The answer we’re looking for is “shunt circuit”, because with that switch closed, as shown in Figure B, no current will flow through the heater. It will all flow through the wire with the switch then through the lightbulb.

In Figure A, with the switch open, the current has to flow through both the heater and the bulb, so those loads are in series in that situation.

Make sense?

[Susan Brown]

Hi Matthew,

The Internetology course doesn’t result in MST Certification, so it doesn’t matter what your scores are on those quizzes. Therefore, we usually don’t do resets on those unless you really want one 🙂

Once you are taking Fundamentals, however, you can request a reset if you really need it, but please pay attention in the Orientation module on how the resets work so you understand them.

Let me know if you have any more questions.

[Susan Brown]

First of all, the best way to know how to do this problem is to re-watch the video at the end of Unit 3. We step you through it, although the resistances are a little different than in the midterm.

Let me know if there’s a step that you don’t understand, and I’m happy to elaborate or explain in a different way. You could tell me the time at which you don’t understand something (for example, “I didn’t follow what he did at the 5:30 point of the video.”)

As far as voltage goes, if you are talking about calculating circuit current, then the value for “E” is the source voltage. You know that by looking at the drawing of the circuit (is it L1-N or L1-L2?).

[Susan Brown]

Hi Brandon,

First note that I moved your topic the the correct Forum. I also renamed the topic because we prefer people talk more generally about the concept, rather than ask for help on a quiz or exam question.

We’ve stepped some other students through this already. Please see the topic below:

https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/calculating-equivalent-resistance/

You should also know the rule of thumb, which we give in unit 5. Do you know what it is? It was one of the quiz questions, for that unit, too.

[Susan Brown]

Glad you figured it out!

[Susan Brown]

Great question. You are asking why the equivalent resistance of loads in parallel is less than the smallest individual resistance.

This is all from the perspective of the power supply. Be Zen-like and become the power supply, which has constant voltage, and think about the affect that having multiple paths for current to flow (parallel circuits vs. just one series circuit) has on the number of electrons you can push out (current). You can push a lot more out when there are multiple branches, even though each branch has a load.

The best way to think this through is to draw out a couple of circuits on paper and play with the calculations.

The first one has two loads in parallel, R1 and R2. Assume a 120vac circuit, so L1 and N. Assign easy values to R1 and R2, say 10 ohms and 20 ohms. Then calculate the current flowing through each load. (I = E/R, and remember in parallel circuits each load gets the full 120 vac.) The total current draw from L1 would be the sum of those two different currents.

Now, let’s think about the “equivalent resistance” of this scenario. The equivalent resistance is taking the resistance of loads in parallel and theoretically combining them into a single load. So, do the math and come up with the equivalent resistance based on the loads above. Now use I = E/R to calculate the current draw from L1. Should be the same as what you came up with before.

Since current is inversely proportional to resistance, for the same amount of current to flow through our theoretical one load that was flowing through the two branches/loads in parallel, the single load, or equivalent load, would have to have a lower resistance.

Does that help?

[Susan Brown]

Hi Bill,

Each quiz and exam is set to automatically allow two attempts, regardless of your score on the first one, but it is up to you if you need to use the second attempt. So, in this case, you’ve already scored 100%, so you can just move on to the next unit. But if you take a quiz later and don’t score 100%, you can study some more, ask a question in the forum, etc., then retake the quiz to improve your score.

Does that answer your question?

[Susan Brown]

Yes – that’s it! I’m going to hide parts of your answers here so we don’t just give it away to others who read this.

[Susan Brown]

You’re welcome! If you are ever having trouble with any of the course material, please post a question. We like to answer them!

[Susan Brown]

Sure!

So – you see that the V for each load would have to be the voltage drop for that particular load, not the supply voltage. That can be calculated, but it’s faster to just use the P=I²x R formula.

[Susan Brown]

Hi Gregory,

P=V²/R can work, but what would you use for V for each load?

Remember – the supply voltage is the total amount being dropped over all loads in the circuit.

P.S. let me know if you need a reset on a quiz!

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