00:01
Hi, in part a of this question they are asking us to find the unknown currents i1, i2 and i3 in this multi -circuit which is shown.
00:11
So if we apply the kirchhoff's voltage loop in the loop edef, so here we have edcf, so in edcf if we apply so we can write 4v -3i2 -5i3 as equals to 0.
00:37
So let this be equation 1.
00:39
Now similarly we will apply kirchhoff's current law at junction c over here.
00:44
So we will get that i1 plus i2 is equals to i3.
00:50
Now this is second equation.
00:52
Now if we substitute the value of i3 in this equation 1, then we will have 4 times minus 3 times i2 minus 5 times i1 plus i2 equals to 0.
01:08
So from here we will get 5i1 plus 8i2 as equals to 4.
01:15
So let this be equation 3.
01:17
Now again if we apply kirchhoff's voltage law in dcfg which is this middle block.
01:25
So we can write from dcfg and we are applying kirchhoff's current law at the junction f.
01:37
So we will get i3 minus i2 as equals to i1.
01:45
Also 8 minus 5i1 plus 3 times i2 will be equals to 0.
01:53
So from here we will get 5i1 minus 3i2 as equals to 8...