I am confused as to the power of the primary and the secondary of a transformer.
An elementary textbook says the power on the primary and the secondary are equal (and therefore there is no power gain or loss, in the ideal case) because \$V_PI_P = V_SI_S\$, but it seems to me that it is ignoring phases. The power is \$VI\$ only when the voltage and current are in phase. In actuality the power on the primary is \$V_P I_P \cos\theta_P\$ where \$\theta_P\$ is the phase difference between \$V_P\$ and \$I_P\$ and the power on the secondary is \$V_S I_S \cos\theta_S\$, where \$\theta_S\$ is defined similarly. So the above equation holds only when the phase difference between the voltage and the current are the same on the primary and the secondary (\$\theta_P = \theta_S\$). Is this always the case?
What I'm also not sure is what is meant by 'power.' Is it 'power consumed'? An ideal transformer cannot consume power because it is purely inductive. If you are talking about a resistive load on the secondary, then there would be another resistive load on the primary, and so the power consumed is doubled if the loads are equal (in the case of 1:1 transformer)? Well, I'm not sure how general the situation I'm talking about, and I might be missing something as well. Clarifications will be appreciated.