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A series circuit is just describing a circuit with one or more loads in it, where they are connected one after another.
The relationship between voltage (E), current (I), and resistance (R) in a series circuit is E = I x R. This tells you how changing one value will affect the others.
In the real world, in a typical circuit, the source voltage will be set – either 120 or 240.
Then the circuit current will be determined by the total resistance in the circuit.
I = E/R
So, if R is increased, I will decrease. That is true regardless of what the source voltage is.
(In Unit 8, you’ll learn that you can calculate the “voltage drop” across each load using E = I x R, which is another way to use that same equation.)
1. Your example is correct IF the two loads have the same resistance. If they don’t, then the voltage drops will be different (proportional to the resistance… E = I x R).
Go back and look at the Midterm Exam in Core, Questions 2, 3, and 4 where you had to calculate a scenario with 3 loads in series with different resistances.
2. Current is electrons hopping from one atom to the next. They are like tightly packed billiard balls in a track – they are either all moving at the same rate, or not at all. So, in a circuit, the electrons are moving at the same rate throughout the wire, regardless of where they are (on the line side or the neutral side).
The difference between the two sides of the circuit is the voltage. The neutral end is always at ground potential. The line (or, “hot”) end is rapidly shifting from +120vac and -120vac, creating the charge that causes electrons to move back and forth towards the more positive side of the circuit.
This motion of the electrons through a load creates a voltage difference (“drop”) from one side of the load to the other. The sum of the voltage drops across the loads that are in series will total the source voltage.
Does this help?
Depends on the model, but yes, they can. Will still be separate tubes.
They are completely separate sealed systems. The tubes may run side-by-side at certain points (e.g., the condenser), but the refrigerant never mingles.
May 9, 2021 at 4:36 pm in reply to: replacing the compressor and filter(unable to move any further into the course ) #21861Okay, I tried adding some more space.
If that doesn’t work, you might have to try a different browser.
Does the last video show up bigger than the others in that unit?
May 9, 2021 at 8:52 am in reply to: replacing the compressor and filter(unable to move any further into the course ) #21856That’s interesting – it’s the first time we’ve heard of this happening. What are you watching it on?
I tried adding a couple of blank lines after the video… try again and see if it has changed.
Hi Thomas,
I also saw your followup post that you read over the info about the Midterm.
Approach the Midterm as part of the learning experience. The open-answer format and a few of the questions we ask will challenge you to apply the basic electricity concepts that we’ve been teaching. It’s very common to need more than one attempt to pass this one, but it helps things to “click” for lots of students.
There are only 9 questions, but some of them may take some time to think about.
May 8, 2021 at 9:27 am in reply to: replacing the compressor and filter(unable to move any further into the course ) #21851Hi Andrew – did you click the button to mark the unit as complete?
I love seeing a student paying attention to details! 😀
We want students to be comfortable with either letter. We say in the lesson:
Voltage is measured in units called Volts and is represented by the letter V or sometimes you’ll see the letter E used.
Keep up the good work!
Hi Darren,
The video played fine for me. More than likely if you closed/reopened your browser, and/or cleared your cache, it would play normally. But you aren’t missing much – he’s just talking about the valve near the bellows that can cause water to back up. You can basically read the text on the photo and understand the point.
Yep! And I see you found an older topic on this same question 🙂
Hi Darren,
Sure, you can always ask for a reset if you want to retake a quiz/exam for any reason. Just do so before moving forward. Do you know that there is a Quiz and Exam Reset Request form in the Campus Support menu?
I reset it for you.
R x I^2 = P (power, or watts). We aren’t looking for work in this question. (In other questions, where we are looking for heat generated by a load or a loose connection, you might use this formula.)
Remember, voltage is E (some people use V, but we usually use E in the course).
Since you are trying to find “E”, you want to look in the section of the pie chart for E. In other words, a you’re looking for a formula that starts with E = …
Hi William,
I’m glad you posted here. I emailed you using the address on your student account, but it bounced back to me. I tried your old gmail address that I had as well.
Please let me know what email address I can use to get in touch with you in the future!
Part 4 of #7 is incorrect.
And here’s a link to our Midterm Help Page.
Justin’s answer is close, but I don’t like to think of voltage as being “consumed.” Voltage measurements are always showing us a difference in charge between two points. Sometimes it’s between a point in our circuit and another known-good neutral point that is not part of the circuit (“wrt N”). Sometimes we are measuring voltage with our probes on each side of a load.
Which electrical property, that we can measure with our meter, is unique to loads?
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