got exam 2morw , instead of wasting time on 1hr+ lectures i stumbled upon your channel and man you are great at explaining all the complex stuff (not complex at all after watching your vids) in less than 10-15 mins! love you!!!!
Have not been able to understand Supermesh or Supenode from my class notes, you explained it so easy in this video and your Supernode one! Thank you!!!!!!!!!!!!!!
Great video, one doubt, At 11:15, how you assumed that current flowing through 4 Ohm resistor is (I3-I2) and through 2 Ohm is (I1-I2); in this comment section actually, there are many people who asked this doubt, and you replied to them but still, I don't understand the reason behind it.
You can assume anything which you want just that the signs would be reversed in that case. He is focusing on the voltage sign convention he had given to that 4ohm resistor (ie that + and -)..so whichever current enters the +ve he took it greater than the one entering the -ve.
Please help sir You've told us in previous videos that.. when apply kvl in a mesh then the current of that should be considered as the higher current than the other mesh... But at 2:55 you write the opposite thing their..... Is it right their or am i just confused ? Please tell me sir
Yes, it's one of the same thing. Here, I have considered voltage V2 polarity. If you go with an earlier convention, then it would come as 3 (I2 - I1). Here it is (-3) (I1 - I2). therefore, it's the same thing. I hope it will clear your doubt.
In the mesh -2, the net current flowing through the 4 ohm resistor is I2 - I3. And it has been assumed that the current is flowing from right to left. So, while applying the KVL, in clock wise direction, the voltage drop acorss it would be - 4 (I2- I3). But here it has been written as 4*(I3 - I2). Same is the case for 2*(I1-I2) I hope it will clear your doubt.
It depends on the circuit. Sometimes it is better to use source transformation to simplify the circuit, while sometimes it is good to leave it as it is.
The sign is negative if it is a voltage drop and it's positive if it is a voltage gain. I don't think it's wrong. It's just a different approach from yours where you consider the polarity of the terminal encountered first as you travel around the loop. Whatever approach you use in deciding which sign to put, you will arrive with the same answers anyway.
I think the equation at Mesh 2 is wrong. It is supposed to be -6I2 - 4(I3-I2) - 2(I1-I2) = 0 -6I2 - 4I3 + 4I2 - 2I1 + 2I2 = 0 -4I3-2I1 = 0 -2I3 - I1 = 0 I got my equation from the direction of the current in each resistors. Correct me, If I'm wrong thanks.
Actually, the equation depends on how you take the sign convention. Here it has been assumed that in mesh 2, the net current is equal to I3-I2. Which is flowing from right to left. And accordingly, the voltage drop is shown across the 4-ohm resistor. Now, if you apply the KVL in the mesh and move around the mesh in a clockwise direction then the voltage across the 4-ohm resistor is increasing as we are moving from lower to the higher potential. So, the positive sign is taken for the voltage drop across the 4-ohm resistor. The same is true for 2-ohm resistor. I hope it will clear your doubt. And stil lif you have any doubt then do let me know here.
Because the current through 2 ohm resistor is 3A. Sine the current source is attached at one end, it means the current entering into 2 ohm resistor is 3A and the same current is also leaving through it. Therefore it is not considered in calculation. I hope it will clear your doubt.
Depending on the direction of the assumed current I1 and I2 (clockwise or the anti clock wise), it can be decided. In fact you can choose any of them. The final output will remain same. Because let's say if you choose i1 -i2, then you will get negative answer. But at the end you will get the same answer. I hope it will clear your doubt.
Actually, this equation or relation is not based on KCL. But just by inspection, you can observe that current I1 is flowing in the clockwise direction while current I3 is flowing in the counterclockwise direction. So, we can say that net current flowing through the 2-ohm resistor is I1- I3, which is flowing in the downward direction. And current through 3A source is also flowing in the downward direction. So, we can say that I1-I3 should be equal to 3A. I hope it will clear your doubt. And if you still have any confusion then do let me know here.
If you are watching on PC then you can simply drag the substitle anywhere on the screen. On smartphone, for some period you can turn of the subtitles in the video setting. I hope it will solve your problem.
Better you assume the voltage across each element first. Let's assume it as V1, V2, V3. Where V1, V2, and V3 are the voltage drop across 6 ohm, 4 ohm and 2 ohm resistors respectively. And if you apply the KVL with the assumed signs, then it can be written as, -V1 + V2 + V3 = 0 Now, V1 = 6*I2, V2 = 4 (I3- I2) and V3 = 2 (I1 - I2) (Here, it has been assumed that the net current I3- I2 and I1 -I2 is larger. ) Now, if you put the value of V1, V2 and V3, then you will get that expression.
Here, we are applying KVL. And 2(I1-I2) is the voltage across 2-ohm resistor. Now, while applying KVL we are moving from higher to the lower potential across 2-ohm resistor in the supermesh. So, the negative sign has been taken for the voltage across 2-ohm resistor. I hope it will clear your doubt. But if you still have any doubt then do let me know here.
You mean to say, while writing I1-I3 = 3A right? Well, actually it is not required. Because current through that 2-ohm resistor will be 3A only. And we are only looking at the current. If we are calculating the voltage then we need to consider the voltage drop across that 2-ohm resistor. I hope it will clear your doubt. If you still have any doubt then do let me know here.
well, you can find the voltages too across each element, once you know all the currents in the circuit. Using KVL, it's easy to find the voltages across each element. I hope you understood.
In that example, the direction of mesh currents I1 and I2 were assumed in the clockwise direction. So, the effective current through 3-ohm resistor will be I1 - I2. And voltage across it will be 3(I1-I2). I hope, it will clear your doubt.
We have assumed that the I3 is greater than I2. So, net current flowing through will be equal to I3-I2. And hence drop across 4-ohm resistor is from left to right. You can also assume that I2 is greater than I3. (i.e drop across 4-ohm resistor is from right to left, and can get the same result.) It's just sign-convention that we have assumed for the calculation.
ALL ABOUT ELECTRONICS Thanks for your response. @10.05 we take I1-I2 for resistor 2 ohm resistor. Hence same current will flow through that branch till it diverts to 4 ohm resistor. So current in 4 ohm should be less than current in 2ohm resistor. Therefore (I1-I2) - I3 is to be consider. If I3 is flowing through current source and 2 ohm resistor. I'M NEW IN THIS. PLZ CLEAR MY DOUBT शुक्रिया 🙇♂️
Here, current I1, I2, and I3 are the mesh currents. And here we are assuming that they are flowing in particular mesh only. So, if you consider the 2-ohm resistor at the top, it will interact with mesh current I1 and I2. While the 2-ohm resistor at the bottom will interact with mesh current I1 and I3 only. I hope it will clear your doubt.
I would not recommend his videos. They might get you the answers but will surely destroy your concepts. He is making the equations all in the opposite direction of the currents he is taking.
As I've replied to the other comment, "The sign is negative if it is a voltage drop and it's positive if it is a voltage gain. I don't think it's wrong. It's just a different approach from yours where you consider the polarity of the terminal encountered first as you travel around the loop. Whatever approach you use in deciding which sign to put, you will arrive with the same answers anyway." The approach in this video will not destroy your concepts. In my opinion, his approach illustrates the concept of KVL better than the approach you are used to.
