00:01
Hello students, let's find out the voltage drop per unit.
00:05
So, the equation for voltage drop is equal to high voltage minus low voltage per unit divided by high voltage per unit.
00:35
This high voltage is also per unit.
00:40
So, now since a high voltage is an ideal positive sequence source, negligible source impedance per unit magnitude of high voltage.
00:49
So, high voltage is equal to 1 .0.
00:54
So, the low voltage per unit equation is the low voltage in kv divided by the low voltage in kv, which will be equal to 1 .0.
01:10
Therefore, let's find out the voltage drop, which will be equal to 1 minus 1 divided by 1, which will be equal to 0.
01:22
So, the voltage drop.
01:23
Now, let's find out the rated current per unit, which is equal to the rated current divided by the hv base voltage.
01:38
Sorry, it's hv base current.
01:42
So, now let's find out the hv base current, that is the equation is hv divided by root 3 into apparent power into 100.
01:54
On substituting the values, we'll get 345 divided by root 3 into the apparent power is 200 into 10 raised to 6, which is equal to 995 .041...