00:01
In this question, solution here is for the a part apply nodal analysis at node v1.
00:17
So v1 minus minus of 30 whole divided by 5k plus v1 divided by 500 plus v1 minus 80 divided by 1k plus 10m is equals to 0.
00:33
So v1 multiplied by 1 divided by 5k plus 1 divided by 500 plus 1 divided by 1k plus 30 divided by 5k minus 80 divided by 1k plus 10m is equals to 0.
00:53
So v1 multiplied by 2 divided by 625 is equals to 8 divided by 125.
01:01
On solving we will get here v1 is equals to 20 volt.
01:09
Now t1 is equals to minus 30 minus v1 divided by 5k which is equals to minus 30 minus 20 divided by 5k which is equals to minus of 10m ampere.
01:30
Now t2 is equals to v1 divided by 500 is equals to 20 divided by 500 which is equals to 40 m ampere.
01:44
Now apply kcl at node 80 volt.
01:48
So we get here minus 10m plus 80 divided by 4k plus i3 plus 80 minus v1 divided by 1k is equals to 0.
02:02
So minus 10m plus 20m plus i3 plus 80m minus 20m is equals to 0.
02:13
Therefore i3 is equals to minus 70m ampere.
02:19
This implies here this is here i1 and this is here i2.
02:30
Now this implies here i1 is equals to minus 10m ampere and i2 is equals to 40m ampere and here this is i3 is equals to minus 70m ampere...