Question
Download Solution PDFShort circuit test is performed on a transformer with a certain impressed voltage at rated frequency. If the short circuit test is now performed with the same magnitude of impressed voltage, but at a frequency higher than the rated frequency then the magnitude of current:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFShort circuit test:
- A short circuit test in the transformer is conducted on the high voltage side and the low voltage side is short-circuited.
- In this test, we need to increase the applied voltage gradually towards rated voltage up to rated load current passing in the secondary short-circuited winding by keeping the frequency constant.
- In HV winding, the rated current is less than LV winding. As rated current is less on the HV side, it is convenient to conduct this test on the HV side by short-circuiting the LV terminals.
- This ensures that the low range of meters can be used for conducting this test as the wattmeter is connected to the primary side (High voltage side).
- Therefore, the primary side is generally chosen as the primary side.
- A short circuit test can be performed on either side but generally performed on the HV side.
Power Factor (cos ϕ) = R / Z
Where,
R is the equivalent resistance
Z is the equivalent impedance
Z = R + j 2π f L
f is the supply frequency
L is the equivalent inductance
⇒ cos ϕ = R / (R + j 2π f L)
Power factor during the short circuit test is inversely proportional to supply frequency
Current (I) = V / Z
Where,
V is the supply voltage
⇒ I = V / (R + j 2π f L)
The magnitude of current during the short circuit test is inversely proportional to supply frequency
Therefore as the frequency of a transformer increases both power factor and current decrease.
Last updated on Jul 2, 2025
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