Electrical Engineering: Ch 4: Circuit Theorems (17 of 35) Thevenin's Theorem Ex. 2
ฝัง
- เผยแพร่เมื่อ 7 ก.ย. 2024
- Visit ilectureonline.com for more math and science lectures!
In this video I will ffind i=? of a load resistor by converting a linear circuit with 1-voltage and 1-current source to a Thevenin's circuit.
Next video in this series can be seen at:
• Electrical Engineering...
![[LIVE] MCHOICE & MINT AWARDS 2024 at TRUE ICON HALL, ICONSIAM](http://i.ytimg.com/vi/wMewUeT5nbY/mqdefault.jpg)
all my classes have been moved online due to corona virus and this one man is singlehandedly saving all of my grades thank you so much, these videos are amazing.
van bizen is better than all my profs combined
I used source transformation to find Vth and it worked very well
Good input. 🙂
Same here
at 1:08 when you combine the resistors, why is it not possible to combine the 4 ohm and 6 ohm resistors as they appear to be in series?
From the 2 end point connectors, (let's label them A (top) and B (bottom)), there are two seperate paths from A to B. Through the two 6-ohm resistors OR the one 4-ohm resistor. That is what makes them in parallel.
Sir, When we're finding Vth, couldn't we use V - V(the voltage that across 6 Ohm resistor)? I mean:
15 - 6 x i_3
And to find the i_3, we could use i_1 + i_2 = i_3.
Since i_2 is given as 2 A, and i_1 can be calculated as (12V - 15V) / 6 Ohm which is -0.5A,
i_3 must be equal to 2A + (-0.5)A = 1.5 A.
Now we have i_3, so:
15V - 6R x (1.5)A = 15-9 = 6V which agrees as the result that you've found but I'm not sure if I get this result in a right way.
Thanks for your time that you've prepared these extremely helpful videos.
There are often different ways in which you can solve the same problem. All are just as valid. A good check is to see if you get the same answer.
He's better than my own professor.
explanation is superb
Awesome video sir! Thank you!
keep it going.
I have solved for (i_1) by k.v.l And it was 2 ampere this means i_3 equals 4 ampere And these values are not consistent with the value of v=15 volt
very well done job professor, thank you :)
Many thanks!
you need to explain more on the node method, confusing.
I wish this dude taught at my university.
What university are you attending?
In the first thevenin example using nodal analysis to find vth, you assigned the node as vth. Here you assigned it as node V and you used it to find the VTH. I'm kinda confused cause I taught when you found voltage at node V, it is already the vth since voltage in parallel circuits are constant. Thank you.
Juan Carlos Raymundo bro, can you please elaborate ? Because I have the same confusion.
since the voltage get divided when it is series so we can not simply say that voltage across 4k resistance is the whole voltage as it is also divided into 6k resistance which is in series with 4k ..thats why we use voltage division rule to find out how much voltage resistances in series are sharing ..
In this example, the 6ohms resistor makes V not parallel with Vth whereas in the first example the 1ohm resistor is part of a open circuit.
Sir pls solve thevenins with only dependent source
aren't all the resistors in series? for when you're simplifying the thevin resistance? I'm confused
Was going to ask the same question.
Eliaziim c very basic mistake, I am moving to other channel.
no, I think the 6+6 and 4 ohms are in paralle, which means 1/((1/12)+(1/4))=3
@@qiangou1400 why? I think the 6+6 and 4 ohms are in paralle, which means 1/((1/12)+(1/4))=3, you shouldn't treat it as a independent circuit, it is a part of a circuit.
@@qiangou1400 they are in parallel because they have 2 common points. You should move yes, but to study the basic lessons
Physics 🔥🔥❤️my dream is to become scientist in physics
Thanks
Studying the material in these videos will put you on track for that. 🙂
@@MichelvanBiezen I will do that 💪💪❤️❤️
The 6 and 4 oh resistor, arent they in connected in series?
yes, they are, found my mistake
@This is my jam I was confused too so I searched up other videos. Thevenin resistance is actually the resistance in respect to the load resistor, going from node A to node B. Let the node between the 6 ohm and 4 ohm be node A, let the node connected to ground be node B.
You can go from node A to B via the two 6 ohm resistors, or just the 4 ohm itself. This makes the 4 ohm resistor actually parallel to the two 6 ohm resistors, not in series.
Here is his other video, it might make more sense: th-cam.com/video/CdRjhqii080/w-d-xo.html
@@ericliu7705 Thanks for clearing that up, I was really confused at first
I have a question: Is Vth always going to be the voltage going through a resistor thats in series with the Vth gap that was made?
Please sir I don’t understand how you got i_3 as V/10 , Explain it to me I’m getting confused. 6 ohm and 4 ohm are in parallel, right?
The 6 ohm and 4 ohm resistor are in series. Thus the total resistance along that branch is 10 ohms. Since current = voltage / resistance. I3 = V/10
They're in series because the junction between the wire of the 6ohm resistor and 4ohm resistor is no longer a node (the current isn't flowing to the end of the wire)
Excellent!!
Love it! Thank you!
You are welcome.
Professor in this example can we write KCL for those two nodes . In one node the voltage is V and in node next to it is Vth . And for current I3 we can write V-Vth / 6 . Can we do in this way or not ?. ( I know that this solution that you have given is in more easy way with voltage devidor but I want to know that maybe this exercise can be solve by doing 2 KCL )
You should always be able to use KCL. (It always works). Try it and see if you get the same answer.
@@MichelvanBiezen yes is the same answer . ( I like to solve those problems with KVL but I realised that also KCL is very powerful tool ) thank you
Usually you want to use both in the same problem.
@@MichelvanBiezen yes is completely true
@@MichelvanBiezen thank you for your advice
good explaination
@ 1:11....Why is the 4ohm resistor in parallel with the two 6 ohm resistors? When you remove the load, it appears to be in series ( one current path)
There are 2 paths for the current from point A (top terminal) to point B (bottom terminal).
@@MichelvanBiezen Thanks. I see now.
Professor, please answer to Eliazim's question!!
Sir how can we assume I3 current is leaving the node, and I1 is entering
We can't be 100% sure, but the left branch has a voltage source with the positive end at the top, and the middle branch has a current source pushing current upward.
How is the equivalent resistor of 12 ohms parallel to the 4 ohms ?
you got the answer? PLEASE TELL ME
IL=0.41A
Exactly!!?!!!!????
can you make more questions but for the cases of dependent sources and independent+dependent sources?
Sir, I'm totally confused . In 1st example you took the Node as Vth and in this example you used voltage division rule for Vth . Why ? Please answer ASAP
Note that in example 1 (video # 16) no current will flow through the 1 ohm resistor, thus Vth is the same as the node voltage.
@@MichelvanBiezen y did multiplied *30 there for vth
Can you please kindly tell me why you the voltage in the node 1 is 15V which is greater voltage source, ia that because of the 2nd source im bit confused, also at the end you multiplied the voltage by 4 ohms over 10 ohms is that V×(R2÷R1+R2) law ? Thanks in advance, i find your videos helpful and easy to understand thanks again.
The voltage at node 1 is found by using Kirchhoff's current rule at that node. Current in = current out.
Michel van Biezen thanks for your clarification
hello
Please why was the node above 4ohms not solved using node analysis
Giving us two a simultaneous nodes equ instead of using voltage divider
There are many techniques available to solve circuits. When we use one of them, that doesn't mean that another technique could not have been used instead. We try to cover all the different techniques.
Hi, is it possible to have used nodal analysis to find voltage at 2nd node?
I can not see when you write while you explaining. oh, it takes a time to hanging on. thank you for your videos.
Yes, that is a problem that we have been working on, by working from right to left. You can also skip ahead to the end to see what is written.
how do you find R load, or is it given?, what do i do when i am given a cirvuit that has no load resistor given?
The load resistor is usually given, as it can take on any value. The load resistor will indeed affect the rest of the circuit once it is connected. It is often the case the load resistor is not given and that is why we use Thevenin's Theorem to solve the problem. We have other examples like that.
Thank you so much sir, I have a question though.
Why did you add the 6 & 4 ohm resistors during the nodal analysis step when finding Vth? I don't understand how they are I series...
Are you referring to the step where I calculate the Thevenin voltage? (Vth = V (4 / (6 + 4)) ? That is the method used to determine the voltage used in that branch. (voltage divider technique)
@@MichelvanBiezen Yes, thank you. I'll definitely research more on the voltage divider method.
@@MichelvanBiezen sir , why we use this equation
great sir..!
Can anyone tell me, when finding R_th, why the 6ohm resistors are parallel with the 4ohm resistor
In relation to that, if that was parallel, why is the 6ohm resistor in series with the 4ohm resistor in the V_th part. Many thanks
Going from the top terminal to the bottom terminal the current has the choice to go through the 4 ohm resistor OR the 6 ohm resistor. By definition that places them in parallel.
@@MichelvanBiezen Thanks for the quick reply! That makes sense. I guess I was just confused because I was looking at it in the wrong direction. Would it be correct to say then that when finding R_th, we must view the current as coming from the terminals A or B?
That is correct.
god bless u
Why did you multiply to 30?
Nice sir
Sir, are there any other ways to find Uth?
Sir. I dont understand how Vth being calculated. As you used Vth=15V x (parallel). Why you used parallel? Please help me sir. Thank you
First you find V at the node using node analysis method which is = 15V Then you use a voltage divider method. (Vth = voltage drop across the 4 ohm resistor)
How about if we assume the load resistor to be the 6 ohms in the middle ,,how do we go about doing it?
You work the problem the exact same way. You remove the load resistor and follow the same procedure.
@@MichelvanBiezen ..Thank u sir for replying to my request....can you explain it further because i face difficulties when trying to find the V(th) for the similar problem. Oh can i get your email so i can send it to u so u can give me an advice on how to proceed please?...Thank you in advance sir.
We are very busy (multiple jobs) and are unable to respond to such requests from the many viewers.
Thank you, very much.
You are welcome!
sir I just wanted to ask again.. why is the Vth is not 15V ?? why do you need to do the 15(4/6+4) to get 6V ??
Think of it as a voltage divider. Part of the voltage drops across the 6 ohm resistor and part of the voltage drops across the 4 ohm resistor. Vth = the voltage drop across the 4 ohm resistor.
Michel van Biezen sir .. does this mean that when you open the 2A the two 6 ohms are in series and parallel to 4 ohms.. and when you added the 2A again the two 6 ohms are in parallel and series to 4 ohms?? why sir??
Sir pls tell why the 4ohm resistor is in parallel
If you take the 2 - six ohm resistors and move them to the left branch (the vertical part) it will look exactly like the parallel circuit.
K now i get it.
You are an awesome teacher.respect from india.
Hello Professor, hope you're doing well. I have a random question: let's say that the 6-ohm resistor above the 12-volt source was removed from the circuit and nodal analysis was used, as you showed. How would you represent i1 in this case? The numerator would be "12 - V" but there is no resistor to divide it by? It seems like an easy answer but I am struggling with the basics. Thank you.
The volt of that node will automatically 12V if no resistor is there
Sir, you did a proof of thevenin's theorem but it is not the proof of the method that you used here. Because in the proof, there must be a current source (or maybe a voltage source) outside of the linear part. But there is only a resistor outside of it. I want to see a proof of this method.
We didn't give a "proof" of the theory. We just illustrated how to use the theorem.
@@MichelvanBiezen But you have a video of the proof
01:26 why are they parallel?
Going from A to B (the two terminals on the right side), the current can either go through the 4 ohm resistor or the 2 - 6 ohm resistors. When the current has two paths they are in parallel.
thank you very much though i think i will not make it to phasors until my exams the videos seem too many
You'll get there. 👍
@@MichelvanBiezen I appreciate the encouragement! If I make it, I’ll let you know and it will be because of you🙂
Keep it going.
The two 6 ohms and 4 ohms resistors are in series?right?
When the load resistor is removed, yes.
but why in the video you calculate it in parallel?
Dear sir, is i1 direction arbitrary? Because thought current flow from high to low potential.
When there is more than one source, it is not always easy to tell which direction the current will flow in any particular branch. But yes, current flows from high potential to low potential.
why are 6ohm resistors in series?
When the 2 A current source is removed and replaced by an open, the two six ohm resistors will then be in series.
does norton theorem and thevenin theorem yield the same answer for the same circiut
There are a number of methods to solve circuits like this including Norton and Thevenin. They will all give you the same answer.
prof. van i dont get what you do in that voltage divider u tell
The voltage drop is proportional to the ratio of the size of the resistor and the total resistance of that branch. Total resistance is 10 ohms and the 4 ohm resistor will therefore have a (4 /10) * 15 V drop.
There are also a number of circuit videos in the physics videos.
Why you use 4 ohm not 6? I dont really get it im sorry im an idiot😢
Don't call yourself that. It is normal not to see something at times. (It happens to me all the time). The 4 ohm resistor is parallel to the gap between Vth and 0 V, therefore the voltage across the gap is the same as the voltage drop across the 4 ohm resistor.
@@MichelvanBiezen hi sir may I know for the voltage divider, why your Rt is 10ohm? Why you didn't include 6ohm as in your Rt?
sir.. why did you take 4 to be divided by 6+4 ?? in the last part when you're getting the Vth?
That is using the "voltage divider" concept. (What is the part of the total voltage drop across the 4 and 6 ohm resistor that is dropped across the 4 ohm resistor?)
Michel van Biezen thank you sir!
Sorry sir, but on finding your equivalent resistance, you said the two 6 ohm resistors are in series. How are they then in parallel to the 4 ohm resistor, since the same current flows through all resistors upon open-circuiting the load?
Sorry I may have been a bother, I can't just help but be inquisitive about things I don't quite understand
The best way to determine if resistors are in series or in parallel, is to see if there is only one path, or mulitiple paths for the current. In the second circuit drawn, all the current flowing through the first 6 ohm resistor MUST also flow through the second 6 ohm resistor, therefore they are in series. The current flowing from a to b, has a choice. It can flow through the 4 ohm resistor OR through the two 6-ohm resistors. They are in parallel.
@@MichelvanBiezen where can point a and b be on the circuit that the current gets divided, it was open circuited at that branch if I recall correctly.
Look at the first sketch under : "To find Rth". Going from a to b there are 2 paths.
I dont see any labelling for a or b....can u please indicate by some means sir 🥺
Sorry sir, can you do an example with dependent sources, I`m getting confused
We have examples with dependent sources on the channel.
when calculating the thevenins resistance isnt the circuit in series
oh got my mistake
Thanks anyways
Shouldn't it have been (12v - v / 6) for current I₁
No, you do need the parentheses. I = (12V - V)/6 (drop in voltage divided by the resistance).
No I mean shouldn't the 12 have the voltage unit in front of it? Your equation has 12 - v / 6. Great series by the way.
I usually omit the units to give the equations a cleaner look.
I think rth should be 16 ohm
No, the video is correct. Thanks for checking.
@@MichelvanBiezen yes i just realised. But thnks for replying
How is it 15? 6ohm + 6ohm + 3ohm?
Joss
✌
Dear Sir ! Why we are multiplying the equation with 30 , what is reason ! kindly reply as i'm getting confused ! thanks
Notice I multiplied both sides of the equation by 30 in order to eliminate the denominators. That makes it easier to solve the algebra equation.
Thank You So much Sir ! ya I got it now !
@Michel van Biezen Hi How can I call You to say by phone ????
We don't have the time to respond to viewers by phone, but we try to respond to question in the comments.
Ooh
hey Sir could i please have your email i have a burning circuit analysis question or i can send it here
If you send it here we try to answer the questions, but we can't get to all of them.
@@MichelvanBiezen which channel should i use to send
so many unnecessary steps...
im confused, why is the equivalent resistance of the two 6 ohms equal to 12 ohms? are they not in parallel?
or, does it mean we neglect other branches that are not resistors when we're determining whether resistors are connected in parallel or in series ?
When you are finding the Thevenin resistance, current sources are removed which places the 2 six ohm resistors in series.