00:01
Hi, in this question we are given with the circuit with the following connections and we have to find the nodal voltage at point 1 and 2.
00:08
So in the first case applying kcl to node a we have v1 by minus j2 plus v1 by 10 plus v1 minus v2 divided by j 4 plus 30 degree is equal to 0 which becomes equal to j into 0 .5 v1 plus 0 .1 v1 minus j into 0 .25 v1 minus v2 plus 20 to 30 degree is equal to 0 so this becomes equal to 0 .1 into or point 1 taking v1 outside so 0 .1 plus 0 .25 j into into v1 plus 0 .25 j into v1 plus j into 0 .1 25 v2 which is equal to minus 20 to 30 degree in the second part so apply kcl to node in so node 2 we have v2 minus v1 divided by j 4 plus v2 divided by minus j 5 plus v2 divided by minus j 5 plus v2 divided by j 2 is equal to 30 degree so which becomes minus j into 05 or v2 minus v1 plus j into 0 .2 v2 minus j into 0 .5 b2 is equal to so minus j into 0 .55 v2 plus j into 0 .25 v1 is equal to that is 3.
02:01
So this is equation 2 and previous equation was equation 1.
02:06
Now writing in a matrix form we have 0 .1 plus 1.
02:10
J into 0 .25 so combining equation 1 and writing into matrix form so j into 0 .25 into j into 0 .25 here it's minus j into 0 .55 into v1 by v2 or v1 v2 is equal to minus 20 30 degree and this is 20 angle 30 degree so now therefore delta becomes equal to minus j into 0 .55 into 0 .1 plus j into 0 .25 which is equal to minus j into 0 .25 into j into 0 .25.
02:54
So which is equal to delta becomes equal to 0 .2 minus j into 0 .055.
03:02
Therefore delta 1 can be written in the form of it is minus 20 30 degree j into 0 .25 so this is determinant from so 20 30 degree j into 0 .55 with negative sign.
03:19
So which becomes so on determinant function using determinant function we have these two are multiplied and these two are multiplied.
03:29
So this on determinant multiplication we have minus 3 plus j into 5 .916 or 196...