# Power transferred in a transformer Power absorbed from the voltage source in a circuit is $$=V_{rms} \ i_{rms} \ \cos(\phi)$$ where $$\phi$$ is the phase factor.

So in the primary circuit power absorbed from the voltage source is$$=V_{rms} \ i_{rms} \ \cos\left(-\frac{\pi}{2}\right)=0$$

So if the power absorbed from the voltage source is $$0$$ so then how can power be transformed to the secondary circuit?

• why do you think that $\phi = -\pi/2$? – hyportnex Apr 9 at 13:30

## 3 Answers

In an ideal transformer (with no resistance), that has an open circuit on the secondary, the current in, is limited by the reactance of the primary, and is out of phase with the voltage. There is no power in or out. If current flows in the secondary, that changes the flux through the magnetic core. That shifts the phase of the voltage in the primary relative to the input current. Now you are putting power in to match the power coming out on the secondary. With resistance, you are losing some power in each coil, including some in the primary even if there is no load.

So if the power absorbed from the voltage source is 0 so how then power can be transformed to the secondary circuit?

The power can be transferred to the secondary circuit precisely because it was not absorbed by the primary! Power that is absorbed is power that comes in but does not leave. In order for the power to be transferred to the secondary the power must come in to the primary and leave it and go to the secondary. Therefore it cannot be absorbed by the primary.

The power absorbed from the voltage source can never be 0 from the circuit that I am seeing because that's a closed circuit. Even when no power is transformed to the secondary circuit, the resistance of the primary circuit is still consuming power from the voltage source as IxIxR losses.

All the mathematics is inconsequential so long as the voltage source is a closed circuit.