00:01
Hello students, here let's assume that the current flowing this branch is i1, here i2 and here it's i3.
00:12
So from kcl kirchhoff's current law we can write that i1 plus i2 plus i3 equals to 0.
00:22
So i1 plus i2 plus i3 equals to 0.
00:35
So this is equation number one.
00:37
This is one of the equation we get from kirchhoff's law.
00:41
Now applying kvl in this loop we have minus i1r minus i1r minus 2 epsilon plus 3ri2 equals to 0 or rearranging it 3ri2 equals to 2 epsilon plus ri1.
01:13
This is equation number two.
01:15
Now again applying kvl in this loop kirchhoff's voltage law we have minus 3ri2 plus 2 epsilon plus 6r6i3r minus epsilon equals to 0 or simplifying we have epsilon plus 6ri3 minus 3ri2 equals to 0.
01:46
Question number three.
01:47
So these are the three equations.
01:49
Now from one we have from one we have i1 equals to minus i2 minus i3...