could someone tell me wich current should flow trough an asynchrone motor if power is 200kw,and thevol.380?

Question

nothing

Answer 1

To calculate I (amps) the equation is kW= IE*1.732

Therefore I = kW / E*1.732

So I at 380V = 304A

Typically the power factor of a 200kW cage motor would be around .85, So 304 / .8 = 357amps

Your 200kW motor would draw approximately 357A at 380V

Answer 2

P = V * I
I = P / V

Not equal to (200000 / 380)

Since this works only if the current is in phase with the voltage, which, given that this is a motor, will not be.

The magnitude of the impedance if the motor will be (V/I) or
(V*V)/P , but unless you know the phase shift that voltage leads current, no further work can be done on this question.

Remember the motor's impedance will be expressed as
(m + ni) ohms, so you will need to know how to divide complex numbers.

Answer 3

P = V*I*cos(phi) where phi is the phase angle between voltage and current.

I = P/Vcos(phi)
I = 526/cos(phi)

Total power is the vector sum of "real" power and reactive power. In the States reactive and total power are given dimensions of va, var and kva, kvar to distinguish them from real power, or wattage. If you are talking about total power, ignore the cos(phi) qualification.