00:01
So here we have the circuit diagram and we have to find out the first thing that is vl.
00:04
Let's find out first.
00:05
So this vl will be equal to vph multiplied by root 3 angle of 30 degree.
00:17
So this is vph which is 220 root 3 angle of 30 degree and from here we will get this vl as equal to 381 .05 angle of 30 degree.
00:36
Now let's go ahead.
00:39
Now load is delta connected so in star that is zy that's equal to i mean z gamma we can say or zy zd divided by 3.
00:52
This is equal to 30 plus j5 divided by 3 and that's equal to 10 plus j1 .67.
01:05
Okay now let's go ahead to calculate the i of aa that means line currents we will find out the first thing is iaa that's equal to vph divided by zy plus 1 plus j0 .5.
01:33
Okay that's equal to 220 divided by 10 plus j1 .67 plus 1 plus j0 .5.
01:52
Okay so from here this i of aa that is equal to 19 .62 angle of minus 11 .14 degree ampere.
02:06
This is what the value of the iaa...