# Power calculation of a transformer using only current and resistance

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.
(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.

$$\text{Efficiency} \overset{\triangle}{=} \dfrac{P_s}{P_p}$$