Question

A 10 kva 400/200 V single phase transformer with 10% leakage impedance draws a steady short circuit line current of

Answer 1

Explanation:

Given A 10 kva 400/200 V single phase transformer with 10% leakage impedance draws a steady short circuit line current of

According to question for transformer rated current is 10000 v A / 200 v

                                                                                                  = 50 A

Drop will be 10% of 200 V = 20 V

Internal impedance = 20 V / 50 A = 0.4 Ohms.

When short circuited the current will be 200 V / 0.4 Ohms = 500 Amps

The current secondary side is 500 Amps.

Primary side will be 0.5 x 500 = 250 Amps.