Using the Short Circuit Test on a Single Phase Transformer?

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SUMMARY

The discussion focuses on performing a short circuit test on a single-phase transformer with primary voltage of 5V and primary current of 2A. The primary/secondary turns ratio is calculated to be 5, while the total winding resistance referred to the primary side is determined to be 2.5 Ohms. Additionally, the total leakage inductance referred to the primary side is calculated to be approximately 9.36 mH. The calculations utilize key equations such as P = IV and Zeq = Vsc/Isc for accurate results.

PREREQUISITES
  • Understanding of transformer principles and ideal transformer equations
  • Familiarity with electrical power equations (P = IV)
  • Knowledge of impedance calculations (Zeq = Vsc/Isc)
  • Basic understanding of inductance and its relation to frequency
NEXT STEPS
  • Study transformer turns ratio calculations in detail
  • Learn about short circuit testing methods for transformers
  • Explore the relationship between resistance and impedance in AC circuits
  • Investigate leakage inductance and its effects on transformer performance
USEFUL FOR

Electrical engineering students, transformer design engineers, and professionals involved in power systems analysis will benefit from this discussion.

mikehsiao789
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Homework Statement


A short circuit test is performed on a small single phase transformer. The results are:
Primary Voltage 5V, 60Hz
Primary Current 2A
Secondary Current 10A
Primary Power 8W
a) Making suitable approximations deduce the primary/secondary turns ratio (single number)?
b) Making suitable approximations deduce the total winding resistance in ohms referred to the primary side?
c) Making suitable approximations deduce the total leakage inductance in mH referred to the primary side?

I have the calculations, but I am not sure if I am doing them correctly. It would be awesome if someone can help me! Thank you so much :)

Homework Equations


Power = (Isc)^2 * (Resistance viewed from the primary)
Zeq= Vsc/ Isc
X= sqrt(z^2-r^2)
(current of the primary) / (current of the secondary) = 1/a *for an ideal transformer*

The Attempt at a Solution


For a) (current of the primary) / (current of the secondary) = 1/a, hence a= 5 // I know this is for an ideal transformer...

b) P = IV = Isc^2*Req => 10V/ 4A = 2.5 Ohms. not sure if I should use the 8W primary power given instead...
c) Zeq = Vsc/ Isc = 5/2 = 2.5 ohms =>X= sqrt(z^2-r^2) => X = sqrt( 2.5^2+2.5^2)=3.53 =X
X= 2*pi*60 *L => L = 9.3636mH
 
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mikehsiao789 said:
b) P = IV = Isc^2*Req => 10V/ 4A = 2.5 Ohms. not sure if I should use the 8W primary power given instead...
V/I gives the magnitude of the impedance, so includes both R and X. Using the power equation limits the result to the real component, i.e., the resistance.
 

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