Question

A 50 Hz single-phase transformer has a turns ratio of 6. The resistances and reactances are 0.90 Ω and 1.5 Ω for HV winding and 0.13 Ω and 0.1855 Ω for low voltage winding respectively.

Calculate the voltage to be applied to the high voltage side to obtain full load current of 200 A in the low voltage winding on short circuit.​

Answer 1

Answer:

330 V

Explanation:

Let say Primary side is High Voltage

& Secondary side is Low Voltage

k = Turn Ratio = N₁/N₂   = 6

Reff1  =  0.90 +  k²0.13 Ω  = 0.9 + 6²*0.13 = 5.58  Ω

Xeff1 = 1.5 +  k²0.1855 Ω  = 0.9 + 6²*0.1855 = 8.178  Ω

Zeff1 = √5.58² + 8.178² = 9.9 Ω

V₁/V₂  = I₂/I₁  = k = 6  

=> I₁ =  I₂/6

I₂ required = 200 A  ( on Secondary side / Low Voltage side)

=> I₁ = 200/6

=> I₁ = 100/3 A

Voltage on HV Side (Primary Side) = I₁ * Zeff1

=  (100/3) * 9.9

= 330 V

330 voltage to be applied to the high voltage side to obtain full load current of 200 A in the low voltage winding on short circuit