Susan Brown

[Susan Brown]

Hi Mario,

If you look at the diagram again, the reading across L1 and L2 is zero. Each end of the element with respect to N is 120vac.

First, let me review what voltage drop is. When current flows through a load (resistance), a voltage drop across that load will be created (E = I x R).

The video states that the element has continuity, so the fact that there’s no voltage drop means there is no current flowing through the circuit. Something is open.

The zero reading across the load also means that these are electrically equivalent points (EEPs). It’s like reading two points along a wire.

So what about those L1 to N and L2 to N readings? Those are voltage potential difference readings (not voltage drop), and now that we know that the element is just acting like a wire, and the ends of the element are EEPs, what can we determine? Are we really reading two different voltage differences?

Lemme know what you think …

[Susan Brown]

Hi John,

Have you seen the newer videos we’ve added to the Basic Electricity module since you first went through it? There are some good ones, particularly in units 4, 5, and 8 that might help you understand voltage drop better.

But the short answer is that in a series circuit, the total voltage dropped across all of the loads will equal the source voltage. Or, put another way, if you take the voltage drop across each load in a series circuit and add them together, the sum will equal the source voltage.

The amount of voltage drop across each load depends on the resistance of that load. So, if one load is very high R and the other very low R, then most of the voltage will be dropped across the high R load.

As to any specific question, such as the ice maker, we’d have to see the schematic to be able to help you on that. I believe you’ve also posted about this over at Appliantology – so one place or another, you can post the schematic if you want more help.

[Susan Brown]

Hi Brian,

You are correct! (I removed the numerical answer you gave, just so we don’t give it away to other students who happen to see this.)

Current is the same at every point in a series circuit, so the current has to be calculated using the total resistance in the circuit. Once you know the current, you can use the Ohm’s Law equation for Power along with the resistance of the loose connection to calculate the heat generated.

[Susan Brown]

🙂

[Susan Brown]

Hi Joe,

If you are asking this because you have another version of the book, such as on Kindle, here’s a pdf version of the book you can use to look up by page number:

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

To answer your question – we don’t have a list of the corresponding topic names.

[Susan Brown]

Wire colors don’t always follow convention, so you have to trace them to know what’s what.

[Susan Brown]

Actually, I have a follow-up!

What we need a load to do is “work”, right? This means we primarily need a certain amount of power (voltage AND current) for these components to operate, P=IxE, or P=I^2xR are two common ways to calculate that. Loads run on power (watts), not voltage alone.

When a component has a voltage rating, it means that in order to get the needed current flow to create the Power it needs to operate, it must drop that rated amount of voltage to achieve that.

So, just to reiterate based on your question, a component rated at 120v would need to drop right around 120v to operate according to spec, thus it could not be in a circuit (AC or DC) that contains another 120v load, unless it is a 240v circuit.

[Susan Brown]

Hi Josh,

Good questions!

I’m going to answer your last question first: Ohm’s Law works the same for both AC and DC voltages. The fact that AC is switching polarity doesn’t affect these basic calculations that we’re talking about.

When you have loads in series, the combined voltage drop will equal the voltage supply. So, you are correct when you said that two loads of the same resistance in series will drop 60v each to total 120v. (In a 120v circuit)

You refer to a “120v door lock” or “120v drain pump”. The “120v” is the rating for that component – the voltage it should be getting to function properly.

If two components both need 120v, would they be placed in series in a 120v circuit? No. They would have to be a different configuration – parallel circuits, or use of switches and shunts – to ensure that they are getting their rated voltage.

Does that make sense? Any more follow up questions?

[Susan Brown]

Hi Colin,

It is possible, but not something we advise doing since it’s not always saving that much money and it can be tricky to determine the exact state of all the various components. The best practice is to replace the board.

[Susan Brown]

Hi Brian,

Your original answer (down) is the correct one! I wonder if you accidentally looked at the wrong question/answer the first time you took it? I double-checked the quiz and it still is set up correctly.

But, your original thinking is correct, that’s the important part!

[Susan Brown]

And by the way, we have lots of students who are completely new to the field! You’ll do fine – it may take a little more work or repeating of some lessons than for those who already have some experience, but if you put in the work you’ll learn what you need to!

And feel free to ask questions when you need to! Just please start topics in the Forum for that particular course and unit. Thanks!

[Susan Brown]

Have you watched the video at the end of Unit 3? He steps you through it.

You are wanting to calculate the heat produced by a loose connection. Heat is a type of work, measured in watts (as described in the unit), and is also described as “power”. (You have to get used to different words meaning the same thing sometimes!)

Using the various Ohm’s Law equations there are actually a couple of ways you could calculate the answer to this problem, but it will always be done in two steps. The most straightforward way is to calculate the circuit current (because the current is always the same throughout a series circuit), then you can use that to calculate the watts of heat generated by the loose connection, knowing the resistance of that connection.

With that in mind, please re-watch the video.

[Susan Brown]

Hi Colin,

Even though it comes early in the course, the Basic Electricity module is often the most challenging one for our students! There’s a lot to comprehend, so don’t get discouraged. When we were developing this course I myself was trying to learn this stuff, and kept asking Scott questions until it started to click in place for me.

I encourage you to spend your study time right now going over each unit quiz and finding any questions/answers that you don’t fully understand, and use the forum to figure it out.

You might want to start with looking at previous conversations in the Forums – we’ve given lots of help to others that might be enlightening to you. For example here’s one CLICK.

And then start separate topics to ask remaining questions that you have (for example, the one you put at the bottom of your post). Read this if you need help on how to start a new topic.

So – please ask as many questions as you need to! That is what we are here for 🙂

[Susan Brown]

Hi Phil,

You always have to make sure you are using Ohm’s Law in the context of the realities of a circuit.

Let’s take a very simple circuit: one with a single load and 120vac supply. Let’s say the resistance of the load is 10 ohms.

If we use P= V^2/R we get 120×120/10 = 1440 watts.

If we want to use P= I^2xR then we first calculate I from the given V and R.
I = V/R = 12 amps.
Then 12x12x10 = 1440 watts.

So, that’s pretty simple when you are dealing with a single load. But how about if there are 2 loads? Let’s take my circuit above and say there are 2 loads, one is 2 ohms and the other 8 ohms of resistance.

The total power produced by the entire circuit will be identical to the calculations above, because the total resistance in the circuit still equals 10 ohms.

But if you want to calculate the power produced by each load individually, you have to think about what is different. We know the resistance of each load, and we know the total circuit current (because the current in a series circuit is the same at any point in the circuit).

So, you could use the P=I^2xR for each load to get 288 watts and 1152 watts, respectively.

But, how would you use P=V^2/R to do this? The mistake a lot of people make is to use the source voltage in this calculation, which would give you strange numbers for the watts produced (7200 and 1800, with the higher resistance load producing less heat, which doesn’t make sense).

So, now you have to think about voltage drop. The 2 loads combined will drop the 120vac, with each individual load dropping a portion of that voltage in direct proportion to its resistance.

So if you wanted to calculate power using the V^2/R equation, you’d first have to calculate the voltage drop across each load using V=IxR.

Try that and see if you get the same numbers as calculated above.

Does that answer your question?

[Susan Brown]

I just sent you an email about this!

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