00:01
In this question we have to find thevenin thevenin equivalent voltage equivalent voltage equivalent voltage in this question we have a diagram in this question in this question v this node is 1, 2 and 3 so here at this node voltage is v1, v2 and v3 now we will assume nodes 1, 2 and 3 as shown voltages at the nodes are v1, v2 and v3 respectively.
00:48
So here v3 is equals to v7a and v2 is equals to 250iδ.
01:04
So now we will apply kcl at node 1.
01:16
We get v1 upon 130 plus v1 minus v2 upon 200 plus v1 minus v3 upon 150 is equals to 0.
01:37
On further simplification we can rewrite it as b1 into 1 by 130 plus 1 by 200 plus 1 by 150 plus b2 into minus 1 upon 200 plus b3 into minus 1 upon 150 is equals to 0.
02:13
Now on further simplification, on simplification we get v1 into 151 upon 7800 plus v2 into 1 upon minus 200 plus v3 into v3 into minus 1 upon 150 minus 1 upon 150 equals to 0.
02:46
This is equation a.
02:49
Now on further simplification we will apply kcl at node 3.
03:04
Kcl at node 3 we get v3 -v2 upon 50 plus v3 -v1 upon 150 is equals to 0...