00:01
Hello students in this portion we have given load that is 3 minus j7 also we have 3 minus j7 minus j5 this will becomes 3 minus j2.
00:18
Now for delta connection v phase will be equal to v line that is 440.
00:27
Now the total impedance per phase will be z this will be 3 minus j2.
00:45
Now the phase voltage phase voltage will be this value.
00:55
Now for phase current we have 440 divided by 3 minus j2 this will be equal to 122 33 .69.
01:17
Now for delta connection line current is equal to line current needs phase by 30 degrees so root 3 into 122 with 30 degree plus 33 .69.
01:42
So we can write line current will be 211 .31 ampere.
01:54
Now line voltage will be equal to 440 volt.
02:04
Now this is line current and this is line voltage.
02:08
Now for power factor we have the here this represent the angle of voltage and this represent angle of current.
02:31
Now substituting the values we get 0 minus 33 .69 therefore the power factor will be 0 .832...