Sam Brown

[Sam Brown]

The schematic will always be the best way to determine this, hands down. There’s no easier way to determine whether it has an inverter.

[Sam Brown]

Assuming we’re looking at the same part of the video (about 14 minutes in), then yes, there would definitely be line present on both sides of the regular speed motor winding.

That’s because, in this scenario, that 32 to 16 switch in timer cam 6 is stuck open. If that’s open, is there going to be any voltage drop across the motor winding?

[Sam Brown]

When someone says that something is failing under load, that’s a very general statement. There are a lot of different possibilities.

In the specific example of a pump motor, you could have a mechanical problem that causes it to fail under load. The motor could test perfectly fine electrically, but when you apply power, it’s physically incapable of turning — for example, if it’s jammed.

Another case for a motor would be if its windings are contacting ground — the windings might check good on ohms, but when you try to run the motor, you’ll have current flowing to ground instead of neutral. You would, of course, be able to check for this one by measuring continuity from the windings to ground, but people don’t always think to do this.

And those are just two examples. There’s not a silver bullet for identifying if something is failing under load, except by simply identifying that there is a proper power supply right up to the load, and yet it’s still not doing what it’s supposed to do.

[Sam Brown]

If you’re only reading 80V of voltage drop across a load that’s supposed to drop the full 120V, you’re most likely dealing with a high-resistance connection in that load’s circuit, not a problem with the load itself.

When you have only a single load in a circuit, it will always drop all the voltage. But if you have two or more loads in series (even if one of those “loads” is actually just a high-resistance connection), then the voltage drop will divide between the loads according to their resistance.

Make sense?

[Sam Brown]

When it comes to testing a hall sensor, you’re very much at the mercy of whether the manufacturer provided specifications and pinouts. There are different ways that hall sensors can be implemented, so you need to know what specific voltages/resistances you should expect, and where.

For an example of a particular hall sensor configuration, see this page: https://appliantology.org/gallery/image/733-wiring-harness-and-resistance-checks-on-the-hall-sensor-in-an-lg-washer/

As for phase rotation meters, you can find plenty on Amazon. Here’s the one we have: https://www.amazon.com/Sequence-Presence-Rotation-Indicator-Detector/dp/B018Q2CMPY/ref=sr_1_7?ie=UTF8&qid=1550438244&sr=8-7&keywords=phase+rotation+meter

[Sam Brown]

The fact that you can safely touch a neutral wire is stated in the first video of module 3 unit 6.

Additionally, with everything you learn throughout units 1 – 8 about the nature of voltage, voltage drop, and the neutral wire itself, you can figure out that neutral would be safe to touch, even before it’s explicitly stated in the text of unit 9.

[Sam Brown]

There you go! Looks like you’ve figured everything out. Anything still not clear?

[Sam Brown]

Looks to me like it’s not hardwired — there’s something between the indicator light and L1. What is that?

[Sam Brown]

Correct! So that’s one way that L1 gets to the hot indicator light. The other way is a lot easier to see. Do you know what it is?

[Sam Brown]

It’s actually any of those three jumpers — doesn’t really matter. The circuit that those jumpers connect to is actually representing three separate, but identical surface element circuits, each with one of those jumpers connecting to it.

So now that you’ve located the jumper connecting to the LF hot indicator light, how does L1 get to that jumper (and by extension the indicator light)?

[Sam Brown]

but I am still having a hard time with question 2.

Well, let’s go through it one step at a time. Did you find the other end of the black jumper wire that comes off of the left front hot indicator light’s circuit? It’ll be marked as BK(J).

[Sam Brown]

Gotcha. So, question 1: What power source does the light run on?

This one’s pretty self explanatory, as long as you’re looking at the schematic (looking at the wiring diagram will just confuse you). Just do the Zen trick from Fundamentals, “become” the light, and see what you’re touching on either side.

Question 2 takes a little more thinking, and as it says, there are multiple correct answers.

One way that the light gets L1 is pretty easy to see — it’s probably the one that you saw when you were doing the Zen trick above.

The key to finding the other way is to look at those little black jumper wires coming off the circuit for each light — they’re marked as BK(J). The jumpers seem to just dead-end, but what they’re actually doing is connecting to somewhere else in the schematic. See if you can find the other ends of those jumpers, and then follow that path back to L1.

Let me know when you figure it out.

[Sam Brown]

There’s only one question in module 11 unit 5 — do you really mean that unit, or are you referring to a different one?

[Sam Brown]

There are two misconceptions that I think are what’s confusing you.

First: Voltage does not flow. It is simply present, or it is not. Current is what flows, and current is present even in parts of the circuit that have no voltage — such as the neutral wire. However, touching a part of the circuit with current but no voltage would never shock you, because there is no voltage present to push the current through your body.

Second: If you have a single load in a circuit, it will always drop all of the voltage, no matter how much voltage there is. You would never have a circuit with only a single load where that load does not drop all the voltage. If you have more than one load, then those loads together will drop all the voltage (the voltage drop being distributed across the loads according to their resistance).

So even in a circuit supplied with a million volts, all of that voltage will be dropped across its load(s). There might be thousands of amps of current coursing through the neutral wire, but touching it would not shock you, because all the voltage has been dropped.

Let me know if anything still isn’t making sense.

[Sam Brown]

The short answer is that neutral and ground have the same electrical potential — that is, 0 volts. Remember that no matter how much voltage is supplied to a circuit, it’s all dropped across the loads. That’s why neutral, by definition, has no electrical potential.

Does that make sense?

[Author]

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