Susan Brown

[Susan Brown]

Here’s what we say about EEPs in the unit, above the video:

Another powerful troubleshooting concept is Electrically Equivalent Points (EEP). EEPs are points that look exactly the same to electrons. The two characteristics of EEPs are:

1) They have electrical continuity with each other. In other words, if you measure the resistance between two points that are electrically equivalent, you will measure zero ohms. It would be like measuring the resistance in a section of wire.

2) There is zero voltage difference between points that are electrically equivalent. If you were to measure voltage across two EEPs, you would read zero, even if both points were at 120vac.

Is there anything about this description that doesn’t make sense to you?

[Susan Brown]

Hi Jose,

First of all – are you talking about the second video in Unit 4 of the Troubleshooting Module?

Second, I’m having trouble following your question. You mention measuring voltage on eep points but then say “u will measure 0 ohms”, which is a resistance (or, continuity) reading. Could you clarify your question, please, and maybe tell me at about what time in the video you are referring to?

[Susan Brown]

Hi Sal,

Thanks for your thoughts on this.

One of the things we learned early on when we started attending national training events is that there is a lot of regional variability in appliance sales (brands/models), and thus variability in recommended parts inventories. So, there’s no way we could offer a list that would be useful and current for most of the MST students.

We began our business 20 years ago with a large investment in a set of “commonly needed” parts. We still have some of those sitting on our shelves.

Not all local parts houses seem willing to be very helpful, so try talking to a large distributor such as Marcone to see what they can do. Set up an account and then ask if they have “most commonly used parts” packages by region.

When you are just starting out, you will naturally have more second trips because of having to get a part. But then you will start to build up a feel for your area and start fine-tuning your inventory. That’s okay – better than spending a lot of money up front on parts you may not use for awhile!

As for your first question, you can do part number research and cross referencing at some of the online parts sites. AppliancePartsPros.com is one that we use for that.

Participating in the tech community over at Appliantology is a great way to get up to speed faster on some of these start-up issues, like parts. Plenty of guys are willing to share their experience and what they’ve learned!

[Susan Brown]

Correct. So, that means that the diode failing open changes the current flow, right?

[Susan Brown]

Hi Jose,

Assume that originally the diode was functioning and therefore closed. Then it fails open. Is there a change in the current flow in that particular branch between those two situations?

[Susan Brown]

Note to any students reading this topic – John successfully answered the question himself, but I’m not showing it here, because I don’t want to ruin the opportunity for you to have the satisfaction of figuring it out the rest of the way on your own! 🙂

[Susan Brown]

Go look at the top of the quiz for that unit. Is there a yellow box with a green “Retake Quiz” button in it? If so, then you can retake the quiz. (Everyone has a total of 2 attempts for each quiz.)

[Susan Brown]

Hi Sal,

Couple things –

When the compressor is off, the start and the run windings are in parallel with each other. Their resistances are given on the tech sheet, so you can calculate their equivalent resistance.

For the condenser fan motor, you can calculate an approximate resistance using the specified current and voltage. This isn’t an extremely accurate way to know the resistance of a motor, but it gives you enough of a ballpark to use in the second calculation. The point is that it is vastly larger than the resistances of the windings, so has very little effect on the equivalent resistance, given the way the math works on that calculation.

Does that help?

[Susan Brown]

Yep! Now, do you know how much voltage each branch is supplied with and how many loads are getting current in each branch?

[Susan Brown]

Hi Sal,

That info is mentioned in one of the videos, but I don’t know which one off the top of my head. Must be in the unit where you saw the questions in the quiz! Which unit is that?

[Susan Brown]

Hi John,

I don’t want to completely give out the answer here (we can finalize this discussion in email, if necessary), but let me start giving you some prompts here…

The key to this question is being able to look and see what kind of circuits you have going on, then to apply what you know about voltage, current, and voltage drop to what you are dealing with.

First of all – we added a few new videos earlier this month, so make sure you’ve seen them (Units 4 and 5). They talk about series and parallel circuits.

Tell me what you think the drawing for this problem shows (what kind of circuits?).

[Susan Brown]

Hi Glenn,

We think this stuff is fun, too 🙂

Okay – first of all, where did you get the 40 amps from?

One more thing – is this a series or parallel circuit? How you determine equivalent resistance is very different between the two.

[Susan Brown]

Hi John – this video is much more about understanding the circuit and power supply rather than how to perform this “half splitting” troubleshooting technique. You’ll see a video of Scott doing that in the Troubleshooting module!

[Susan Brown]

Hi Chris,

1. ANY circuit must have a load for it to be a circuit, so having loads doesn’t determine if it is parallel or series.

Have you seen the new videos in Module 3, Unit 4? They may have been added after you did that unit. Go check ’em out! This question comes from the first video, where he talks about parallel circuits in the second half.

https://my.mastersamuraitech.com/module-3/basic-electricity-circuit-components/

2. This question also comes from those videos, although the info is in other parts of the module as well. When you have L1 and L2 being supplied to a circuit, what’s your total supply voltage?

[Susan Brown]

Hi Chris,

First of all, thank you for taking the time to explain your thoughts on this! It’s very valuable for us to be able to see where any points of confusion are so we can get better and better at communicating these things.

As to the example you discussed of a series circuit with two resistors, one is twice the resistance of the other. (And you are correct – a tech sheet would never say that, but often times when learning concepts it’s helpful to use streamlined scenarios like this, before adding the various complications and details that come with using actual tech sheet info.)

You can answer this question without doing “E=IxR”, but you need to *know* E=IxR. In other words, you need to know that voltage drop is directly proportional to current and resistance.

Then there are some fundamental truths about circuits that you need to keep in mind. One is that the supply voltage will completely drop across the circuit, with the drop being divided up across the loads based on their resistance. The second is that in a series circuit, the current will be the same at every point in the circuit.

So, knowing those two things and that E=IxR, you can think through this particular problem without having to actually plug any numbers into E=IxR.

The loads in the circuit need to drop 120vac total. Since the current is constant, the voltage drop across each load will be completely determined by the resistance of each load. If you know the percentage of the total resistance that each load represents, then the voltage drop will divide up the same way. If there was only one load, it would drop 120v. If two equal loads, they would each drop 60v, etc.

You could be given numbers for R, so you could then calculate the total resistance in the circuit, calculate the current, then calculate the voltage drops. But this was more about getting the foundational concepts of what goes on in a series circuit.

Also my perception of voltage drop is a bit fuzzy, I am getting the math (I think) to figure it out. But what it actually IS I dont quite get. I thought of it has a measurement of power(volts) being absorbed by a load. Which would leave me to believe that if a single bulb in a circuit absorbs 120v, then adding a second bulb further down on the neutral side would do nothing(which I know is wrong)

Be careful not to confuse power and voltage! Power is work being done. Voltage is basically like “pressure”. The equations for power show its relationship to current, resistance, and voltage.

As to the lightbulb question – keep reading…

I am having difficulty reconciling these two perceptions I have. I thought of voltage drop as a sponge. for example: If a sponge could soak up 2 cups of water, and if the quantity was 2 cups to start with then a second sponge added to the line up would have nothing to absorb.
But from your lesson and Kirchhoffs law, I am finding out that this is not true, that voltage drop is not a set amount for any given load but is completely dependent upon resistance and current.

The thing that is set for any load (light bulb, motor, element, etc.) is its resistance. It is what it is. The voltage that is dropped across that load is dependent on the current that flows through it. The power that it can produce is also dependent on the current that flows through it. And the current that flows through it is dependent on the total circuit configuration (supply voltage, total resistance of all loads).

Equations are essential to understanding electricity, but you have to use the equations in the context of understanding some fundamental physical concepts of electricity and circuits.

Does this all help you get any closer to understanding?

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