00:01
So in this question, r1 is given as 3 .8 ohms, r2 is given as 7 ohms, r3 is given as 2 .7 ohms and r4 is given as 8 .9 ohms.
00:20
Epsilon 1 is equal to 12 volts and epsilon 2 is equal to 12 volts as well.
00:31
So if we apply kirchhoff's current law at point a, we are applying kirchhoff's current law at point a, so we get i1 plus i2 that is equal to i3 is the first equation.
00:49
Now we are applying kirchhoff's voltage rule in loop 1, the first loop from the left.
01:01
So we get minus epsilon 1 plus r1 i1 plus i3 r3 plus i1 r4 is equal to 0.
01:13
This implies that r1 plus r4 multiplied to i1 plus i3 into r3 equals epsilon 1.
01:25
Now if we put in the values r1 plus r4, 3 .8 plus 8 .9 multiplied to i1 plus i3 multiplied to 2 .7 that is equal to 12.
01:40
So this is our second equation.
01:41
Now if we apply the kirchhoff's voltage rule in loop 2 that is the right loop or the second loop, we get the value minus epsilon 2, this is for loop 2 and this was for loop 1.
02:16
So minus epsilon 2 plus r2 i2 plus r3 i3 is equal to 0...