4:40 You mention that for any current entering the node will have a negative value, but the next term you use (representing the current passing through the 120 resistor) also goes into the node, but you left it as a positive addition. How come?
Oh holy fuck all semester I've been working on this and I've watched thousands of tutorials on it but I didn't get jack shit until I watched this video. Fuck, you saved my life man I'm actually in tears
KCL says if a current enters a node it equals it when it leaves, THEN shouldn't it be : 0.2A +i1-i2=0? i1 is = v-12/120 i2 is = v/45 Please someone explain it to me! i don't understand why he did it like that!
In the node voltage method it was said that current entering a node would be negative but in the calculations it was added for the first fraction ((v-12)/120). Why would it be added instead of subtracted in this instance?
Sum of input currents are equal to output current(s) i1+i2=i3 (i1->over 120ohm....i2->current source... i3->V point to ground-known voltage) i1=(12-V)/120 i2=0.2 i3=V/(20+25)
why do we replace the voltage source with short , and current source with open ? Also, why did he add 20 Ohms to 25 ohms to get 45 Ohms, they are both connected in parallel right? I don't see how they are connected in series. I really want to understand it! thank you in advance.
An ideal voltage source has no resistance to current, and a current source ideally has infinite resistance of current. For the sake of calculating a Thevenin equivalent, you assume that the voltage and current sources are ideal so that you can treat them as shown in the example. Those resistors are not in parallel-- there is one wire that enters the 20 ohm, the same single wire exits the 20 ohm into the 25 ohm, and then continues to the rest of the circuit. Since the resistors are on one continuous wire, they are in series.
@@ozymandiius since the 25, 120, and 20 ohms resistors are in series, why not calculate the voltage divider like this : 9.82*20/(120+20+25). Instead of what he did 9.82*20/(20+25).? Isn't the 120 ohm resistor is also part of the series of resistors in the circuit? Why not include it.... Thank you again!
Awesome video! There is just one gap that had me a bit confused. When calculating the resistance toward the end you added the 120 and 25 ohm resistors. These would be in series because the 20ohm resister should be removed at this step. I believe 145 ohms is the Thevenin equivalent resistance and 20ohms is the load.
So, you need to remember that nodes A and B are your start and end points, and that all of your equivalents are from the perspective of A and B. Basically, if there was something between A and B, what would that thing see? Also, 145 and 20 are in fact parallel, he just forgot to add nodes A and B to the drawing.
What Jayson said makes sense to me. I was confused why he was merging the RL and Req as well. In the thevenin equivalent circuit, my teacher wants us to show the RL and Req seperate.
Overall a good video. There is an error around the 3 minute mark. You mention that you are adding the two resistors because "they are in parallel," but they are in fact in series since they share the same node. But I think you knew that since you used the proper equation. Perhaps making a little memo portion in the video would help. Thanks though.
firs thank you for all... and I want to ask you please,when we calculate Rth ,we dont use RL=20 ohm.. how you calculate thisRth=17,6 ohm .I think Rth =145ohm..isint it?
Based on the diagram, the left side of 120 Ohm and left side of 25 Ohm are connected with a continuous wire nothing in between so they are connect in series. After you add them up to 145 Ohm then the 145 Ohm resistor is now parallel to the 20 Ohm, just imagine they are connected to make a loop. Sometimes, it is hard to see but if we do lots of practice then it will become easier.
Why he uses in the end the voltage divider? (in order to use the voltage divider you have to assume that there is a small current that leaves from point b)
After a little research I've found the answer. Actually, according to Thevenin theorem, the Thevenin voltage is measured considering a and b terminals open so there's no current flowing from terminal b so using the voltage divider is appropriate.
why not 12 - V instead of V - 12? I thought the arrows started from the voltage source so that would be the start of the Ohm's Law. Correct me if I'm wrong.
2kΩ. Is not short-circuited if you put hypothetical voltage source between the gap terminals you will notice that the current is splitted in two parts between the 1k and 2k resistance!!!
Why do you write V-12, which V, why 12, why over 120Ohm , why can you just assume a ground wherever, why can you just add the last two R without having the whole system changed when you put a ground
Current division works in that way. Using voltage divider, one must use the resistor of which voltage he/she would like to know, as the numerator to the equation.
4:40
You mention that for any current entering the node will have a negative value, but the next term you use (representing the current passing through the 120 resistor) also goes into the node, but you left it as a positive addition. How come?
It's because i2 = (12-V)/120, so since you're subtracting i2, you get -(12-V)/120, which is +(V-12)/120, which is what he writes.
Oh holy fuck all semester I've been working on this and I've watched thousands of tutorials on it but I didn't get jack shit until I watched this video. Fuck, you saved my life man I'm actually in tears
Bruh, useful af. Methods are explained simple and clean. Helped a lot
I'm sorry but the intro music almost had my ears bleeding (this is exaggeration)
The screeching of the marker is equally as bad
at least it was groovy
KCL says if a current enters a node it equals it when it leaves, THEN shouldn't it be :
0.2A +i1-i2=0?
i1 is = v-12/120
i2 is = v/45
Please someone explain it to me! i don't understand why he did it like that!
In the node voltage method it was said that current entering a node would be negative but in the calculations it was added for the first fraction ((v-12)/120). Why would it be added instead of subtracted in this instance?
I think this was a great example. Very clear and easy to follow. Thank you!
Sum of input currents are equal to output current(s)
i1+i2=i3 (i1->over 120ohm....i2->current source... i3->V point to ground-known voltage)
i1=(12-V)/120
i2=0.2
i3=V/(20+25)
law of currents is i1 + i2 = i3, i1 = 0.2A. So +0.2+i2-i3=0 not is -0.2+i2+i3=0
Why didn't you put your ground between the 25 and 20 ohm resistors on the bottom right?
You said currents that enter the node are with minus sign, but you wrote (V-12)/120ohm instead of -(v-12)/120ohm though it enters the node.
As my opinion, V=12-(120i1) originally, so i1=(12-V)/120
After adding the minus sign, i1=(V-12)/120
Please tell me if I am wrong
@@monkeylin7256 you are right, it's what he did. I think he should have said that so nobody would get confused
Great explanation, way better than most on youtube
Is he drawing backwards?
He's probably writing normally, the total video is just mirrored
@@tommysaeys mindblown lol
best example of thevenin yet, and i finally know an actual use case for it too
Can someone please clarify why he is using RL in the calculation Rth? I thought those were kept seperate...
Why cant you make the node directly under the 20ohm resistor your ground ? I tried this and got a Vth of 5.14volts
Aren't the 145 and 25 resistors in series?
why do we replace the voltage source with short , and current source with open ?
Also, why did he add 20 Ohms to 25 ohms to get 45 Ohms, they are both connected in parallel right? I don't see how they are connected in series. I really want to understand it!
thank you in advance.
An ideal voltage source has no resistance to current, and a current source ideally has infinite resistance of current. For the sake of calculating a Thevenin equivalent, you assume that the voltage and current sources are ideal so that you can treat them as shown in the example. Those resistors are not in parallel-- there is one wire that enters the 20 ohm, the same single wire exits the 20 ohm into the 25 ohm, and then continues to the rest of the circuit. Since the resistors are on one continuous wire, they are in series.
Thank you so much! Your explanation helped me understand!
@@ozymandiius since the 25, 120, and 20 ohms resistors are in series, why not calculate the voltage divider like this : 9.82*20/(120+20+25). Instead of what he did 9.82*20/(20+25).? Isn't the 120 ohm resistor is also part of the series of resistors in the circuit? Why not include it.... Thank you again!
Thank you so much I was looking for something like this
hi wouldn't the 120 and 25 be parallel ? why did you add them
thanks for helping me understand why you disregard a resistor/ path !
Current takes the path with least resistance. The path with more resistance is disregarded.
Awesome video! There is just one gap that had me a bit confused. When calculating the resistance toward the end you added the 120 and 25 ohm resistors. These would be in series because the 20ohm resister should be removed at this step. I believe 145 ohms is the Thevenin equivalent resistance and 20ohms is the load.
So, you need to remember that nodes A and B are your start and end points, and that all of your equivalents are from the perspective of A and B. Basically, if there was something between A and B, what would that thing see? Also, 145 and 20 are in fact parallel, he just forgot to add nodes A and B to the drawing.
What Jayson said makes sense to me. I was confused why he was merging the RL and Req as well. In the thevenin equivalent circuit, my teacher wants us to show the RL and Req seperate.
Overall a good video. There is an error around the 3 minute mark. You mention that you are adding the two resistors because "they are in parallel," but they are in fact in series since they share the same node. But I think you knew that since you used the proper equation. Perhaps making a little memo portion in the video would help. Thanks though.
There is a sign mistake while writing the Kirchoff current law (4:43), otherwise nice and clear.
firs thank you for all... and I want to ask you please,when we calculate Rth ,we dont use RL=20 ohm.. how you calculate thisRth=17,6 ohm .I think Rth =145ohm..isint it?
same question here
I think that 20ohm is not the load...load is connected across ab
@@sohambhattacharya8975 yup! exactly
OH MY GOSH! That was a bit tough!but with your explanation it became much easier:)
At 3:54, why is it that 120 and 25 is series and they together are in parallel with 20 ohms resistor?
That’s what I’m saying!
I think this guy is wrong
Based on the diagram, the left side of 120 Ohm and left side of 25 Ohm are connected with a continuous wire nothing in between so they are connect in series. After you add them up to 145 Ohm then the 145 Ohm resistor is now parallel to the 20 Ohm, just imagine they are connected to make a loop. Sometimes, it is hard to see but if we do lots of practice then it will become easier.
pretty crazy how he draws stuff inverted for the viewer. Thanks
during the rTh part, why aren't 25 ohm and 20ohm counted together instead of 25 and 120?? aren't they in series as well??
You look like McLovin lmao
better lookin than McLovin!
hahahahahha
aaaaaaaaand now im reading about how someone tried to use mclovin's ID as a fake instead of studying circuits like i should be, thanks! 😂😂
is the correct answer for vth not 5.143??
3.2
Why he uses in the end the voltage divider? (in order to use the voltage divider you have to assume that there is a small current that leaves from point b)
After a little research I've found the answer. Actually, according to Thevenin theorem, the Thevenin voltage is measured considering a and b terminals open so there's no current flowing from terminal b so using the voltage divider is appropriate.
How you do norton equivalent for this circuit
What's the Rth if there is no 0.2A current source (shorted)?
Great thank you so much for the awesome explanation!
Smooth and easy explanation. Well done Dan!
May you explain more about the current (V/45 ohm)? Why can't we just write as (V/20) as the current across the 20ohm resistor?
it is a voltage divider formula.
you can also write V/20 there is no problem but in your way, you have to use two unknown like Va and Vb
This is amazing! I solved that by using nodal analysis and mess analysis which was quite lengthy bt the VDL could be used I learnt from you..
why not 12 - V instead of V - 12? I thought the arrows started from the voltage source so that would be the start of the Ohm's Law. Correct me if I'm wrong.
ty v good video :] i finally learned about node voltage and was super confused by Thev but I know where to start now :]
how to solve step by step the 9.82v?
Thank you so much! This is really helpful :)
F-the haters. This has been the more useful videos for Thevenin's and the light board thing was actually just a great idea.
2kΩ. Is not short-circuited if you put hypothetical voltage source between the gap terminals you will notice that the current is splitted in two parts between the 1k and 2k resistance!!!
what if we're going to find the current across 20 ohm?
I believe thats why he did voltage division since the V=9.82 is the suppy voltage to that part of the circuit. So across the 20 ohm is 4.36V
Thank the online teaching gods for this video. Lifesaver!
Excellent video !!
This video could not be more Uconn
great explanation
Thank you so much it's nice tutorial
why is it 145//20 and not 45/120?
same question
bro u should know first how to calculate parallel resistance
Thank you for the lesson!
this man has a rip n dip shirt on. what a chad!
I always know I can trust a man with a cat in his pocket.
Great video but the sound of your marker is annoying 😬😬😬😬 but thank you! 😊
This dude can so well write backwards
Why do you write V-12, which V, why 12, why over 120Ohm , why can you just assume a ground wherever, why can you just add the last two R without having the whole system changed when you put a ground
very good to understand :)
Holy shit, that lecturer is amazing!
Is he writing backwards or is this some sort of magic
We use a Lightboard so it gives off that illusion lol
Great presenter!
I think your wrong about the 120ohms and 25ohms adding up to 145 because it's parallel, so it would be the 1/R equation.
Anakuya1 even though they seem as they are parallel they share 1 node exclusively therefore they are in series.
very nice video
This is incorrect the sum of the current have to equal each other meaning i1 plus i2 equals i3
yeah i think the same , two currents entering one leaving so sign should be the opposite
The tension divider is wrong, isnt it?
Is the Vin*25/45, bc Vin*rdown/rdown+rup.
Correct me if Im wrong
Current division works in that way. Using voltage divider, one must use the resistor of which voltage he/she would like to know, as the numerator to the equation.
Dan! What is in your pocket??
if 120 and 25 are in parallel, they wouldnt just be added together? did you mean to say in series?
120 Ohm and 25 Ohm are in series. When he says "Parallel" combination, he really means series parallel.
No theyre in parallel. If you draw the a and b points out you can see it.
did u forget the super mesh
Excellent
I'm more curious about what you're writing on and if you just flipped the video? It ends up being a distraction for me lol
You're a legend
this is so goood!!
Intro music is far too loud!
Why did you use nodal analysis without explaining what is is?
Tanmirt
love you from india
Thanks sir ji
thanks a lot..pleasure!!!!!!!
I got 3.29V across the 20ohm ffs
lmao rip n dip
I LOVE YOU
100th comment!
you just can't be a teacher. I can't get a single idea how you solved this.
stop reading the comments and watch the video. This can't be that hard.
@@nathanoosterhuis6232 Maybe not for you; but it is definitely hard for me.
this guy gives me a super bad type character vibes
Wrong
shouldn't you ignore that 20 om resistor when u r calculating the Thevenin resistance?
OH MY GOSH! That was a bit tough!but with your explanation it became much easier:)