A transformer winding is generally more inductive than resistive thus there is a considerable phase difference between voltage and current, so wouldn't the supply power factor be very very poor in case of residential transformers where no means for power factor corrections are used?
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asked Feb 7, 2018 at 17:12
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Unloaded transformers look like inductors. However, a resistive load on the secondary makes the primary look more resistive too. For any decent transformer, the primary looks mostly resistive when the secondary is loaded with a resistance near the power limit of the transformer.
So the power factor is very poor when a transformer is drawing little power, and gets better as the power increases. Fortunately, you care more about the power factor at high power levels.
All that said, the grid does generally look partially inductive. Large industrial motors generally present a more inductive power factor than that added by typical transformers.
Large electric customers that pay extra for a low power factor usually have a system of compensating for the inductive power factor. This is usually banks of capacitors that are switched across the power line to cancel out the inductance.
answered Feb 7, 2018 at 17:26
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\$\begingroup\$ but how does a resistive load on the secondary side has anything to do with the inductance of primary , as there's no direct connection between the primary and secondary windings & the current drawn by primary in any case lags voltage by a considerable amount,as cosΦ =R/Z and R of primary is very very less than Reactance(inductive) \$\endgroup\$ Commented Feb 7, 2018 at 17:37
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\$\begingroup\$ @SubratBavarianBastola, when there's a load on the secondary, energy from the generator is delivered to the load rather than being stored as magnetic flux in the core. \$\endgroup\$ Commented Feb 7, 2018 at 17:40
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1\$\begingroup\$ "how does a resistive load on the secondary side has anything to do with the inductance of primary" By the basic principle of what a transformer does. Resistance is reflected across a transformer by the square of the turns ratio. For example, a ideal transformer that drops 115 V to 12 V with a 3 Ohm load on the secondary will make the primary look like a 276 Ohm load. \$\endgroup\$ Commented Feb 7, 2018 at 17:43
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\$\begingroup\$ okay, so you mean , if a resistive load is connected to the secondary of the transformer then the secondary current lags its secondary voltage by a considerably lesser phase angle Φ ? For example, resistive loads are connected across the secondary side of the transformer, here the secondary side of transformer acts as a localized energy source for the loads , so the power supplied by transformer to loads is V(secondary)*I(secondary) cos Φ , so if the secondary winding is considered more of an inductor than a resistor, won't the active power supplied to the resistive loads be insufficient ? \$\endgroup\$ Commented Feb 7, 2018 at 17:55
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\$\begingroup\$ @Sub: It might help to think of the load presented by the primary as having two parts. One is the inductor it is, and the other is the impedance of the secondary reflected by the square of the turns ratio. At no load, you only have the inductive part. At significant load, the reflected impedance dominates. Due to how the transformer works, the inductive component actually goes down on its own (energy is transferred to the secondary, not stored in the core), in addition to becoming a smaller fraction of the whole due to the reflected load. \$\endgroup\$ Commented Feb 7, 2018 at 18:00