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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 🙂
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.
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.
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?
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)
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?
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?
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.
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.
If there is an open on L2, where is that 120vac coming from in Measurement #2? (L2 wrt N)
Which measurement makes you think there’s an open on L2?
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?
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.
February 11, 2021 at 8:36 pm in reply to: LOOSE CONNECTION — Why does “P = E2 / R” yield incorrect answer? #21466You are in the right place for that! 🙂
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.
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