Sam Brown

[Sam Brown]

Hi NJ Appliance Tech,

Something that’s important to know about these carborundum ignitors is that their resistance lowers as they heat up. This is why the amps going through an ignitor increase as it heats up — lower resistance means higher current. It’s expected that the resistance of an ignitor at room temperature would be higher than its resistance at its running temperature.

If you want to learn more about ignitors (and every other kind of oven, range, and microwave technology), I recommend you check out our Cooking course. It covers all of these things in depth.

[Sam Brown]

Hi Donald,

This is a reignition system that we’re dealing with — the easiest way to tell is because it keeps sparking continuously after you turn on the faulty burner. And the return current path for reignition systems is in fact usually not shown on the schematic, since it simply travels through the burner head, then the chassis, then back to the spark module.

And yes, you’re correct that if you hear a snap, that means there is a spark. But since we’re not seeing a spark at the burner head, we can assume that the spark is being grounded out before it reaches the burner. This is further confirmed by the fact that the sparking sound is muffled.

[Sam Brown]

If you take a measurement from a good source of L1 to a good source of L2, then yes, you would read 240 VAC. However, that may not be the situation we’re dealing with in this question. Take a good look at the readings you get on your meters, especially in figure 2, and figure out what that tells you about the circuit.

[Sam Brown]

Hi John,

The point of taking that voltage measurement is to check for a good neutral. One probe is on line at A12, and the other is on neutral at A1. When you measure across those two points with a loading meter and read 120V, then you know that you have a valid neutral all the way from A1 to the outlet.

Sounds like you might be confused about how A1 is connected to neutral. The path is a little circuitous (pun intended), but the neutral comes to A1 through the A3 contact (remember that the question states that A1 to A3 is closed). See if you can trace out neutral’s path from the outlet to A1. If you can’t figure it out, let me know.

[Sam Brown]

All that depends on where you, the technician, are relative to the board. Which is why it’s best to avoid trying to think about whose right is whose left and so on. If you imagine that your face is the clip, regardless of whether it’s attached to the connector or the board, you should be able to find pin 1 of the connector by looking to your left.

[Sam Brown]

Hi Grasshoppah,

The trick is to become the harness connector clip and face away from the harness connector.

So, depending on which side of the connector the clip is located, you may or may not end up facing away from the entire board. It doesn’t really matter. The important thing is that you “become” the clip and face away from the connector, then find the pin that’s farthest to your left.

[Sam Brown]

Power companies are sending power over the transmission lines, not simply voltage or current. The Ohm’s law equation showing the relationship between power, voltage, and current is P=I*E. According to this equation, you can jack up the voltage and reduce the current and still transmit the same amount of power in watts.

This is desirable because it reduces the required ampacity of the transmission line, meaning smaller diameter wires (lower cost, easier to work with). And it reduces I^2*R losses (actually called “I-squared R” losses) due to heating the conductors.

There are a lot of different Ohm’s law equations that show a lot of different relationships, so you need to make sure you’re using the correct equation to express what you’re thinking about since, in the end, electricity can only be understood through the math.

[Sam Brown]

Yes, that’s right. To illustrate this further, you can go back to the I = E / R equation that you brought up earlier. If you plug in the circuit’s source voltage for E and the circuit’s total resistance for R, then you get the circuit current.

[Sam Brown]

Hi Boyd,

You’re certainly allowed to ask about your own appliances, but it must be phrased as a proper troubleshooting question, with a schematic provided. You’ll learn about this and all kinds of troubleshooting strategies that will really help you out in the Troubleshooting module of the Fundamentals course.

Once you have worked through the course to the end of the Troubleshooting module, please feel free to ask this question again as a new topic in the Troubleshooting forum, using all the tools you have learned. We would be more than happy to work through it with you!

[Sam Brown]

Hi Grasshoppah,

That’s a pretty common question. To understand the answer, it helps to visualize how current works. Current is electrons physically moving through a conductor from one side of the power supply to the other. They’re like a bunch of billiard balls being pushed through a tube, all crowded together and knocking each other forward. If you think of it that way, it makes sense that current can’t lessen as it progresses through a circuit — if the electrons earlier in the circuit are moving, then the ones later in the circuit have to keep moving as well, because they’re still being pushed along.

Essentially, the current in a series circuit is decided as soon as power is applied to it. The power source instantly “sees” every load in the circuit and how much resistance it has, then supplies a certain amount of current accordingly.

[Sam Brown]

Sorry for the confusion — I misread the table. Yes, that table is correct, and in fact would be regardless of which side is line and which is neutral. I edited my previous reply to avoid any further confusion.

[Sam Brown]

I think this video from the Fundamentals course explains the formula better than I could through text:

If it still doesn’t make sense after watching that again, let me know and I’ll try to explain it.

[Sam Brown]

Yes, that’s correct.

[Sam Brown]

If you use the normal VAC function on your multimeter, you would read 120V across that push to start switch — even though you have no connection to neutral because of the open door switch.

A quick correction to what I said here — because the door switch is open and therefore not providing a path to neutral, you would most likely not read a full 120 VAC across the push to start switch with your standard VAC function. This is because, with that wire not being connected to neutral, it’s just floating there with some amount of ghost voltage sitting on it. So when you make your non-loading measurement, you would get a weird, indeterminate voltage reading — less than 120, but more than 0, probably with some fluctuation.

Here are a couple videos by the Samurai that illustrate this concept very well:

[Sam Brown]

Hi George,

We’re dealing with a pretty simple circuit here: just the two switches in series with each other. Let’s say that you’re measuring across the push to start switch with it not actuated (and therefore open), and the door switch has failed, going permanently open.

In this case, the reading you would get depends on the meter you use. If you use the normal VAC function on your multimeter, you would read 120V across that push to start switch — even though you have no connection to neutral because of the open door switch. However, if you make the measurement with a loading meter, you would read 0V across the push to start switch, since even with the 120V present at one side of the switch, there is no valid path to neutral. That’s a much more useful measurement, since it shows you that something in the circuit is interrupting your power supply.

This is one of many cases where using a loading meter instead of a normal voltage meter is essential. The standard VAC function would lead you to believe that everything about the circuit is fine. But the loading meter clearly shows that there is something wrong with the circuit.

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