Given Information:
Primary secondary voltage ratio = 2500/250 V
Short circuit voltage = Vsc = 100 V
Short circuit current = Isc = 110 A
Short circuit power = Psc = 3200 W
Required Information:
Series impedance = Zeq = ?
Answer:
Series impedance = 0.00264 + j0.00869 Ω
Step-by-step explanation:
Short Circuit Test:
A short circuit is performed on a transformer to find out the series parameters (Z = Req and jXeq) which in turn are used to find out the copper losses of the transformer.
The series impedance in polar form is given by
Zeq = Vsc/Isc < θ
Where θ is given by
θ = cos⁻¹(Psc/Vsc*Isc)
θ = cos⁻¹(3200/100*110)
θ = 73.08°
Therefore, series impedance in polar form is
Zeq = 100/110 < 73.08°
Zeq = 0.909 < 73.08° Ω
or in rectangular form
Zeq = 0.264 + j0.869 Ω
Where Req is the real part of Zeq and Xeq is the imaginary part of Zeq
Req = 0.264 Ω
Xeq = j0.869 Ω
To refer the impedance of transformer to its low voltage side first find the turn ratio of the transformer.
Turn ratio = a = Vp/Vs = 2500/250 = 10
Zeq2 = Zeq/a²
Zeq2 = (0.264 + j0.869)/10²
Zeq2 = (0.264 + j0.869)/100
Zeq2 = 0.00264 + j0.00869 Ω
Therefore, Zeq2 = 0.00264 + j0.00869 Ω is the series impedance of the transformer referred to its low voltage side.
