00:01
Hello students, the increasing values of, sorry, the initial values of ei, p and q are, the equation is ei is equal to vt plus j into i into xb.
00:17
So, substituting the values, we will get 1 .0 less than 0 plus j into 1 .7241 into 0 .8 less than cos inverse of 0 .9.
00:33
Here we will get the answer as 1 .0 plus j 1 .3654.
00:40
Now, finding out p value, p is equal to mod i plus mod ei plus cos of ei minus i.
00:57
So, substituting the values, here we will get 0 .8 into 1 .0 into cos of 1 .3654 minus 25 .84.
01:12
Here we will get the value of p as 0 .72273 p.
01:18
Now, to find out the q value, q is equal to mod i plus mod ei plus sin of ei minus i.
01:34
So, here we will get the value on substitution that is 0 .8 plus 1 into sin of 1 .3654 minus 25 .84.
01:50
Here we will get the q value as minus 0 .3175 mu.
01:55
So, the decreasing excitation by 20 percentage that is a new excitation equation is new excitation will be equal to 0 .8 into excitation and the equation for finding out delta is equal to ei minus vt and to find out the reactive power input that is q, q is equal to q plus that here it is q new and q equal to q plus 0 .8 into excitation into mod ei into sin of del del 1.
02:47
Now, the equation for the increasing real power input is equal to that is new power will be equal to p plus 0 .2 into p.
03:01
Now, for the first case that is the decreasing excitation by 20 percentage.
03:10
We need to find out the values for all these...