Thank you, sir, 🫡I've been looking for this explanation and answer. I had a misunderstanding in analizing the individual mesh, but now I know the correct way. I was writing the equation for mesh 3 [2(i1 - i3) + 2i3 + 4(i2 - i3)]. Turns out you have to make the (i) a priority for the mesh you are analyzing. So, the correct equation like in the video [2(i3 - i1) + 2i3 + 4(i3 - i2)] 🤗
The condition is that a super mesh is formed if two meshes have a common current source between them, (dependent or independent). So regardless of a resistor in series with the current source, so long at there is a common current source that is fine.
@@SkanCityAcademy_SirJohn I think He is referring to to mesh i3 where you have two 2 ohms resistor. One of the 2 ohms resistor is shared with i1 and i3 but the other is solely for i3 only.
4:34 Why Can't we write for 1st loop : -20+6i1-6(2)=0 For 2nd loop : 10i2+4i2+6(2)=0 We are evaluating voltage drops across resistors in a loop soo why can't we apply that above equations instead of using super mesh
Well, there are a number of theorems available to solve this problem, only that in this case we want to use mesh analysis. And this is how we go about it when we have a super mesh.
i1 and i2 are the current in the super mesh. After creating the super mesh, there are only two meshes present. The super mesh and mesh 3. So going round each mesh, you can select one as equ 1 and the other as equ 2 and vice verse.
The explanation here is that, we are moving sequentially, and we take our reference direction to be clockwise to be +, and i2 is also taking the clockwise direction, so it can never be i3-i2. Else we have altered our notation.
The reason is: according to the concept of super mesh, if a current source is found between two meshes, a super mesh is formed by eliminating the current source in series with any circuit element (in this case, the 1ohm resistor). That is why the 1ohm is not considered. Thank you, hope you got it.
It is so because you know: For mesh analysis, we take the clockwise movement. In that regard, referencing the direction of i3, and that is clockwise hence positive. Now i1 is meeting i3, which means it's opposing i3 hence it will be negative (anti-clockwise)
thank you so much for the neat hand writing, lecturers dont realise that poor hand writing makes it more confusing. You are a great explainer too!!
Awww thanks so so much.
Where do you watch from?
@@SkanCityAcademy_SirJohn Colorado
@Electromagic27 oh great. Thanks so much for watching
Sitting for my exam on Monday, feeling relaxed after watching your videos. I hope I don't mess it up. Thankyou for your efforts
Nice feeling right.....
God will surely guide you through. Good luck
Thank you, sir, 🫡I've been looking for this explanation and answer. I had a misunderstanding in analizing the individual mesh, but now I know the correct way. I was writing the equation for mesh 3 [2(i1 - i3) + 2i3 + 4(i2 - i3)]. Turns out you have to make the (i) a priority for the mesh you are analyzing. So, the correct equation like in the video [2(i3 - i1) + 2i3 + 4(i3 - i2)] 🤗
you are most welcome.
please where do you watch from?
Thank u so much sir 😊 I am searching for clear explanation, now I found it with same question what I am doing right now😊
Aww that's Great
@@SkanCityAcademy_SirJohn thanks sir
Thank you so much sir!
Most welcome
Thnk u sir for ur help i really appreciate it ❤
You are most welcome. Where do you watch from
really helping🎉
Great
does there need to be a resistor with the current source to make it a supermesh or can we do it even if its just the current source by itself
The condition is that a super mesh is formed if two meshes have a common current source between them, (dependent or independent).
So regardless of a resistor in series with the current source, so long at there is a common current source that is fine.
Nice one
Incredible
Thank you very much
Pls for the eqn(1), you did not include the 2i3…. Kindly check or that is how it is done pls
At which time specifically, so I check and explain to you better
@@SkanCityAcademy_SirJohn I think He is referring to to mesh i3 where you have two 2 ohms resistor. One of the 2 ohms resistor is shared with i1 and i3 but the other is solely for i3 only.
4:34
Why Can't we write
for 1st loop :
-20+6i1-6(2)=0
For 2nd loop :
10i2+4i2+6(2)=0
We are evaluating voltage drops across resistors in a loop soo why can't we apply that above equations instead of using super mesh
Well, there are a number of theorems available to solve this problem, only that in this case we want to use mesh analysis. And this is how we go about it when we have a super mesh.
you are the best
Thank you
Please why did you solve i1 and i2 as equation one and i3 as equation 2....please what's the reason
i1 and i2 are the current in the super mesh. After creating the super mesh, there are only two meshes present. The super mesh and mesh 3. So going round each mesh, you can select one as equ 1 and the other as equ 2 and vice verse.
Please ehy not 4(I3 - I2) for the super mesh?
The explanation here is that, we are moving sequentially, and we take our reference direction to be clockwise to be +, and i2 is also taking the clockwise direction, so it can never be i3-i2.
Else we have altered our notation.
Please why don't we involve the one ohm resistor in the 3rd equ
The reason is: according to the concept of super mesh, if a current source is found between two meshes, a super mesh is formed by eliminating the current source in series with any circuit element (in this case, the 1ohm resistor). That is why the 1ohm is not considered.
Thank you, hope you got it.
Why is it 2(i3-i1) in mesh 3
It is so because you know:
For mesh analysis, we take the clockwise movement.
In that regard, referencing the direction of i3, and that is clockwise hence positive.
Now i1 is meeting i3, which means it's opposing i3 hence it will be negative (anti-clockwise)
Please sir do you have any WhatsApp line or email address which i can send my circuit problems for help