00:01
Let's solve the question.
00:02
Here a circuit is given to us having 3 resistances r1, r2 and r3.
00:17
This is r2, this is r1 and this is r3 and this is r4.
00:24
It is 1 .2v, this is 1v and here the emf is e.
00:29
R1 is 5 ohm, r3 is 1 ohm and r4 is 4 ohm.
00:36
R2 equals to 2 ohm and here current i3 equals to 50 ma it is given to us.
00:44
We write this point a, b, c, d, e and f.
00:50
We take this loop as loop 1 and this loop as loop 2.
00:56
Now we apply kcl at d point.
01:04
So we can write i1 plus i3 plus i2 is 0 and we denote here current is i1 and here current is i3 and here current is i2 or we can write i1 plus i2 equals to 0 .05 a.
01:28
This is equation 1...