Let us consider a resistive load R is connected across the secondary of transformer whose impedance is r+jX, so in calculation of the power factor of the supply (here in this case it is the secondary of the transformer ), Will the supply power factor be : cos theta = R+r/Z or will it be r/Z only ? Does supply power factor of the system has anything to do with load impedance ?

  • \$\begingroup\$ It certainly does have a lot to do with load impedance. \$\endgroup\$ – Andy aka Feb 8 '18 at 11:46

The power factor measured at the input side of a transformer is mostly determined by the load. Unless the load is very small, the effect of the impedance of the transformer itself is almost negligible.

The power factor measured at the primary of the transformer is the result of applying the source voltage to the combined impedance of the transformer equivalent circuit and the load.

The power factor at the secondary of the transformer is determined only by the load circuit. If the load changes, the power factor changes according to the change in the components of the load circuit that are responsible for the change in load.

  • \$\begingroup\$ So what in case of the secondary side of the transformer ? Does the power factor measured at secondary side of transformer (which supplies the load) also increases with load ? \$\endgroup\$ – Subrat Bavarian Bastola Feb 8 '18 at 2:27
  • \$\begingroup\$ and what is the formula of the supplied power factor to the load ?? Is it cos theta= Resistance of secondary side of transformer/Impedance of the transformer ? Or is the supply power factor equal to Resistance load and secondary side combined / equivalent impedance. ?!? \$\endgroup\$ – Subrat Bavarian Bastola Feb 8 '18 at 2:29
  • \$\begingroup\$ I am not quite certain what you are asking, but I have added to my answer. \$\endgroup\$ – Charles Cowie Feb 8 '18 at 3:04

See the equivalent circuit of a transformer:


simulate this circuit – Schematic created using CircuitLab

For the usual mains transformer, copper losses R_p and R_s' and stray inductance L_sigma_p and L_sigma_s' are so low they are negligible aside from secondary short circuit condition.

What means your load impedance Z_l is in parallel to iron losses R_Fe and L_µ, the main inductance of the transformer.

Again, for the usual mains transformer, R_Fe and L_µ are so high their influence on primary current is negligible for any condition but secondary open.

  • \$\begingroup\$ so here in this case, if i had to calculate the supply power factor which i believe here is the power factor supplied by the secondary side of transformer, should I calculate it as supply pf - Total Resistance(load+secondary of transformer) /Total impedance or should it be - Resistance of secondary of transformer/Impedance of Transformer secondary ?!? And does- the pf increases with load mean that the current lags the secondary voltage of transformer by considerably lesser phase angle when load is increased ? \$\endgroup\$ – Subrat Bavarian Bastola Feb 8 '18 at 2:35
  • \$\begingroup\$ When you have neither an open nor a short circuit on the secondary, all of the transformer's internal resistances and inductances are negligible. All you have to do is transform your secondary impedance to the primary according to the winding ratio. \$\endgroup\$ – Janka Feb 8 '18 at 9:20
  • \$\begingroup\$ The power factor of a transformer with an open secondary depends on R_Fe||L_µ (it's near zero), the power factor of a transformer with a shorted secondary depends on R_p+R_s'+L_sigma_p+L_sigma_s (it's near one). For all other cases, these are negligible. \$\endgroup\$ – Janka Feb 8 '18 at 9:23

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