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If I use \$P = I^2R \$ for primary and secondary will that tell me the effective power being dissipated? Uf I divide secondary power by primary power and multiply by 100%, would that tell me the efficiency of the transformer?
I haven't considered the voltage or phase angle of the primary but since the inductive impedance is allowing the power to return to the source is it okay that I do not consider it in my efficiency calculation?
Assuming that the magnetizing current on the primary side is negligibly small enough, first define:
Pp: Primary side power applied to the transformer.
Vp: Primary side voltage.
Ip: Primary side current.
Ps: Secondary side power applied to the transformer.
Is: Secondary side current.
Rs: Secondary side load resistance.
(All values are effective values.)
If I use P = (I^2)*R for primary will that tell me the effective power being dissipated?
What is R doing on the primary side? Do you mean the reflected R seen from the primary side? If so, the reflected load is \$R_{ref} = \left(\frac{N_p}{N_s}\right)^2R\$; where Np and Ns are number of turns on the primary and secondary windings. So, the primary side power consumption will be (for an ideal transformer):
$$ P_p = V_pI_p = I^2_pR_{ref} $$
If I use P = (I^2)*R for secondary will that tell me the effective power being dissipated?
Yes. \$P_s = V_s I_s\$ is true.
if I divide secondary power by primary power and multiply by 100%, would that tell me the efficiency of the transformer?
Exactly. That's the definition of the transformer efficiency.
$$ \text{Efficiency} \overset{\triangle}{=} \dfrac{P_s}{P_p} $$
inductive impedance is allowing the power to return to the source
You should consider that, transformers work differently than inductors. An idealistic transformer whose primary side inductive reactance is very high, does not behave as inductive at all. Because, the transformer is designed such that the primary side inductor drains only a tiny amount of current from the supply at its rated frequency when the secondary side is open.
