Sam Brown

[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?

[Sam Brown]

We covered it earlier in the Oven and Range course — Click here to review that unit.

[Sam Brown]

This question uses terms that are all introduced in the video for this unit. Before we go any further, I recommend that you rewatch the video, since it essentially gives you the answer to this question.

If, after rewatching, you’re still not clear, let me know and we’ll go over it some more.

[Sam Brown]

Flame rectification is an important part of how reignition systems detect the flame at the burner. We have a video that goes into it in depth in module 7, unit 4 of the Oven and Range course (which I think you’re just about to start on). So stay tuned for that!

[Sam Brown]

Correct! Since it’s not trying to sense the presence of a flame, there’s no need for the spark current to return to the spark module in a regular ignition system.

[Sam Brown]

You’re correct that, in the first question of this unit’s quiz, we’re just talking about an ignition system, not a reignition system.

With that in mind, does it even make sense to talk about a “return path” for the spark current?

[Sam Brown]

For the exam unit 5 question #5 asks about the test point I choice L1 to N because there’s a fuse in series with that side of that capacitor so I would measure those 2 points first to indicate I’m still getting my line voltage input then if I do I can precede to check L2 to neutral to see if the noise filter is putting out the voltage I just find it better to follow sequence in order.

Remember that in question #4 it’s stated that all those components of the noise filter (including the fuse) are contained in a box — and therefore inaccessible. You’ve also been told that you have a good power supply to the noise filter. Your next logical test should be to measure the output of the noise filter.

Also for the video in this unit 5 the half splitting master Samuria was able to demonstrate getting 120v L1 to ground chassis with the dryer running and 0v through L2 to chassis and motor not running did he do this demonstration having the door open ? And then closed it to get the motor to run and get that 120v.

That’s most likely what he did, yes. He started up a heated cycle, then opened the door to switch off the motor while he checked for voltage at L1.

[Sam Brown]

so if we measure after R For example bk to orange we will get a voltage reading since there is a open?

Yes, with the thermal fuse open, you would read 240 volts between BK and OR (assuming the machine is set to run). This is because you’re simply measuring the difference between L1 and L2.

lets say we measure across the heating element on this circuit with nothing open we will get 240v?

Yes, we would also measure 240 volts in this situation. If all the various switches and controls are closed, that means we have a good circuit with current flow. Since current is flowing, all of our 240 volts will drop across that heating element, and so your meter will show a voltage difference of 240 volts when you put your probes on either side of the element.

but for instance let’s say the thermofuse is open or the contacts between bk and R is open we wilL read 120v across the element because we will only we getting 120v from L2?

This is where it gets a little bit tricky. With the thermal fuse or the timer contacts open, you would not read any voltage difference when you put your probes on either side of the element.

The reason for this is that with the circuit open, there’s no current flow, and therefore no voltage drop across the element. Assuming you have the machine set to run, you would in fact have 120 volts present at that element (from L2 through the centrifugal switch). But you won’t measure any difference across the element, and therefore would read 0 volts on your meter.

Things making more sense now?

[Sam Brown]

An excellent observation!

Keep in mind that, whenever you make a voltage measurement, you’re measuring the voltage difference between two points. This means that if you measure across a closed switch, you’re going to read 0 volts no matter how much voltage is present there, since your two measurement points are electrically equivalent. That’s why we could make that test regardless of whether L2 was present or not.

Let me know if it still doesn’t make sense.

[Sam Brown]

Both of these questions are actually addressed in the video on electronic commutation, so I recommend that you give it another watch. Make sure to take notes so you don’t forget any important bits!

If, after rewatching it, anything that’s covered in the video still isn’t clear to you, by all means post it here. Make sure to include what timestamp in the video your question pertains to.

[Sam Brown]

Yes, simple as that! It’s the thing that’s not doing its thing, so it’s our LOI. Now that we’ve identified that, we can move onto figuring out why it’s not working properly.

Make sense?

[Sam Brown]

The short definition of an LOI is just “the thing that’s not doing its thing.” We’re not trying to conjecture about potential causes of the problem at this point — that’s getting ahead of ourselves. Right now, we just need to focus in on the load that’s not doing what it’s supposed to be doing.

Given the information from the previous two units in this case study, which load would that be?

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