Engineering Using the Short Circuit Test on a Single Phase Transformer?

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A short circuit test on a single-phase transformer yielded primary voltage of 5V, primary current of 2A, and primary power of 8W. The primary/secondary turns ratio was approximated to be 5 based on the current relationship. The total winding resistance referred to the primary side was calculated as 2.5 ohms, though there was uncertainty about using the primary power in the calculation. The total leakage inductance was derived to be approximately 9.36 mH using the impedance and frequency. The discussion emphasizes the importance of distinguishing between real power and impedance in calculations.
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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