Your tutorials are great. 20 years ago, the resources online were not so good and many filled with people not so kind! lol Before I had done any college for E.E., just finding someone online in those days to provide the explanation and simple circuit to produce a virtual bipolar supply from a single ended only took me about 2 *YEARS* before someone was decent enough to help me without demanding I build a *shrine to them lol.*
Very well-made content. I think your channel fills a spot on youtube, most of the other equivalent channel's contents are showcasing (not teaching). The only major thing that sets you back is the upload frequency.
At 11:55 you say the resistance is in parallel, but at 12:40 you calculate resistance in series. Could you explain why it changes from my understanding you're pulling the same values from the same circuit but combining them differently ?
During the second step of removing the load resistor in the second example, what is the issue with simply removing the 50-ohm resistor and keeping the voltage sources in the exact same spots? Obviously it's wrong and you'd get 10-ohm and 20-ohm resistors in series rather than in parallel--but what's the concept error that I am missing here?
That's a good question and actually, you CAN remove the 50 ohm resistors and keep the voltage sources in the same spot. But then the challenge will be to remember that your node A is between the two resistors. As mentioned in the video, we get comfortable looking at circuits in a certain way and, by moving it so that nodes A and B jut out to the right, it's a more "comfortable" way to look at it. But if we just removed the 50 ohm resistor and labeled node A BETWEEN the two resistors, you'll get the same result. I hope that helps!
Since the voltage sources are shorted while finding the Thevenin resistance, it puts the 10 and 20 ohm resistors in parallel. Their equivalent resistance is 6.667 ohms and is the resistance between nodes A and B, so that is the end equivalent resistance. Hopefully that's what you're looking for!
At 2:57 you mentioned that the 3 resistors will be in parallel, but how? If you place a short at the two voltage sources, wouldn't the first resistor get shorted out since it is in parallel with R=0? Then the Req as seen by the load should be the two other resistors in series with each other.
I'm assuming you're referring to 3:47 or so? I reviewed it and you are absolutely correct. That far left resistor will have the two ends shorted to each other so, in this case, that first resistor would be taken out completely. Good catch, thanks!
Sure! A load is a very generic term because it could mean a lot of things. In general, though, it's the thing that we're trying to get the correct amount of power to. So, in a lot of example coursework, we model a load as a resistor. But in reality, it could be a motor (which could be a fan, motor on a car, blender, whatever), a transistor, a heating element, an antenna, etc.
In the past we've sometimes just used the transcript from the video for the written tutorial or just read the written tutorial verbatim to make the video but we're trying to make the written tutorials more complementary now, so that there's a slightly different style in presentation and different examples that should help out. www.circuitbread.com/tutorials/thevenin-theorem-finding-a-thevenin-equivalent-circuit-circuitbread-circuits-1
Hi Kamil, I'm not certain what you're referring to. In the second example, the 10 and 20 ohm resistors are in parallel and the Rth equivalent comes out to be 6.667 ohms. Could you give me a time stamp for reference?
@@CircuitBread 12:36 you calculated the thevenin voltage, you did 20/30 instead of 20/6.66 so the current was wrong and lead to a different voltage drop across the 10ohm resistor
Which current source are you referring to? Could you give me a time stamp? There shouldn't be a current source at the end, though - the Norton Equivalence has a current source as part of the end result. Is that what you're talking about?
If the two nodes of the resistor are attached together, it's kinda like putting a 200 ohm resistor in parallel with a 0 ohm resistor. You do the calculations, you find that the equivalent resistance is 0 ohms. So you can just act like the 200 ohm resistor isn't there at all.
Electricity Takes The Least Resistive Path That's Why When There Wire And A Resistor In Parallel , The Electricity Will Flow Through Wire As It Has Less Resistance 😊
Please figure out how to animate the stuff you draw on paper. Believe me, you will get more views. I mean the background is perfect. You just need to make other things like your thumbnail and animation fancier.
Thanks for the feedback! We're trying to figure out a way to do that animation faster, as I agree, it would look cleaner and be more attractive to not have my scribbles all over the place.
Your tutorials are great.
20 years ago, the resources online were not so good and many filled with people not so kind! lol
Before I had done any college for E.E., just finding someone online in those days to provide the explanation and simple circuit to produce a virtual bipolar supply from a single ended only took me about 2 *YEARS* before someone was decent enough to help me without demanding I build a *shrine to them lol.*
Very well-made content. I think your channel fills a spot on youtube, most of the other equivalent channel's contents are showcasing (not teaching). The only major thing that sets you back is the upload frequency.
Love these videos. Perfect for practice.
Great video! Thanks
Very good video, thank you.
At 11:55 you say the resistance is in parallel, but at 12:40 you calculate resistance in series. Could you explain why it changes from my understanding you're pulling the same values from the same circuit but combining them differently ?
you're angelic. thank you.
During the second step of removing the load resistor in the second example, what is the issue with simply removing the 50-ohm resistor and keeping the voltage sources in the exact same spots? Obviously it's wrong and you'd get 10-ohm and 20-ohm resistors in series rather than in parallel--but what's the concept error that I am missing here?
That's a good question and actually, you CAN remove the 50 ohm resistors and keep the voltage sources in the same spot. But then the challenge will be to remember that your node A is between the two resistors. As mentioned in the video, we get comfortable looking at circuits in a certain way and, by moving it so that nodes A and B jut out to the right, it's a more "comfortable" way to look at it. But if we just removed the 50 ohm resistor and labeled node A BETWEEN the two resistors, you'll get the same result.
I hope that helps!
Bro you are awesome please keep going
Funny explanation ❤
Hi. For the second example. How did you get 6.66 for thevenin equivalent ?
Since the voltage sources are shorted while finding the Thevenin resistance, it puts the 10 and 20 ohm resistors in parallel. Their equivalent resistance is 6.667 ohms and is the resistance between nodes A and B, so that is the end equivalent resistance. Hopefully that's what you're looking for!
Why do we replace voltage sources with shorts and current sources with opens?
This guy is like Paul Rudd if he pursued engineering instead of acting
Tonns of thanks to you sir. It was a reallly helpful video.
Very well made video I subscribed :)
At 2:57 you mentioned that the 3 resistors will be in parallel, but how? If you place a short at the two voltage sources, wouldn't the first resistor get shorted out since it is in parallel with R=0? Then the Req as seen by the load should be the two other resistors in series with each other.
I'm assuming you're referring to 3:47 or so? I reviewed it and you are absolutely correct. That far left resistor will have the two ends shorted to each other so, in this case, that first resistor would be taken out completely. Good catch, thanks!
As someone new to the study of electronics, can you clarify what you mean by "load"? Can you give me an example of a load?
Sure! A load is a very generic term because it could mean a lot of things. In general, though, it's the thing that we're trying to get the correct amount of power to. So, in a lot of example coursework, we model a load as a resistor. But in reality, it could be a motor (which could be a fan, motor on a car, blender, whatever), a transistor, a heating element, an antenna, etc.
Thank you Sir ., 🙏
You're welcome!
Is there another video of yours I can watch to better understand this because I am having a hard time following
In the past we've sometimes just used the transcript from the video for the written tutorial or just read the written tutorial verbatim to make the video but we're trying to make the written tutorials more complementary now, so that there's a slightly different style in presentation and different examples that should help out. www.circuitbread.com/tutorials/thevenin-theorem-finding-a-thevenin-equivalent-circuit-circuitbread-circuits-1
In example 2, when you were finding Req you did 20+10 but aren't the resistors in parallel ? wouldn't it be 200/30 ?
Hi Kamil, I'm not certain what you're referring to. In the second example, the 10 and 20 ohm resistors are in parallel and the Rth equivalent comes out to be 6.667 ohms. Could you give me a time stamp for reference?
@@CircuitBread 12:36 you calculated the thevenin voltage, you did 20/30 instead of 20/6.66 so the current was wrong and lead to a different voltage drop across the 10ohm resistor
why did u not draw out the current source in the sample
Which current source are you referring to? Could you give me a time stamp? There shouldn't be a current source at the end, though - the Norton Equivalence has a current source as part of the end result. Is that what you're talking about?
At 7:00 why does the short circuit cut out the 200 Ohms
If the two nodes of the resistor are attached together, it's kinda like putting a 200 ohm resistor in parallel with a 0 ohm resistor. You do the calculations, you find that the equivalent resistance is 0 ohms. So you can just act like the 200 ohm resistor isn't there at all.
Electricity Takes The Least Resistive Path That's Why When There Wire And A Resistor In Parallel , The Electricity Will Flow Through Wire As It Has Less Resistance 😊
Vth can not be 5V based on one source ......as other source also contribute ...
Cheers.
Sourcing and sinking. Tutor us on these concepts, please.
That's a good one, we already have that in the list as part of this Circuits 1 series!
Please figure out how to animate the stuff you draw on paper. Believe me, you will get more views. I mean the background is perfect. You just need to make other things like your thumbnail and animation fancier.
Thanks for the feedback! We're trying to figure out a way to do that animation faster, as I agree, it would look cleaner and be more attractive to not have my scribbles all over the place.
shush]