Question

When no power is drawn in the secondary coil of an ideal transformer, The power factor of the primary coil of an ideal transformer is:​

Answer 1

The power factor of an ideal coil is 0

Explanation:

An ideal transformer is an imaginary transformer which has

- no copper losses (no winding resistance)

- no iron loss in core

- no leakage flux

An ideal transformer will give output power exactly equal to the input power. So, its efficiency is 100%. But this situation cannot be achieved in real. It is used only for modeling purposes.

For an ideal transformer, E1I1 = E2I2 where E is the emf and I is the magnetizing current of the two coils, respectively.

So power factor of an ideal transformer is exactly that of its load. The power factor of an ideal coil is 0 since phase difference between the current and the applied voltage is - π/2.