00:01
We have been given that r1 is equal to 1 kω which is equal to 1000 ω.
00:10
R2 is equal to 2 kω which is equal to 2000 ω.
00:17
R3 is equal to 3 kω which is equal to 3000 ω.
00:23
E1 is equal to 10 v.
00:29
E2 is equal to 6 .75 v.
00:34
The diagram given in the problem is this is a, b, c, d.
00:58
I1, i2, i3.
01:03
This is battery e1, this is battery e2.
01:14
This is r1, this is r2 and this is r3.
01:24
We will consider these two loops in which ix current is flowing in clockwise direction and iy in this current is flowing in clockwise direction.
01:41
Therefore, applying kvl to loop acba 3000 ix minus 2000 iy is equal to 10.
02:12
This is equation 1.
02:13
Applying kvl to loop adca minus 2000 ix plus 5000 iy is equal to minus 6 .75...