00:01
In this question here, so here, source transformation is applicable only when current source is connected in parallel with register or voltage source is connected in ced with register.
00:13
So here transformation is not applicable.
00:17
So here in a part, source transformation is not applicable.
00:24
So here from loop 1 here i1 equals i1.
00:33
To 1 ampere.
00:37
Here kvl to loop 2.
00:44
So here 6 plus 250 plus 4 plus 260 plus 40 multiply by i2 minus 260 i3 minus 40 i1 is equals to 0.
01:06
So here.
01:07
It will come 560 i2 minus 260 i3 equals to 40.
01:16
And here 14 i2 minus 6 .5 i3 is equal to 1.
01:25
So here this is our first equation.
01:30
So here now kvl to loop 3.
01:40
So here 260 i3 minus 260 i2 is equals to minus 520...