00:01
In this problem consider the node voltage as v.
00:07
Node voltage as v.
00:13
So here i can draw a diagram like this.
00:17
Just look at it carefully.
00:21
This is the diagram which i am drawing here.
00:32
And here it is of 6v.
00:35
Here it is of 3k.
00:37
And here it is of 2.
00:43
Here it is of 2k.
00:46
And here it is of 12v.
00:54
And here current i is flowing.
01:00
Now going forward and the equation for the node voltage is given by v -12 by 2k plus v plus 6 by 3k plus i is equal to 0.
01:20
So the value of v is 0.
01:23
So the current equation will become 0 minus 12 by 2k plus 0 plus 6 by 3k plus this value is equal to 0.
01:38
Finally you will get this value as 4 ma.
01:43
Now the current through the inductor cannot change instantaneously.
01:49
So i can write this value as it is equal to this which is equal to 4 ma.
01:59
Now on further simplification i can draw another diagram like this.
02:08
This is the diagram which i am drawing here.
02:14
Just look at it carefully.
02:19
Here it is of 6v.
02:21
Positive, negative.
02:25
And here it is of 3k.
02:29
And here it is of 2k.
02:38
Here it is of 12v.
02:43
And here it is of 6mh.
02:50
Here it is i -t.
02:53
Here it is of 2k.
02:56
Now going forward and redrawing the circuit...