Susan Brown

[Susan Brown]

Hi David,

No worries – we’ll discuss it here and I can hide certain portions afterwards. You answered “current”, which is not entirely wrong, but it is not the best one of the answer choices we gave you. “Power” is the best answer. Think about the Ohm’s Law equations. We use “P” (for Power) for work being done. P = E x I.

Does that make sense?

[Susan Brown]

At 70 deg. F it was 2000 ohms
At 32 deg. F it was 15,000 ohms

The temperature decreased, but the resistance increased.

[Susan Brown]

Hi Rigo,

The question is:

Question #6 – You’re testing an unknown sensor. You measure its resistance at 70F at 2 k-ohms. You then dunk the sensor into a glass of ice water and then measure its resistance at 15 k-ohms. From these test results, you conclude that the sensor’s type is:

Your answer on the second attempt was “PTC”

Do you think that is the correct answer?

[Susan Brown]

Good question!

The difference occurs because of the decrease in current once you have additional resistance in the circuit.

The current goes from 7.5 amps to ~6.5 amps when we add in the 5 ohm loose connection.

Knowing that the voltage dropped across the entire circuit will be a total of 240vac in either case (Kirchhoff’s Law), you can see that losing an amp of current will decrease the wattage by 240 watts.

P = I x E = 7.5 x 240 = 1800
P = I x E = 6.5 x 240 = 1560 (~1350 from the element and ~ 210 from the loose connection)

Does that make sense?

[Susan Brown]

Hi Jason,
Your initial one got lost in the mail. I put your name on the list for the next batch of Certificates that will be printed. I should have those next week and will put it in the mail. Sorry for the delay!

[Susan Brown]

Hi Samori,

I assume you are talking about the video at the end of Unit 6. The measurements “wrt N” would be taken with a multi-meter. One probe would be on a heating element terminal, the other would be put on a known-good neutral point that would be identified by looking at the schematic. You’ll be learning more about this as you go along!

[Susan Brown]

Hi Julian,
Some of the presentation slides are available upon request. I’ll see if I can get my hands on this one and if so will get it to you.

[Susan Brown]

So, is it accurate to say that any load that performs work in any kind of electrical circuit will always experience a voltage drop?

yep!

As far as learning this stuff goes – just keep at it! It takes awhile to nail down, but you’re doing great.

P.S. I’m going to hide parts of your answers above just so we don’t give it away to other students 🙂

[Susan Brown]

Hi Julian,

It’s great that we are working on this together, as you are clarifying your understanding of these critical electrical concepts.

So, you’ve go the right view of the detector switch shunting the main coil, which leaves the other 3 loads in parallel.

We’ve just got to clear up what that means in terms of voltage drop for those 3 loads.

Being in parallel means that they are each tied to the same power supply, but on their own circuit. So, they each have access to full source voltage rather than dividing it up, as happens with loads in series.

You can think of each of these three circuits as three independent series circuits with one load in them that happen to be connected to the same power supply. (We show this in the videos in Units 4 and 5.)

These loads have Line voltage on one side, and neutral on the other. Will current flow through them?

If so, remember what we said about voltage drop above: it is created when current is flowing through a load.

In other words, for a load that is the only load in a circuit, if it has current flowing through it, it will drop the source voltage.

[Susan Brown]

You are mostly right. It is not “potential” difference, however, because current is actually flowing.

That’s the big difference between voltage and voltage drop.

Just plain voltage is usually referring to the potential for current to flow. A lot of measurements “wrt N” (as in Question 9) are just looking for potential, or source voltage. (For example, if you measured 120vac across an open switch, you’d be measuring the potential for current to flow – which means current will flow when that switch closes. When the switch closes, then you would no longer measure any voltage across it because it would be just like measuring two points next to each other on a wire.)

But voltage drop is created by current flowing through a load, which means work is being done.

So – back to this scenario.

There are 4 loads (ignitor, booster coil, safety coil, main coil) that we want to know the voltage drop across. So the first thing to decide is if all of the loads are receiving current with the circuits as drawn, and how they are arranged in relationship to each other – series or parallel.

The “Zen trick” is helpful with that.

One thing you have to consider is that closed detector switch, and what effect it has, if any, on the current.

[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.

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