Scott Brown

[Scott Brown]

How is 120V measured from L1 (or L2) wrt neutral when there seems to be no visible circuit from the heater to neutral?

Hi Steve,

In a 240vac, L1-L2 circuit, there is no current path to Neutral! Nor do you need a current path to Neutral (or anywhere else) to measure voltage!

In a normally functioning 240vac circuit, you will measure 120vac from each Line to Neutral even though there is no path for current from either Line to Neutral. You will also measure 240vac from Line 1 to Line 2. That’s just the way split phase 120/240 vac household power supplies work. The video on household power supplies in this lesson explains how that works: https://my.mastersamuraitech.com/module-3/basic-electricity-breaker-panel-basics/

[Scott Brown]

Look at the ladder diagram on the LHS of the tech sheet and tell me which igniter is getting power when the switch is closed for P21-1 to P1-N.

Now look at it again and tell me which ignitor is getting power when the switch is closed for P20-2 to P1-N.

Now lemme hear you say, “AH HAH!” 🙂

[Scott Brown]

However, I’m not real clear on how 120V AC is measured from either heater terminal to neutral. I don’t see a circuit for current to flow from L1 to neutral or L2 to neutral to allow for 120V at the heater. I’m guessing this is because the heater is grounded (mounted to chasis) and ground is connected to neutral?

Hi Steven,

Remember that voltage is always measured with respect to (wrt) a reference point. It doesn’t matter what that reference point it. The reference point does not need to be in the same electrical circuit.

Neutral is just as valid a reference point as is L1 or L2.

So, if both L1 and L2 were present at the element, we would measure 240 vac at L1 wrt L2. Similarly, we would measure 240 vac at L2 wrt L1. But in both cases, the voltage at either L1 or L2 wrt Neutral would be 120 vac. Also, in this case, we would measure 240 vac ACROSS the element. We say that the element is dropping 240 vac or that 240 vac is being dropped across the element. Different ways of saying the same thing.

Now if only L1 was present at the element and L2 were absent, we would still measure 120 vac from L1 to Neutral. In fact, if both wires were connected to the element, we would measure 120 vac at either end of the element wrt Neutral. If we were to remove the L2 wire to the element, we would measure 120 vac wrt Neutral at the the L1 side of the element AND at the the other side of the element (with the L2 wire disconnected). The dangling L2 wire would measure 0 vac wrt to neutral. In this case, we would have determined that L2 is missing and we would troubleshoot that side of the circuit.

Also, since L2 is absent, there is no return path for current– no complete circuit– so there is no current flow through the element and hence the voltage drop across the element is 0 vac.

The point of the question is to get you to realize that you can measure 120 vac on either side of the element wrt Neutral but 0 vac across the element (and the element won’t get hot). When this happens, it means you’re missing one of the Line voltages and you need to half-split the circuit to figure out which one is missing.

Does that make more sense?

[Scott Brown]

It’s not for testing the PWM signal– that’s done with your DMM using the volts and frequency functions.

I think you’re referring to a phase rotation tester that I’ve mentioned on occasion. These are simple voltage testers made for use with three-phase voltage. This just means that instead of two leads like on a single-phase tester, it has four. Other than that, works the same way– low impedance meter that places a load on the circuit (albeit very small) and turns on a light if the voltage is present. Here’s the one I use: http://amzn.to/1RQsx4E

[Scott Brown]

And, to answer your question about the good connection, you are exactly right: the resistance, voltage drop, and power dissipated in a good connection should all be zero (or really close to it!)

[Scott Brown]

Glad it helped!

Ohm’s Law rules! It is not a useless academic exercise. It is the means by which we understand circuits. There is no other way.

Since we can’t see electricity, we can understand it in only two ways: Math and Measurements. Both are required. The math explains how electricity behaves and measurements prove it to us or make it real to us.

[Scott Brown]

An excellent question! As with all questions pertaining to voltage, current, resistance, and power, we turn to the governing law on these matters: our dear friend, Ohm’s Law, whom we met and played with waay back in Mod 3, Unit 3…

loose connection can generate heat and burn up

In a loose connection, oxides and other things can build up on the surfaces of the wire ends that connect to each other. This causes the resistance between these wires to increase. Increased resistance means increased power (as heat) is dissipated across that connection. The increased resistance in the connection causes the connection to get hot.

We can actually calculate the amount of heat in watts generated by a loose connection using Ohm’s Law! Oh yeah, it’s crazy! Dance with me now…

Look in the Ohm’s Law pie chart at the equations for power. We can use any of these simple equations to calculate the amount of heat given off by a loose connection if we know something about the circuit.

Let’s assume we’re dealing with a 240vac circuit with a standard bake element load of 32 ohm PLUS a high resistance connection in series with that load. For simplicity’s sake, let’s also suppose that we know or have measured the resistance of this loose connection and it is 5 ohms.

Ohm’s Law defines the relationship between voltage, current, resistance, and power. If we know any two of those, we can figure out the other two.

In our example circuit above, we have a total circuit resistance of:

32 ohms (bake element) + 5 ohms (loose connection) = 37 ohms

So the total circuit current is:

E = I * R ==> I = E / R = 240vac / 37 ohms = 6.4 amps

Since, in a series circuit, the current is the same at everywhere in that circuit, we know that’s also the current flow through the loose connection. So we can use this.

Since we also know the resistance of the loose connection, we can use this this, too.

Now its plug n’ chug! We pick a convenient equation from the Ohm’s Law Pie Chart. Since we know current, I, and resistance, R, then we would use: P = I^2 * R

P = I^2 * R = (6.4 amps)^2 * 5 ohms = 205 watts

So with a quick, simple calculation, we see that a loose connection with a resistance as little as 5 ohms in a typical application can generate an amount of heat equivalent to a 200 watt light bulb. Have you ever touched a 200 watt light bulb after it’s been on a while? YOWZAH!

[Scott Brown]

There are thousands of variations on the procedure for entering diagnostic or service modes for icemakers and other appliance components. Aside from being impractical to catalog all of them, it’s also unnecessary. As Master Samurai Techs, we understand the technology we’re dealing with and the principles of operation (or we discern it from the schematic). And we also know where to go to get specific test procedures for a particular appliance we’re working on: Appliantology.org

[Scott Brown]

Alright, let’s put the question here:

What component controls which speed winding is energized for a particular cycle?

The question is technically precise– that is to say, it conveys exactly and precisely what I want it to convey and it does so with technical precision (using the correct technical vocabulary).

To answer the question requires that you first have an understanding of how multiple speeds are achieved with single phase motors. The screencast itself explained that but it’s also covered in more detail in the Fundamentals course, which is the recommended pre-requisite for this course (Advanced Schematic Analysis and Troubleshooting).

How would you re-word it and ask the same question?

[Scott Brown]

There’s really not so much to remember if you just keep in mind that we’re talking about two different labels for the same things. The difference is how they’re used: in math equations or as a unit of measure.

Current: Represented by the letter I in equations. But the units of measure of electrical current is Amps, or A.

Voltage: Represented by the letter E in equations. But the units of measure for voltage is Volts, V.

Resistance: Represented by the letter R in equations. But the units of measure for resistance is Ohms, represented by the Greek letter, Omega: Ω

Power: Represented by the letter P in equations. But the units of measure of electrical power is Watts, W.

[Scott Brown]

There is a tendency among uninformed techs to think of electric current as this mystical, magical thing that somehow has a mind of its own. Don’t be one of those guys.

You have to start with the understanding that electrons are just dumb, negatively charged particle. The only thing an electron “likes” is a difference in voltage potential between two points. It will always go to the voltage with the more positive (or less negative) charge. Opposites attract. Other than they, electrons are too stupid to know anything about “paths.” They do what they’re told by voltage. If a more negative voltage is pushing electrons to a more positive voltage, the electrons will go there.

In the shunting example you raised, the best way to understand that– and EVERYTHING ELSE about electricity– is to let the math keep your thinking straight. You simply can’t understand electricity otherwise because you can’t see electricity. So we understand electricity through two ways, BOTH are required: Math & Measurements.

The equation for parallel resistance explains why all current will flow through the shunt and that a shunt is NOT an example of parallel circuits. I spend time considerable time explaining that in the screencast on shorts and shunts. Take a moment to re-watch and let me know if you have any other questions.

[Scott Brown]

This video is like mechanical timer porn! Complete teardown tour inside a mechanical washer timer:

https://www.youtube.com/embed/pYg0TE5gW6U

[Scott Brown]

Okay, you’re unblocked so you can take the quiz again.

Let me know if you have any other questions!

[Scott Brown]

Hi Ron,

The questions you marked as “answer later” will be pulled back up at the end for you to answer before you click the final “Submit” button.

[Scott Brown]

Right, so being the INVERTED sum of the INVERSES of the resistors values is not the same thing as simply adding the resistances together.

The quiz question is trying to see if you understand how this affects the value for the equivalent resistance of parallel resistances: will the equivalent resistance being less than, equal to, or greater than the smallest resistor in the parallel circuits?

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