Copper Loss in Transformer
This question (and answer) covers both copper losses and iron losses in a power transformer. Here is a slightly modified picture taken from my answer to that question: -



It can be a little difficult to distinguish between copper losses and iron losses because the equivalent circuit of a transformer has the primary copper loss component (Rp) in series with the iron loss (Rc). In effect with no load, Rp is in series with Rc across the primary supply voltage.

However, a technique is used that can pretty much split the two apart. Iron losses are measured at full supply voltage with the secondary open circuit and Rc can be derived. In fact you are deriving Rc + Rp but Rc will dominate because the primary current (secondary open circuit) is so small on power transformers and probably in the region of a hundred milli amps. Compare this with full load current for a 1kW transformer on 220VAC - primary current will be \$\frac{1kW}{220V}\$ = 4.55A. 

To make copper-loss measurements even more exact, a much smaller input primary voltage is used and the secondary winding is shorted. Care is taken to ensure that the reduced primary voltage does not cause the shorted secondary current to exceed its normal full-load rating. For the sake of this question, we could say that the reduced primary voltage is going to be more like 11VAC (i.e. 5% of normal).

This would make the primary current (due to iron losses) more like 5mA and this can be ignored. 

Now, a wattmeter measures power into the transformer (shorted secondary) at 11VAC and the power it measures can be assumed to indicate copper losses. Remember, the currents in and out of the transformer are representative of full load because the secondary is shorted.

If you know the primary current (shorted secondary) and input power (wattmeter) you can derive \$R_P + R_S(\frac{N_P}{N_S})^2\$ because power = \$I^2\times R\$. So you arrive at an equivalent resistance that represents copper losses.