Susan Brown

[Susan Brown]

Again, this is just testing for basic electricity/circuits knowledge that we teach in Module 3, and doesn’t require specific knowledge of how gas dryers operate.

Let’s start with this. Can you define what voltage drop is?

[Susan Brown]

Hi Julian,

Happy to help. Even though we’re using a real-world scenario, you don’t have to know much about how a dryer operates. We’re just going over basic electrical concepts here and practicing the interpretation of electrical measurements.

Key concepts for this problem:
How an L1-L2 circuit functions (from Unit 6)
The difference between measuring for voltage drop (across a load) vs. voltage potential (in this case, from L1 or L2 wrt N).
What is necessary for current to flow in a circuit? (voltage and a complete or closed circuit)

A good place to start is to think about what measurements you would expect to get in Figure 1 if the circuit were functioning properly. What do you think? (It’s helpful to label them as reading 1, 2, and 3)

[Susan Brown]

It’s very important to keep in mind what type of voltage measurement you’re doing. Are you measuring across a load and looking for voltage drop, or measuring from one point wrt N, looking for potential?

The measurements of L1 or L2 wrt N are not looking for voltage drop – so current isn’t coming into play there. Those are just looking for voltage potential, or “source” voltage. Where you have this voltage, you have an unbroken wire back to the power source.

Your last statement is correct.

If this were a 120vac L1-N circuit, then you would know from the initial measurements (without having to disconnect one side) which side the open was on. If you measured voltage from either side of the element wrt N, then you would know that Line voltage was present at the element, thus the L1 side was intact and the open was on the neutral side. If you measured 0 vac wrt N at the element, then you’d know L1 was open somewhere.

In an L1-L2 circuit, however, you have to disconnect one side and then do the measurements wrt N in order to figure out which side is faulty. The side that goes to zero is the side with the open.

Does that make sense?

[Susan Brown]

I’m working with you! I am deliberately asking little questions and trying to step you through. Try to focus on the questions I’m asking.

(BTW – I asked if you had rewatched the videos in Unit 6. Do you feel like you understand those, or do you have specific questions about what we present there?)

Getting back to our exercise –

So, we’ve got 3 meter readings in each Figure.

If the circuit were functioning properly, you’d read:

Meter 1: 240vac (L1 wrt L2)
Meter 2: 120v (L2 wrt N)
Meter 3: 120v (L1 wrt N)

You’ve got both L1 and L2 coming in with 120vac. Because they are 180deg. out of phase, the combination of them results in 240vac from L1-L2 (as we showed you in Unit 6).

Now, I’m assuming this makes sense to you so far.

But, now we have a failure in the circuit and suddenly the element is not longer heating. We rule out the heater itself (it has continuity). The only option for why no current is flowing is that there is an open somewhere else in the circuit – either on the L1 side or L2.

Now we have these readings (still in Figure 1). The first one makes sense – there’s no voltage drop across the element because there’s no current.

But why do we still read 120vac on both the other two meters?

Meter 1: 0vac (L1 wrt L2)
Meter 2: 120v (L2 wrt N)
Meter 3: 120v (L1 wrt N)

Can you answer that?

[Susan Brown]

Okay so here is what i know. We are seeing a series circuit, also we know there is an “open” in this circuit. We know if there is a break in a series circuit, current flow will be interrupted to the entire circuit. We need a closed circuit for current to flow. I understand if we put our leads 1 in neutral and 1 on L1 we see 120v. or 1 on L2 and neutral we see 120v. But i just don’t understand if the heating element is “good” and L2 is also good then why are we getting 0v? Where is the open at?

First of all, which 0v reading are you talking about? We read 0vac across the element (in both Figures) because there is no current going through the element which would create a voltage drop.

And you know what would be helpful – think about what the measurements would be if the circuit was working properly and the element was heating.

[Susan Brown]

Ack! How does neutral “send” hot voltage? Neutral is at ground potential. Have you reviewed the material in Unit 6 while working on this question?

Also, this is not a neutral line. We’re dealing with an L1-L2 circuit, but using a neutral point for some of the electrical measurements as reference. In other words, for measurements 2 and 3 we’ve got the red probe of our meter on either L1 or L2, and the black probe on a known-good neutral point. That neutral point is not part of this particular circuit.

[Susan Brown]

If there is an open on L2, where is that 120vac coming from in Measurement #2? (L2 wrt N)

[Susan Brown]

Which measurement makes you think there’s an open on L2?

[Susan Brown]

Both of those Figures have the same type of measurements. We just get a different reading at Meter 3 (L1 wrt N) in Figure 2 compared to Figure 1.

This is a series circuit. We just concluded above that there is an open somewhere in the circuit. Do we have current anywhere?

[Susan Brown]

Remember that current is electrons moving through a wire at a certain rate.

Parallel circuits share a power supply. The electrons (current) coming from that power supply is what we mean by “total circuit current”.

Total circuit current equals the sum of the currents in the parallel branches.

When one branch fails, its current goes to zero, but the other one is unaffected. So, less current is needed from the power supply.

[Susan Brown]

You are in the right place for that! 🙂

[Susan Brown]

Also, we have a two types of voltage measurements in this problem. There are the measurements across the element (L1 wrt L2) where we would expect a voltage drop reading IF current were flowing through the element.

(Remember, only current “flows”. Voltage is just “present.”)

We also do some measurements from L1 and L2 wrt N, just using a neutral point somewhere as reference to see if we have any potential voltage.

[Susan Brown]

So, since there is no current (despite the presence of voltage), we know there is an open in the circuit.

The element has continuity, so what would you conclude?

[Susan Brown]

P.S. We haven’t taught a lot about what “voltage drop” is yet at this point in the course. We’ll be teaching that more as we go along!

[Susan Brown]

There are several different equations for P. How you use them depends on the scenario and what you are trying to find.

The “E” in the equations is the voltage drop of the resistance that you are trying to find the heat for. (NOTE: we haven’t yet taught a lot about voltage drop, but you’ll keep learning more about this as we go along.)

Our scenario is two resistances in series with each other. If you want to just find the heat generated by the entire circuit, then you could do
P = E squared over R
Where E is the source voltage (because the total voltage drop will = source voltage) and R is the total resistance in the circuit.

But if you only want the heat generated by one of the resistances, then E has to be the voltage drop across just that resistance, and of course R = the ohms of that load.

That’s doable for this scenario, but the technique we show in the video at the end of the unit is easier. Find the circuit current first (using the total resistance in the circuit, because that’s what determines current), then P = I squared * R (where R is the resistance of the load you are interested in)

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