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Okay – so if the ignitor [answer hidden] what is its voltage supply? What is its voltage drop?
Here’s how I get that:
1/30 + 1/50 = 0.0333 + 0.02 = 0.0533
Then 1/0.053 = 18.75
Hi Kyle,
Check out these two previous forum topics and see if they help you. Feel free to ask more questions after looking over these:
Hi Kaleb,
You will be encountering similar calculations as you move along, so it is awesome that you are asking for help understanding this, even though you technically passed the quiz!
First of all, we step you through a similar calculation in the last video of this unit. That’s helpful to review. And read the caution under that video about using the P = E(squared) / R equation. That only works if you calculated the voltage drop across the loose connection first – you can’t use the entire source voltage. Since we haven’t yet covered voltage drop in detail at this point in the course, there’s a better equation to use for P that involves I and R.
Rewatch that video and see if that helps you. If there’s something you can’t quite follow, let me know!
correct – if a load is by itself in a circuit, then we know it will drop the source voltage across it.
Okay – now onto the other loads in the gas valve circuit that you are having trouble with. The next step is to make sure you know which loads are getting current, and what their relationship is with each other (series or parallel).
1. Which loads are getting current? Which are not?
2. Do the “Zen trick” on the loads that are getting current. Which ones (if any) are in series?Correct!
So, if you have only one load in a particular circuit, do you need to know its resistance in order to know what its voltage drop is?
Jordan – where’d you go? 🙂
Correct!
One way you can think of it is that “voltage” is a cause – it causes electrons to flow. “Voltage drop” is an effect – the result of electrons flowing through a load.
And you are correct that if we have loads in a series, the voltage drop across each load is based on the resistance. We can calculate that using E = I x R.
What is true about the sum of all of those voltage drops (if you have two or more loads in series)?
Hi Jordan,
Thanks for posting a topic! I’m going to step you through with some questions to help you with this, so we’ll go back and forth a bit. Be sure to keep an eye out for my replies!
Let’s first just make sure you understand voltage drop. The general term “voltage” means a difference in electrical charge between two points. Voltage drop is a more specific type of voltage – can you describe in just a few words what voltage drop is?
Correct. Current flowing through a load will create a voltage difference, which we call a “voltage drop”. The main reason we have a slightly different term for that is that there’s a big functional difference between “potential” voltage and voltage drop.
You can think of voltage as a cause, and voltage drop as an effect.
Voltage causes current to flow (if there’s a complete circuit for the electrons)
Voltage drop is a result of current flowing through a load/resistance.When you have voltage drop, you will have work being done (power is voltage AND current)
Well, what I was talking about is just a wire. Two points along a wire, when there is no component between the two points you are measuring – you’ll measure 0 volts because there is no difference in electrical charge between those points.
If a switch is open, there will be a difference in charge because of the open in the circuit.
If a switch is closed, it will be just like you are measuring two points along the same wire, with no difference in charge.
Well, that’s not always the case. It depends – sometimes you do want to measure voltage at a certain point with respect to N. But sometimes you are measuring voltage with your probes at two points in the circuit, such as across a load to check for voltage drop, or across an open switch.
Voltage is simply the difference in electrical charge between two points.
If you put your meter probes at two different points along the same wire, and there’s no open switch or load between those two points, what will you measure for voltage?
That is talking about measuring voltage across a switch that’s either open or closed. If a switch is closed, it acts like a wire, correct? Do you measure a voltage difference at two points along a wire, when there is no load between those two measurement points?
Hi Darron,
Unit 5 is describes various characteristics of series circuits and parallel circuits.
When two or more loads are in series, you can add the resistances of the loads together to get a “total” resistance. This can be useful when calculating the circuit current, since the “total resistance” is what determines that.
However, when loads are in parallel with each other, we talk about “equivalent resistance”. It’s a similar concept – it’s a way of describing the overall resistance that is present from all of the loads combined. It’s more complicated than simply adding them together, however, since they are not in series.
We show the formula for calculating equivalent resistance in Unit 5. But we also give the rule of thumb in the 3rd video in that unit (you can even see it in the thumbnail for the video!). That’s perhaps more important to know than doing the actual calculation, and it will help you to answer both of the questions that you posted
Take a look at that information again in Unit 5, and see if you understand it. Let me know!
Hi Luke,
Good question!
This may be a 120/208v commercial system. You see this a lot in apartment buildings (you can do an internet search for 208 voltage supply to read more about this). If that’s the case, then your readings are to be expected.
A reminder – we always recommend you do measurements with respect to Neutral, not Ground. They should be the same potential, but that’s making an assumption that you may not know for sure.
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