2
\$\begingroup\$

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?

\$\endgroup\$
2
\$\begingroup\$

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.)
\$\endgroup\$
2
\$\begingroup\$

In an open circuit test you can be sure that about 90% of the current taken is to magnetize the transformer's core. That current taken is due to the inductance of the primary (as if it were an inductor with no secondary). There are some resistive effects due to eddy current losses in the laminates and hysterisis loss too but these are relatively small compared to the dominant current of the magnetization inductance.

This means the dominant component is an inductor and this makes the power factor quite small.

In a short circuit test, it is the leakage components in the windings that are measured and, these can be quite similar in impedance i.e. dc resistance of secondary is of the same order as leakage reactance of secondary and ditto for the primary. This roughly (and with some hand-waving) makes the power factor round about 0.5 but could be as low as 0.2 and as high as 0.9.

The open circuit test will show a power factor that is likely to be below 0.5 and probably in the realms of 0.3 or less.

I am generalizing here and clearly some transformers (especially those used in flyback converters) have very low magnetization inductance but, for the average 50Hz/60Hz power transformer, the above will be the case.

\$\endgroup\$
  • \$\begingroup\$ I want to know if this reason is right or wrong--> that in open circuit there is current flowing ie magnitizing current in a single coil but when the transformer is short circuited the current now flows from both the coils ( primary and secondary) making power factor more lagging?? \$\endgroup\$ – Sohaib Imran Bhatti Apr 23 '15 at 19:18
  • \$\begingroup\$ No that isn't true - the current that flows on a short circuit test is more balanced between the reactive leakage in the windings and the dc resistance of the windings. \$\endgroup\$ – Andy aka Apr 23 '15 at 19:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.