00:01
Determine the voltage v1, v2 and v3 for the circuit given.
00:05
Let the node voltages be v1, v2 and v3.
00:12
Now finding the values for v1, v2 and v3 using the nodal analysis.
00:20
First applying the nodal analysis at v1, i0 equals 4v2.
00:27
Let it be first.
00:30
Applying the kvl v1 by 1 by 2.
00:35
Plus v1 minus v2 by 1 by 1 plus 3 i0 plus v1 minus v3 by 1 by 4 equals to 0 the equation becomes 2 v1 plus v1 plus v2 plus 3 i0 plus 4 v1 minus 4 v3 equals to 2 replacing the value of i not into this equation so we have the reduced form of equation equals to 7 v1 plus 11 v2 minus 4 v3 equals to 2.
01:19
Now applying the nodal analysis at node 2, that is v2 1 into v2 minus v1 plus v2 plus v2 minus v3 equals 0.
01:37
From here the equation becomes 6 v2 minus v1 minus v3 equals 0.
01:42
Let this be the third equation.
01:44
Applying the nodal analysis at node 3, we have 1 into v3 minus v2 plus 4 into v3 minus v1 minus 3 i .0 plus 2 v3 minus 4 equals 0...