00:01
So this problem, we're focusing on bus admittance matrix.
00:05
So that's the y, called the y bus.
00:13
And that is for the four bus system.
00:33
So the given data, the transmission lines are lossless.
00:38
So the r is zero.
00:43
Series impedance, this is our, i use jx.
00:50
Values are provided.
00:55
And then we have the shunt.
00:57
The shunt admittances are given for lines connecting buses two to four and three to four.
01:15
So line impedances between bus one and bus two, that is our jx is equal to j.
01:34
So that is between bus 1 and 2.
01:54
Then we have jx, which is equal to j .3.
02:12
This is between bus 1 and 3.
02:24
1 and 3.
02:27
Then we have j x equal to j 0.
02:43
And this one is between bust two and four and this includes with shunt edmontnesses j 0 .12 at each like and then we have j x x and then j .5 this one is between bus three and four and that's with shunt admittances j .07 at each like all right so i have that mapped out so we have to go with starting at calculating the admittances we'll just say admittances so we have y12 this is equal to negative 1 over j point 2 so that's negative j 5 then we have y 13 this is negative 1 over j 0 .3 so this is negative j 3 .33 then we have y 24 this is 1 over j .4.
06:01
So that's negative j 2 .5.
06:08
Then we have y 34.
06:14
This is negative 1 over j .5.
06:28
So hence, negative j2.
06:35
So then, shunt admittances.
06:40
This would be shunt admittances.
06:48
So this is for boss 2 to 4.
07:04
We have a y shunt 24 equal to j.
07:16
0 .12 plus j .0 .12.
07:26
So we have j is 024.
07:30
And then for bus 3 to 4, we have y shunt 34.
07:49
So this is j .07 plus j .07, and this is j .07.
07:58
And this is j.
08:01
0 .14.
08:06
So that's admittances there.
08:08
So then we go to part two...