2. Given a 50KVA single phase distribution transformer with voltage 2400/240VAC (center tapped on the secondary) and with 3.0% impedance. Using the low voltage side, solve for the following: a) Ibase, the base current, b) Zbase, the base impedance, c) Zact, the actual impedance, d) Isc, the short circuit current if a short circuit would be directly across the low voltage terminals Assume the entire impedance consists of reactance.

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batolisis batolisis · Answer 1 · 2021-04-03T00:19:26+02:00

Answer:

a) 208.33 A

b) 1.152 Ω

c) 0.03456 Ω

d) 6943.64 A

Explanation:

Transformer ratings :

Base power ( Sb ) = 50 KVA ,

Transformer voltage = 2400 / 240 V ,

Impedance ( Zp.u ) = 3.0 % = 0.03

phase = single ( 1 ) phase

Base Voltage on low side ( Vb ) = 240 V

a) Calculate the base current ( Ib )

Ib = Sb / Vb = ( 50 * 10^3 ) / 240

= 208.33 A

b) calculate base impedance ( Zb )

Zb = Vb^2 / Sb = ( 240 )^2 / 50000

= 1.152 Ω

c) Calculate the actual impedance ( Zact )

Zact = Zp.u * Zbase

= 0.03 * 1.152 = 0.03456 Ω

d) Calculate the short circuit current ( Isc )

Isc = Vp.u / Zp.u

= 1 / 0.03 = 33.33 p.u

therefore Isc = 33.33 * Ib

= 33.33 * 208.33 = 6943.64 A