Electronic – Transformer short circuit test


Why is the power factor obtained by performing a short circuit test on a transformer higher than the power factor obtained in an open circuit test?

Best Answer

Answer by friend @Andy is nice. Just to brief it:

  1. Short Circuit

    • In short circuit test we have to pass rated current through the short circuited side.
    • Therefore we select High Voltage (HV) side normally for short circuiting, because it's rated current is less and therefore it is easy to short.
    • Since the winding are short circuited it takes very less voltage to let the rated current flow through the secondary side, i.e., short circuited winding
    • When the voltage is less, the flux produced due to it, is less
    • Therefore, flux linking with core, 'phi' is very less
    • Hence, cos(phi), that is the power factor, is more. This is also the reason for "When we connect watt-meter in SC test it gives us approximately Copper loss only and Iron loss is negligible."

Since, flux linkage is negligible therefore, iron loss is negligible and hence whatever power is consumed, almost all of that is used to meet Copper loss.

  1. Open Circuit

    • In open circuit test we have to apply rated secondary voltage
    • Therefore, we generally select Low Voltage side for secondary
    • In Transformer, Primary Current Ip, is the sum of no load current I0 and the current due to effect of ‘load on secondary' I2".
    • I0 is primary no load current and is approximately 5% of rated primary current, which is very small value
    • I2" is dependent on secondary current I2 and since, secondary is open circuited, therefore I2 = 0 and this implies I2" = 0
    • Thus Ip, which is sum of I0 and I2", is very small
    • Therefore, Copper loss is negligible.
    • And since we have applied rated voltage, therefore normal flux linkage takes place and inductance of circuit increases.
    • This gives low power factor and also Iron loss is shown on the watt-meter (approx.)