Using the Short Circuit Test on a Single Phase Transformer?

In summary, a short circuit test was performed on a single phase transformer with primary voltage of 5V and primary current of 2A. The secondary current was found to be 10A and the primary power was 8W. Using suitable approximations, the primary/secondary turns ratio was deduced to be 1/a, and the total winding resistance referred to the primary side was calculated to be 2.5 ohms. The total leakage inductance referred to the primary side was found to be 9.3636mH. Further calculations were done to determine the impedance and resistance using the power equation.
  • #1

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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  • #2
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.
 

1. What is the purpose of conducting a short circuit test on a single phase transformer?

The short circuit test is used to determine the equivalent circuit parameters of a transformer, such as its resistance and reactance, which are necessary for performing accurate load and no-load tests.

2. How is the short circuit test performed on a single phase transformer?

The test involves shorting the secondary winding of the transformer and applying a reduced voltage to the primary winding. The resulting current is then measured and used to calculate the equivalent circuit parameters.

3. Can the short circuit test be used to determine the efficiency of a single phase transformer?

No, the short circuit test is only used to determine the equivalent circuit parameters and cannot be used to determine the efficiency of a transformer. For efficiency testing, load and no-load tests are required.

4. What are the advantages of using the short circuit test on a single phase transformer?

The short circuit test is a simple and quick method for determining the equivalent circuit parameters of a transformer. It also does not require any additional equipment, making it a cost-effective option.

5. Are there any limitations of the short circuit test on a single phase transformer?

Yes, the short circuit test does not take into account the effects of magnetic saturation and core losses, which can affect the accuracy of the results. Therefore, it should be used in conjunction with other tests for a more comprehensive analysis of the transformer.

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