You are an excellent teacher. No one can teach all these topics better. I love the way you present these complex techniques simply. I've been following all your videos. I appreciate you. I want you to minimize the way the board is covered during lectures. I don't know how you can do that but I want the board to be visible at all times.
Thank you for the feedback. We have been struggling with the same question and we think we have found a solution. We now tend to work the board from right to left such that most of the material remains visible.
Hey Michel, I really appreciate your lectures, and have been watching them for a little over a year now. I was wondering, when you disconnected a current source and replaced it with an open source, what if there happened to be a resistor that ceased to allow current to run through it? How would that affect finding Rth? Thanks!
Out of curiosity, if we had a wire which began between the 16 Ohm resistor and the end-point A which ran down and connected to the end-point B branch, then the 8 Ohm and 16 Ohm resistors would be in parallel, correct? This seems to be something a lot of beginners, including myself, struggle with. There are so many definitions of parallel resistance that seem inconsistent. I usually go by "if the current splits between multiple resistors and end up back on the same branch then they are in parallel".
A wire from A to B would make a short circuit and no current would flow through any of the resistors. If there is more than one path for the current to flow then the branches are in parallel, but there needs to be resistance on each branch.
I am having a hard time understanding why the I1 is equal to 2 Amps. i get that there is a current source there that says 2 amps but what about the 8 ohm resistor and 8v voltage source?
In the circuit on the bottom left side of the board, we removed the current source and shorted the voltage source. At that point the two resistors are in series.
How do you know I-1 of the Mesh Analysis is equal to 2 Amps? It does not appear to account for the I-2 traveling in the opposite direction through the 8 V and 8 R elements. Someone please clarify this point for me.
I1 must be 2 A because there is a current source on that branch. Since the branch with the 8 ohm resistor carries I1 downward and I2 upward, you have to subtract I2 from I1 to find the current in that branch.
The load resistor is whatever the load is that is connected. It is not something we solve for (unless a specific value is given for the current), but the current becomes a function of whatever load you connect.
Great explanations! But I have a question: when you are writing the equation in the loop 2, doesn't the equation have to be 8 + (8)(i2 - i1) + 16(i2) = 0 which would give a current of 1/3 Amp
M4kkMester Yes, this also works. He just started at the positive side of the voltage source which is why his symbols are opposite of yours, but it means the same thing.
Just like in algebra there are many techniques to solve simultaneous equations. With electric circuits, there are many techniques to solve them and all are good techniques. We just teach different techniques as some work better on some circuits than others and it helps in the total understanding.
It is the technique Norton invented to solve a circuit. Learn the method first by going through the steps systematically. Then you'll see how and why it works.
@michel van Biezen l respect you thanks so much you have helped in many courses ...............wait soon l will be a millionaire l will definitely reward you
You are an excellent teacher. No one can teach all these topics better. I love the way you present these complex techniques simply. I've been following all your videos. I appreciate you. I want you to minimize the way the board is covered during lectures. I don't know how you can do that but I want the board to be visible at all times.
Thank you for the feedback. We have been struggling with the same question and we think we have found a solution. We now tend to work the board from right to left such that most of the material remains visible.
EXCELLENT PROF FOREVER
Perfect! Among my favorite online lectures! Thank you!
Glad you like them!
Hey Michel, I really appreciate your lectures, and have been watching them for a little over a year now. I was wondering, when you disconnected a current source and replaced it with an open source, what if there happened to be a resistor that ceased to allow current to run through it? How would that affect finding Rth? Thanks!
You leave the resistors in place and just removing the current source. Then you solve for Rth with the remaining circuit.
Out of curiosity, if we had a wire which began between the 16 Ohm resistor and the end-point A which ran down and connected to the end-point B branch, then the 8 Ohm and 16 Ohm resistors would be in parallel, correct? This seems to be something a lot of beginners, including myself, struggle with. There are so many definitions of parallel resistance that seem inconsistent. I usually go by "if the current splits between multiple resistors and end up back on the same branch then they are in parallel".
A wire from A to B would make a short circuit and no current would flow through any of the resistors. If there is more than one path for the current to flow then the branches are in parallel, but there needs to be resistance on each branch.
So well presented. Thank you so much.
Excellent video sir! Thank you!
Glad it was helpful!
I am having a hard time understanding why the I1 is equal to 2 Amps. i get that there is a current source there that says 2 amps but what about the 8 ohm resistor and 8v voltage source?
There is a current source forcing a current of 2 AMPS through branch 1.
where was I to find a good teacher like you😭😭😭
At least you can get help with the videos. 🙂
Why did we add the 16ohm and 8 ohm resistor if they were in parallel?
In the circuit on the bottom left side of the board, we removed the current source and shorted the voltage source. At that point the two resistors are in series.
@1:38 you say the Norton resistance is the "resistance between A and B"... WHY?. If the circuit is open between A and B the resistance is infinity.
Resistance on the path between A and B. (It is not infinite).
what the value of RL ?
why does current I1 flow through only 8 ohm resistance, but not through 16ohm? sir please explain...
In order for current to flow, there must be a complete circuit. No current will flow if there is not a continuous path.
@@MichelvanBiezen no sir .....i was asking about the circuit when A&B are connected for finding norton current. Why does I1 not split?
oh i got it sir....it was mesh analysis. Thank you sir.
@@MichelvanBiezen If there is no continuous path then how I-2 will flow through 16 ohm?
How do you know I-1 of the Mesh Analysis is equal to 2 Amps? It does not appear to account for the I-2 traveling in the opposite direction through the 8 V and 8 R elements. Someone please clarify this point for me.
I1 must be 2 A because there is a current source on that branch. Since the branch with the 8 ohm resistor carries I1 downward and I2 upward, you have to subtract I2 from I1 to find the current in that branch.
Thank you! That was killing me.
A video on Thermodynamik , Gas - Dampf. Feuchte Luft is needed. Best Regards KQKR Kristian
We have hundreds of videos on thermodynamics on this channel.
Hi Michel, wondering what will be the value of Load R.,I think you did not find that...
The load resistor is whatever the load is that is connected. It is not something we solve for (unless a specific value is given for the current), but the current becomes a function of whatever load you connect.
@@MichelvanBiezen Superb!
Great explanations!
But I have a question: when you are writing the equation in the loop 2, doesn't the equation have to be 8 + (8)(i2 - i1) + 16(i2) = 0 which would give a current of 1/3 Amp
The equation in the video is correct. Remember that if you go across a resistor in the same direction as the current, you have a voltage drop.
@@MichelvanBiezen thanks!
If you use the equation of "- 8 + (8)(i2 - i1) + 16(i2) = 0" it would result i2=1A. You have to put a minus sign in front of 8V.
@@MichelvanBiezen please explain
M4kkMester Yes, this also works. He just started at the positive side of the voltage source which is why his symbols are opposite of yours, but it means the same thing.
????If We assume RL is 1ohm
Than equation Must be IL=(1/1+24)*1
No, the equation in the video is correct. Thanks for checking. 🙂
thank you
Please sir how did the voltage source(8v) goes to zero? I’m confused
That is the technique used for Norton's Theorem. Take a look at a few examples.
just curios what would be the Rth if there was another resistor after the 2A
A current source will adjust to ensure that it produces the current it is set to. Adding a resitor on that branch should not make any difference.
@@MichelvanBiezen so the rth wouldn't change in value to add that additional resistors value as well?
I would not expect Rth to change.
In this case, can we use Thevenin theorem?
Just like in algebra there are many techniques to solve simultaneous equations. With electric circuits, there are many techniques to solve them and all are good techniques. We just teach different techniques as some work better on some circuits than others and it helps in the total understanding.
Thanks, Sir.
im so confuse . at lesson 16, we have the same situation by they are parallel. why are these resistor are in series now ? ???
It is the technique Norton invented to solve a circuit. Learn the method first by going through the steps systematically. Then you'll see how and why it works.
thank you professor I will go over it again. BTW, this is the most helpful series i could find on youtube. thank you so much for these video
Great teacher. Huge back.
so thanks
Most welcome
Thankss 🤗😊☺️
@michel van Biezen l respect you thanks so much you have helped in many courses ...............wait soon l will be a millionaire l will definitely reward you
bho here
life-saver
Glad to be of help.
Michael Biezen is the bill nye of UA-cam
Bilale anlatır gibi anlatmış helal olsun
Glad you liked it. 🙂
what the value of Rl ?
I think it's 0 in this case since there is no resistance between a and b