In an ideal transformer, the mutual induction caused by the secondary load current in the primary coil generates an additional current that has a non-zero power factor with the main AC voltage source. That's the mathematical explanation of the power transfer from the main AC source to the primary and then to the secondary load resistance. But what is the physical intuitive explanation?

For example If I replace the AC voltage source with a capacitor to make the primary an LC oscillatory circuit, even if the power factor be still there, how is energy being lost to the secondary without a resistance in the primary?

  • $\begingroup$ The power factor has nothing to do with the workings of a transformer. Power in electric circuits is always transferred by the electromagnetic field and it is being described by the Poynting vector (named after Mr. John Henry Poynting): en.wikipedia.org/wiki/Poynting_vector. There are several papers that analyze the geometry of transformers using this concept, e.g. "The Poynting vector field and the energy flow within a transformer", F. Herrmann and G. Bruno Schmid. $\endgroup$ – CuriousOne Apr 12 '16 at 19:31

Power factor will be still there due to formula P=VI even if the resistance is zero. Every inductor has some resistance. The effect of this resistance is to introduce a damping effect on the charge and current in the circuit and the oscillations finally die away. Even if the resistance were zero, the total energy of the system would not remain constant. It is radiated away from the system in the form of electromagnetic waves. In fact, radio and TV transmitters depend on this radiation.

| cite | improve this answer | |

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.