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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?

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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.

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  • \$\begingroup\$ Thanks for the response. For primary side, I was referring to the coil resistance. Because if you use P= I^2 * R on the primary, you get a different value than P=VI for the primary. I am thinking that P=VI is the overall power but the power actually being dissipated (not returning to the source) would be P=I^2 * R. Which is the effective portion of the power since it does not return to the source. Is this correct? \$\endgroup\$
    – dude
    Commented Aug 4, 2014 at 6:28
  • \$\begingroup\$ The coil resistance can be modeled as resistance in series with the primary winding. If the primary side current is Ip, the voltage drop on this resistance will be IpR. If the source voltage is Vs, the primary voltage will be Vp=Vs-IpR. The power IpIpR will be lost on the winding as heat. Thus, if you take Vp and Ip as primary side voltage and current, and rest of the things I mention in my message does still hold. Therefore, IpIpR is not the power transferred through the transformer, but it is the winding loss (copper loss). It is not related with the action of transforming. \$\endgroup\$ Commented Aug 4, 2014 at 6:51
  • \$\begingroup\$ Everything you say so far makes sense. I am still unclear about one part though. On the Primary side isn't there a power factor that must be calculated in order to know how much 'real' power and 'reactive' power is being transferred? I may have been using incorrect terminology but I was referring to 'real' power when I said 'effective' power. So what if you take efficiency measurements using just the 'real' power instead of the 'apparent' power? Is it correct since the reactive power is returned to the source (ideally)? Thanks. \$\endgroup\$
    – dude
    Commented Aug 4, 2014 at 14:39
  • \$\begingroup\$ There are two different powers here. One is being transferred through the transformer. The other is being wasted on the primary winding before it is transferred. You are confused at this point. Consider a transformer with no load at secondary. It drains only a tiny amount of current at primary due to its finite primary inductance. So, no power is transferred, almost no power is wasted. Now we connect a load to the secondary. The load drains some current; this current causes also current drain at the primary (Ip). (continued...) \$\endgroup\$ Commented Aug 4, 2014 at 18:54
  • \$\begingroup\$ This Ip passes from the copper wire and it causes some power loss on it. But it also contributes power transfer from primary to secondary. These two things are different matters. Yes, some power loss occurs, but it has no direct relation to the power being transferred. If you make your design with thick enough wires, this power loss will be minimal while the transferred power still remains the same. Check this and this questions. \$\endgroup\$ Commented Aug 4, 2014 at 18:58

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