Nope! an easy way to see this is to look at the circuit with the sources set to zero. We can then see those two resistors share equipotentials nodes! ~andrew
This guy explained this concept better in 12 minutes than my professor did in a week. I've been going to my 8 AM lecture for weeks, listening to my thick-accented professor mumble for an hour when I could've just watched these videos...
If anyone is having issues following any of these videos, your goldem ticket to understanding is to learn about the principle of superposition of linear circuits. Organicchemistry tutor did a video on this. In a nutshell, its the idea that you can sum up the current contributiins of each individual power source. Makes it so much easier and you can forget about all of this norton and thevenin equivalent stuff which is offputting
i'm still trying to figure out if he's looking at us from behind a glass wall and writing backwards on it or what. that alone is too distracting for me to pay attention to the lesson of the video....
Our biggest clue to what is going on is that he appears to be writing with his left hand. Since there is a ninety percent chance he is right-handed, we can assume the technology being used is flipping the original image into a mirror image, and he is in actuality writing normally, and writing with his right hand,
To everyone wondering, it looks like he’s writing on a glass and the video is flipped horizontally, so all writings appear right and not back to front.
my proffessor explained the thevenin much better. The only think we need is the logic behind what we do. Not all the calculation. And everyone do basic calculations (summing, dividing ; of course we can do these things!?) and dont tell what is the logic. I think this video is not different!
Great insight about removing voltage & current sources by removing the effects they would generate, stop a current with an open yet eliminate a voltage with a short
Great video! I think the part where you explain what happened to the 20 ohm resistor when doing your first Thevenin could have used a bit more explanation as to why you can ignore it. Really well done though. I like your idea of mirroring the image so you're not writing backwards. :-)
7:42 That's not quite right. When you have two impedances (like resistors) in parallel, you would say that the "path of least resistance" is through the smallest impedance, and yet the current flows through both. The actual reason you can ignore the shorted resistor, is that you have a current divider built from the 40-ohm resistor and the 0-ohm short. By the formula of a current divider, the resistor gets no current and the short gets all the current.
So real world use age. Let’s move on to line load impedance matching, or implementing the termination resistor, same thing remember that. Great stuff keep it up.
So if it becomes too difficult to find the equivalent resistance, but you can find the voltage and current buy Noodle and mesh analysis, you can calculate the resistance from those.
Why does i2= 4A? Is it because: Mesh currents are made for every non inclusive loop. So there would be two non inclusive loops for this circuit. Loop 1 would contain the 20 ohm and 10 ohm resistor. Loop 2 would contain the 40 ohm and 10 ohm resistor. Because Loop 1 has only resistors in series, the current around loop 1 is a constant 4 A. (**** note: the 10 ohm resistor is in both loops) Is my reasoning correct?
Hello! Thanks for commenting! This isn't entirely right. A loop current can be assigned to any of the smallest closed loops in the circuit. This means loop one consists of the loop with the voltage source, 10 ohm resistor, and 40 ohm resistor. Loop 2 consists of the current source, 10 ohm resistor, and the 20 ohm resistor. The current in any element is then the sum of all the loops currents that have the element in the loop. Since the current source only exists in loops 2, the current in the current source is the same as the second loop current. THis is why the second loop current is the value of the current source. I hope this helped! ~Andrew
Hi Guys, not sure if you still answer questions in the comments or not, but I was wondering. Why isn't IN 14 amps(the 4 amp source plus the new i1 of 10 amps)? Doesn't KCL say that all currents must equal at a junction?
Great teaching skills that takes me back to my high school days, my question is when do we use Norton and Thevenin theorems in real life application ? in Small-signal modeling for example I'm I right ? because in real life there is no such Norton souce of intensity (pure Amperes) or Thevenin source of potential (Pure Voltage) but the source is always a hybrid Source (Volt + Ampere)
Well the real life is always a little different from the theoretical. We use Thevenin and Norton models in small signal applications, yes. We also use them when providing loading for complex systems (i.e. a speaker is 8 ohms , this is the thevenin impedance). As for your question about the different sources, the models differ depending on the capability of the source and its loading. If voltage is maxed out but we can still provide all the current we need, use a voltage source model. If the opposite is true, use a current source model. If there is a combination of limiting factors, you need to be creative to get the same type of source you need. It is all about exploiting theory to closely match the reality, which differs from situation to situation. ~Andrew
It's outside the test points the test points are the same 2 nodes as across the 40-ohm resistor. With the same nodes, they are by definition in parallel and thus must have the same voltage potential.
I was hopeing someone could explain. In the first example It's not clear to me why the R_thevenin is equal the equivalent circuit of 10ohms parrallel to 40ohms as opposed to in series with.
Thanks for commenting! When we eliminate sources we replace current sources with an open circuit and voltage sources with a short circuit. We then look for the equivalent resistance BETWEEN points A and B. In this case the 10 ohm and 40 ohm resistors are in parallel between A and B when sources are eliminated. Also remember that if you are having trouble finding that equivalent resistance, you can instead find the open circuit voltage and short circuit current between A and B. Then R_eq = (V_open)/(I_short).
Hi There. Thanks for commenting! As, @Zac Sutton said above, we need to turn off all of the sources and look at the resistances between A and B. When we do this, we see that the nodes the two resistors are connected to are one and the same, implying that they are in parallel. The current in them may be the same, but this would be by coincidence only, not because of they are in a series configuration (they are not).
When you leave A you can take two paths to get to B. You go straight or turn left. At this point you should notice tge tips and tails are touching for bith resisters making them parallel.
This garbage is easy, but it is the type of math that, if you have not studied it for months, you have to review it again. Good job. I like how you did this video. What did you do? Were you standing behind glass and writing with some kind of bright marker? It was really done nicely. I like how you told us that the 8 ohms was both the Norton and Thevanin equivalent resistance. Frank The Master of Science
When solving for for Norton, when the current flows around from B and onwards, is it not possible for the current to flow upwards throw the 40 ohm resistor. All three branches have resistors?
In this case you would simply find the thevenin resistance the same way, just ignore the step with the current source. if you have a more targeted question we will do our best to explain it better! ~Andrew
could be either or if you keep the signs consistent.resistors usually drops while to voltage negative to positive is a voltage jump if you multiply everything by a negative would fix the sign.
what if i put a resistor between points a and b? can i short those? (eliminating the resistor) and when i draw the norton equivalent should i draw the resistor too between points a and b?
Why is i2 4A? Wouldn't the 60v source create some further current through there? Like if you follow the path of the positive charge from the voltage source it first has the option of no resistance vs 10 ohms, so at least some current should go to i2. Or do voltage sources treat current sources as open circuits?
Aren't the 10 and 40 ohm resistors in series, not parallel?
Nope! an easy way to see this is to look at the circuit with the sources set to zero. We can then see those two resistors share equipotentials nodes!
~andrew
No because the wire with the 20ohm and 4a above it
excellent response.
Yes, for the mesh loop, but not for outputs a and b.
@@UConnHKN I dont understand they still look in series to me :/
You explained this concept better in 12 minutes than my school in 2 hours.
Watching this 15 minutes before my final exam in circuit 1. THANK YOU SO MUCH
This guy explained this concept better in 12 minutes than my professor did in a week. I've been going to my 8 AM lecture for weeks, listening to my thick-accented professor mumble for an hour when I could've just watched these videos...
I would never come over this concept.untill I watched this..Thanks a lot
Your video is rad, and you should feel rad. Thank you.
If anyone is having issues following any of these videos, your goldem ticket to understanding is to learn about the principle of superposition of linear circuits. Organicchemistry tutor did a video on this. In a nutshell, its the idea that you can sum up the current contributiins of each individual power source. Makes it so much easier and you can forget about all of this norton and thevenin equivalent stuff which is offputting
for anyone wanting it ua-cam.com/video/EX52BuZxpQM/v-deo.html
That is for ideal circuit analysis. In the real world you must deal with losses (ie bleed current, parasitic reactance, etc.)
Thanks Sir!😊
Loop1:
60=10(i1-i2)+40i1,50i1=100,i1=2a
Loop2:
I2=4a
and then Rth=10//40=8
Vth=40*i1=80v,Isc=Vth/Rth=80/8=10A
i just got 99% on my test thanks
wow tats amazing
Amazing! Keep it up!
hes good at writing backwards too
I think they film him writing normally then just mirror the video when editing. Would be far easier than writing it all backwards I would think haha
that's not possible.
Some people are just backwards thinkers! ha! sorry, I just couldn't....resist ha!
video is flipped sideways so we can read
It was a joke honey
yall explained this so well, your the reason I passed EE 213 last quarter
from Vth and Rth norton current can be easily done...just divide vth/rth i.e 80/8 =10 amps...just using source transformation...
okay im glad i wasn't the only one who thought this
i'm still trying to figure out if he's looking at us from behind a glass wall and writing backwards on it or what. that alone is too distracting for me to pay attention to the lesson of the video....
i know right me too
Our biggest clue to what is going on is that he appears to be writing with his left hand. Since there is a ninety percent chance he is right-handed, we can assume the technology being used is flipping the original image into a mirror image, and he is in actuality writing normally, and writing with his right hand,
me too :(
It's a black reflective surface! Kinda like mirror...
It probably is backwards for the camera but it's just reversed in post
The product over the sum trick to find parallel eq resistance... GENIUS
He's right handed. I can't believe the comments on this video, the video is simply mirrored. This is not a difficult concept.
he is very much left handed...
Nicolas Palmeiro mirror it the other way not the way you think
If the video was mirrored he'd have to be writing backwards right handed.
Thanks. Needed a quick refresher for my FE exam.
To everyone wondering, it looks like he’s writing on a glass and the video is flipped horizontally, so all writings appear right and not back to front.
Best explanation! Got myself ready for my EE finals later! Thank you!
my proffessor explained the thevenin much better. The only think we need is the logic behind what we do. Not all the calculation. And everyone do basic calculations (summing, dividing ; of course we can do these things!?) and dont tell what is the logic. I think this video is not different!
Thank you so much. I will be using this video for the rest of this semester. I really appreciate you putting this up
Wow thanks, this really helped to explain the fundamentals as well. Great for revision. Cool way of doing videos also.
First video to actually make me understand this
Great insight about removing voltage & current sources by removing the effects they would generate, stop a current with an open yet eliminate a voltage with a short
7:45 just made my life so much easier. Great video!
Thanks bro..
why?
Thanks Zac, your teaching is very clear! Helped a lot
Thank you 4 saving me 3 hrs of revision
Great video! I think the part where you explain what happened to the 20 ohm resistor when doing your first Thevenin could have used a bit more explanation as to why you can ignore it. Really well done though. I like your idea of mirroring the image so you're not writing backwards. :-)
Samwhell he ignores it because it’s an open circuit and no current flows there
7:42 That's not quite right. When you have two impedances (like resistors) in parallel, you would say that the "path of least resistance" is through the smallest impedance, and yet the current flows through both. The actual reason you can ignore the shorted resistor, is that you have a current divider built from the 40-ohm resistor and the 0-ohm short. By the formula of a current divider, the resistor gets no current and the short gets all the current.
i chilled tf down then i watched it for the second time ..now i understand it better ..thank you dude
Wow that last part about being able to use 2 of the values to solve for the 3rd is awesome. Thank you!
you make circuits a cakewalk, your awesome, keep being awesome.
Love these analysis things. Interesting to see how it all works out.
To find the Norton (short circuit) current, couldn't you just short A to B in the thevenin circuit, then find the current there? 80v/8ohms = 10A
So real world use age. Let’s move on to line load impedance matching, or implementing the termination resistor, same thing remember that. Great stuff keep it up.
So if it becomes too difficult to find the equivalent resistance, but you can find the voltage and current buy Noodle and mesh analysis, you can calculate the resistance from those.
Very clear and comprehensive review! Thanks for making this vid 👍🏼
Why does i2= 4A?
Is it because:
Mesh currents are made for every non inclusive loop. So there would be two non inclusive loops for this circuit. Loop 1 would contain the 20 ohm and 10 ohm resistor. Loop 2 would contain the 40 ohm and 10 ohm resistor. Because Loop 1 has only resistors in series, the current around loop 1 is a constant 4 A. (**** note: the 10 ohm resistor is in both loops)
Is my reasoning correct?
Hello!
Thanks for commenting! This isn't entirely right. A loop current can be assigned to any of the smallest closed loops in the circuit. This means loop one consists of the loop with the voltage source, 10 ohm resistor, and 40 ohm resistor. Loop 2 consists of the current source, 10 ohm resistor, and the 20 ohm resistor. The current in any element is then the sum of all the loops currents that have the element in the loop. Since the current source only exists in loops 2, the current in the current source is the same as the second loop current. THis is why the second loop current is the value of the current source.
I hope this helped!
~Andrew
because anything in series the current is the same across
Hi Guys, not sure if you still answer questions in the comments or not, but I was wondering. Why isn't IN 14 amps(the 4 amp source plus the new i1 of 10 amps)? Doesn't KCL say that all currents must equal at a junction?
he looks so nervous lol
Not just he looks, really is.
but the dude did his thing
Omg I was just about to say that. Well, he actually delivered.
but he really did his thing perfectly imo
😂😂👌
that's a really unique way to prepare a video.. very well done. Useful video aswel. Thumbs UP!
Finally clicked how this all works in this video XD Thanks man
nice methode presenttion and very clear. good luck
this guy teaches way better than my college professor
Thank you, you helped me a lot,,, I request you from Iraq
what a fantastic explanation! Great video :3
Huge respect for being able to write all that backwards
He did not, lol. It's on glass and the video is simply mirrored horizontally after filming.
Great teaching skills that takes me back to my high school days, my question is when do we use Norton and Thevenin theorems in real life application ? in Small-signal modeling for example I'm I right ? because in real life there is no such Norton souce of intensity (pure Amperes) or Thevenin source of potential (Pure Voltage) but the source is always a hybrid Source (Volt + Ampere)
Well the real life is always a little different from the theoretical. We use Thevenin and Norton models in small signal applications, yes. We also use them when providing loading for complex systems (i.e. a speaker is 8 ohms , this is the thevenin impedance). As for your question about the different sources, the models differ depending on the capability of the source and its loading. If voltage is maxed out but we can still provide all the current we need, use a voltage source model. If the opposite is true, use a current source model. If there is a combination of limiting factors, you need to be creative to get the same type of source you need. It is all about exploiting theory to closely match the reality, which differs from situation to situation.
~Andrew
Thank you so much, sir. You made it seem so easy.
Clear as day! Thanks!
awesome video, helped a lot. I understood the concept clearly
thanks for your generosity
When finding Vth at 6:03 why did you only multiply it by the 40 ohm resistor and not add the 10 and 40 ohm resistors together?
Thank you
I really appreciate your work :)
Why isn't the current through the 40 Ohm resistor i1 + i2 and just i1? What happened to i2?
I think because the i2 current was eliminated through the mesh analysis.
Nice video! Thank you.
This guy writes backwards like a champ!
Thank you so much. This helped me
Thank you very much! Very good video, really helped me!
thank you for the explanation .. I understand now
Thank you. How about the 10 ohm resistor and 40 ohm resistor?
if 10 and 40 ohm resistor are in parallel than the current should be divided?
You seem awfully nervous but have explained the theorems better than most channels on yt. Be confident!
Holy shit this actually helped me get it. Thanks a lot!
Why didn't we take the 20 ohn resistor for Rth? Because no current flows through it?
Legend
Why is I(N)= 10 ...but not I(2)+I(1)= 14
Wow thank you! Awesome explanation!!
Nicely done.
why does he didn't subtract the 20omhs resistance when he's searching for the voltage between A and B?
It's outside the test points the test points are the same 2 nodes as across the 40-ohm resistor. With the same nodes, they are by definition in parallel and thus must have the same voltage potential.
I was hopeing someone could explain. In the first example It's not clear to me why the R_thevenin is equal the equivalent circuit of 10ohms parrallel to 40ohms as opposed to in series with.
I guess I am kind of confused about how we are supposed to find an equivalent resistance after we eliminate our sources.
Thanks for commenting! When we eliminate sources we replace current sources with an open circuit and voltage sources with a short circuit. We then look for the equivalent resistance BETWEEN points A and B. In this case the 10 ohm and 40 ohm resistors are in parallel between A and B when sources are eliminated.
Also remember that if you are having trouble finding that equivalent resistance, you can instead find the open circuit voltage and short circuit current between A and B. Then R_eq = (V_open)/(I_short).
hi, thanks for the video. so i think that 10 and 40 ohm is in series. they have the same current I !!!
Hi There. Thanks for commenting! As, @Zac Sutton said above, we need to turn off all of the sources and look at the resistances between A and B. When we do this, we see that the nodes the two resistors are connected to are one and the same, implying that they are in parallel. The current in them may be the same, but this would be by coincidence only, not because of they are in a series configuration (they are not).
When you leave A you can take two paths to get to B. You go straight or turn left. At this point you should notice tge tips and tails are touching for bith resisters making them parallel.
Instead of doing the meshes you could have done the 4 amps that are above multiplied per the 20 ohm resistor. It is easier and faster :)
Super helpful, good explanation
134 people think the 10 and 40 ohm resistors are in series, not parallel
I am having a tough time understanding why 4A didn't return to 40 Ohm along i1..
should the loop analysis of mesh 1, treat the voltage source as positive because it is going from neg to positive
this was incredibly helpful
This garbage is easy, but it is the type of math that, if you have not studied it for months, you have to review it again. Good job. I like how you did this video. What did you do? Were you standing behind glass and writing with some kind of bright marker? It was really done nicely. I like how you told us that the 8 ohms was both the Norton and Thevanin equivalent resistance.
Frank
The Master of Science
Very nice video thankyou
very nice explanation
its so much easier to note I Short Circuit 'Isc" = Vth/Rth
When solving for for Norton, when the current flows around from B and onwards, is it not possible for the current to flow upwards throw the 40 ohm resistor. All three branches have resistors?
Shouldn't the 8 ohm and 80 V battery of the simplified circuit be in parallel not series? There's a difference yes?
A very helpful video.
awesome video thanks!
hi, what if you are not given that 4A and instead just given another resistance?
In this case you would simply find the thevenin resistance the same way, just ignore the step with the current source. if you have a more targeted question we will do our best to explain it better!
~Andrew
when doing the first loop, shouldn't it be -60+10(40+i1)+40i1?
omg how are people able to study electrical engineering and not able to figure out they flipped the video so the writing gets corrected
WHY DO YOU CHOOSE I1 TO SOLVE FOR THE THEVENIN`S VOLTAJE?
Why did he put the voltage as negative 60 instead of positive?
why was the 40 ohm resistor used in getting the thevenin voltage?
Hey thanks for the video man good explanation
This is great great review, thank you :)
thank you, I think you've tried hard to write in an opposed way, which deserves to be respected!
can we just simply get In by diving Vn/Rn ? which is the same answer, 10 ohms
Thank u for this video , but why the sign of voltage 60 is negative?
could be either or if you keep the signs consistent.resistors usually drops while to voltage negative to positive is a voltage jump if you multiply everything by a negative would fix the sign.
Why the 20 ohms resistor is not included tho? Shouldn't it be 20 times 4A?
what if i put a resistor between points a and b? can i short those? (eliminating the resistor) and when i draw the norton equivalent should i draw the resistor too between points a and b?
Why is i2 4A? Wouldn't the 60v source create some further current through there? Like if you follow the path of the positive charge from the voltage source it first has the option of no resistance vs 10 ohms, so at least some current should go to i2. Or do voltage sources treat current sources as open circuits?
The independent voltage source will apply the same voltage regardless of the current source. Hope that makes sense.
very clear. thank you
Thanks a million!
why is the upper part not included