00:01
In this problem, we have been given the following circuit and we have to use mesh analysis in order to figure out the current is shown here.
00:10
So basically we have to determine the value of i1, i2, and i3 here.
00:16
And for this purpose, we're going to make use of mesh analysis.
00:20
So for that purpose, we will use different meshes.
00:23
And in those meshes, we will apply kirchav's voltage drop.
00:27
So for mesh one, we will get one equation.
00:30
We'll get another equation for mesh 2 which is let's name it here a b c d e f g and h so mesh 1 that will be basically the loop g a e f mesh 2 that will be the loop a b c h and and finally we will have mesh 3 because we have three unknown variables i 1 i2 and i 3 to be found and we require three equations.
01:04
For that purpose, we are observing three different loops.
01:08
For mesh three, we will observe the loop h -c -d -e -h.
01:14
And now we observe the different currents in different branches as we can observe in mesh 1, the current is i -1.
01:21
In mesh 2, the current is i -2.
01:22
So i -2 will be leaving this junction a, and else we can definitely conclude that the current which will be flowing away from this point a through the resistor 2 oom that will be i1 minus i2 but we also observe here additionally that across 2 oom resistor the voltage is we not so we can just use omslaw and say that v0 is equal to i1 minus i2 into the resistance that's 2 so let's mark it as equation 1 and now we observe that at this point h, the total current which is entering this junction, that is i1 minus i2.
02:04
And out of this i1 minus i2, even i1 has to flow out from here.
02:11
But still we observe that the current that's leaving this is i3.
02:17
So we can definitely conclude that this i3, that's the current in this third mesh, this i3, that will be equal to i2 plus 3, because if we just apply kerchiefs, junction rule here we observe i3 comes out to be three plus i2 let's mark it as equation two and now we observe in mesh one applying kirchhoff's voltage lot directly we're going to get the equation 12 minus two times of i1 minus i2 minus the current that will be flowing through this four -home resistor that will be i1 minus i3 and as we can see minus four times i1 minus i3 that's equal to zero.
03:05
Let's mark it as equation three.
03:07
And for mesh two, we observe that applying kirchhoff's voltage law once again, the equation comes out to be.
03:20
So let's start at this point to get the equation.
03:23
Let's say this point has zero voltage...
