For the second calculation , why are we adding the 20 Ohm to RT ' , because current takes the easy route , so there wont be current flowing in 20 ohm , since its connected parallel to IN' ?
No, that is not the case, IN' is not in parallel with the 20ohms due to 4 ohms you see there. Assuming the 4ohms was absent, then you claim will be true. At the point you should realize that the 20 and 4 are parallel connected hence 20 should be factored on the calculation for RT'
Okay, so we are asked to find the nortons resistance that is being viewed from ab. So you assume a voltage source between the terminal ab, the current that is produced by v is divided at the node where 8 and 6 are connected, hence they are not in series. Mind you the current that flows through 6 is same through 2, (series) . Don't make the mistake to view from the extreme right, with that 8 and 6 will be in series.
Also, now that I have a Norton equivalent of my multi-source project, I would like to see what the Thevenin equivalent is. If you were to explain it as well as you did here, it would build my confidence.
Sir you not add super position theorem in problem 1 and add super position theorem in problem 2 , how wo detect in which problem we add super position theorem??? Kindly clear me please 🙏
@@SkanCityAcademy_SirJohnThen why are these circuits solved separately unlike the problems in the first and second questions sir? And is superposition also used in solving thevenin's theorem problems as well?
@@oluwatobioyedokun5039 yes, you can combine superposition with thevevin or norton's theorem. the issue is, you can solve superposition theorems without involving thevenin or norton, however, we choose to combine them when the circuit looks complex. but if you can do it without involving thevenin or norton, then its fine.
Hi sir if i solve norton's theorem exactly how you taught us how to solve thevenin i'm still correct right i mean in terms of finding vth?? i don't know if you get my question.
I don’t know if my method of approach is okay, but I used; In = 3 + I (kcl) Using kvl; first loop 25 = 5*I + 20*I Solving this should result to 1 From kcl, In= 3+1 = 4
Sir in problem 2 , according to problem 1 if the current direction is downward 👇 we take it negative but in problem 2 both currents IN'and InN" is downward 👇 but you take it positive why sir ?
The issue is, in problem 1 we have current in both direction, so you can either choose one direction to be + and the other to be -, its not compulsory to choose a specific direction for positive or negative. But for problem 2, both are in the same direction, so you can either choose positive or negative for both of them.
Perfect explanation sir keep it up
Thanks so much, Clever. Where do you watch me from?
Thanks for the help. Very few examples out there with more than two sources. I needed this.
You are welcome
Thank you!
Thanks for the good work🙏.
Request: Could you make a video on Millman's Theorem.
For the second calculation , why are we adding the 20 Ohm to RT ' , because current takes the easy route , so there wont be current flowing in 20 ohm , since its connected parallel to IN' ?
No, that is not the case, IN' is not in parallel with the 20ohms due to 4 ohms you see there. Assuming the 4ohms was absent, then you claim will be true. At the point you should realize that the 20 and 4 are parallel connected hence 20 should be factored on the calculation for RT'
good job on electronics
Thanks so so much
clear explanation
great!!!
Thanks so much for watching
Please record a video on Reciprocity Theorem
Thanks Again❤
You're welcome 😊
Good work but try explaining some parts in detail,,Kindly🙏🙏🙏
Thank you very much for your comments. Concern noted.
Thanks for watching
Please I have a question......
In example 1, why did you choose 2 and 6 to be in series over 8 and 6 when solving for the Norton's Resistance?
Okay, so we are asked to find the nortons resistance that is being viewed from ab. So you assume a voltage source between the terminal ab, the current that is produced by v is divided at the node where 8 and 6 are connected, hence they are not in series. Mind you the current that flows through 6 is same through 2, (series) . Don't make the mistake to view from the extreme right, with that 8 and 6 will be in series.
@@SkanCityAcademy_SirJohn Well understood
Gratitude
You are most welcome
Also, now that I have a Norton equivalent of my multi-source project, I would like to see what the Thevenin equivalent is. If you were to explain it as well as you did here, it would build my confidence.
Kindly check the playlist on applied electricity you will see lessons on thevenins theorem
Thanks alot dear am helped
You are most welcome. And thanks for watching.
the second example what if the sources are more that 2 like 4 can i still use the superposition
Yes, you can still use superposition
@@SkanCityAcademy_SirJohn so i will let out alone all the sources and add them together to get IN
Yes please
Thank you ❤❤❤
You are most welcome
Thanks
Sir you not add super position theorem in problem 1 and add super position theorem in problem 2 , how wo detect in which problem we add super position theorem??? Kindly clear me please 🙏
Superposition theorem was used for all two questions, superposition theorem is used when you have more than one source in the circuit.
@@SkanCityAcademy_SirJohnThen why are these circuits solved separately unlike the problems in the first and second questions sir? And is superposition also used in solving thevenin's theorem problems as well?
@@oluwatobioyedokun5039 yes, you can combine superposition with thevevin or norton's theorem. the issue is, you can solve superposition theorems without involving thevenin or norton, however, we choose to combine them when the circuit looks complex. but if you can do it without involving thevenin or norton, then its fine.
@@SkanCityAcademy_SirJohn thank you sir
Sir, why did you say I1 = positive 0.75 instead of -0.75, as it is moving in the oppositive direction.
No please i1 and I joins up and move in the same direction as IN hence it's positive, and not negative.
Hi sir if i solve norton's theorem exactly how you taught us how to solve thevenin i'm still correct right i mean in terms of finding vth?? i don't know if you get my question.
That's correct, but you still need to do it how I taught you
pleae can you just use the norton theorem procedure alone without SUPERPOSITION
Yes, you can.
On problem number 1 how to get the load resistance (Rl) pls answer my question thank you:(
The load resistance is already in question, it's the resistor you removed before starting the whole solution process.
@@SkanCityAcademy_SirJohnI checked it's 2ohms not 4ohms how to get 4ohms Rl
How to get 4ohms Rl?
@joshuabumanglag9199 please check the question, the load resistance is 4
@@SkanCityAcademy_SirJohnafter 8 ohms? series? is that the ohms resistor? right?
I don’t know if my method of approach is okay, but I used;
In = 3 + I (kcl)
Using kvl; first loop
25 = 5*I + 20*I
Solving this should result to 1
From kcl,
In= 3+1 = 4
Equation for mesh 1:
25 = 25 i1 - 20 i2
Equation for super mesh
0 = -20 i1 + 20 i2 + 12 i3
Applying kcl at Supermesh
i2 = i3 - 3
Values for i1-3, respectively
0.2A, -1A and 2A
@@SkanCityAcademy_SirJohn thanks I made a mistake and still got it correctly
Oh okay. Follow my step, and see if you understand.
Where do you watch from?
@@SkanCityAcademy_SirJohn your videos. Super mesh
Why didn’t you take the last loop but the second one on the first example
Can you be more precise by stating the exact time in the video for easy reference?
boylsted or sadiku
Which one is good
Both of them are quite good...you can complement one with the other
cant we solve problem 2 directily
You can
Sir in problem 2 , according to problem 1 if the current direction is downward 👇 we take it negative but in problem 2 both currents IN'and InN" is downward 👇 but you take it positive why sir ?
The issue is, in problem 1 we have current in both direction, so you can either choose one direction to be + and the other to be -, its not compulsory to choose a specific direction for positive or negative. But for problem 2, both are in the same direction, so you can either choose positive or negative for both of them.
@@SkanCityAcademy_SirJohn And sir can our answer of current be in negative??
Yes, you can have a negative current. Negative basically means, the actual current moves in the opposite direction
Are you okay with the answer?
@@SkanCityAcademy_SirJohn yes sir I'm okay with your answers thanks a lot 😌😀
Geez Louis, use source transform dude. Your method is like the second most tedious method next to node and mesh analysis.
Thank you, but that willl be for another video
Question of reciprocity
noted
I used super mesh to solve number 2. Why isn’t it possible
It is possible, I guess you have an error in your calculation
Please why didn't you use the first way you solved the first question for the second question
No issue, just wants you to see the variety... It doesnt change anything
@@SkanCityAcademy_SirJohn okay sure thank you
@@najatal-hassan8420 you're welcome
you metod is wrong why using kirckkoff's law instead of norton
When you are asked to solve using Nortons Theorem it doesn't mean you can't used fundamentals like ohms law and kirckoff in your solution.