Electronic – Powering a modem over extreme distance (~2k feet)

You can use a higher frequency than the standard 50 or 60 Hz to reduce the volume and weight of the transformers (for example the aircraft industry use 400 Hz as the standard in airplanes).

Personally I'll don't go beyond a few kHz to avoid all kind of problems but it's just a rough estimation (and you might need a higher frequency than that to pass 2 kW with a 2 kg transformer), do the math to be sure. Also, if the frequency is relatively high for the wire thickness, think about using litz wire for the transformers windings, you lose on the volume but you win on the weight.

To drive the first transformer you can use a H-bridge (or even just a half-bridge with a center tapped primary on the transformer) and a crude astable since the efficiency isn't a concern. You can do better by approximating a sinus with multiple taps on the primary but it gets far more complicated.

On the other side, after the second transformer, just be careful about the switching speed of the diodes if you want to rectify the current (classic rectifying diodes might not be fast enough), use schottky diodes if you want to be sure to have no speed problem ever.

You don't want contact with the mains. OK, so is your project going to be battery powered?

If you buy a low voltage, AC output wall-wart, with UL, CE, DIN and every other safety marking you can think of, you can regard its output as at least as safe as any other appliance in your house. Then, in the low voltage output, you have a reasonable facsimile of the mains voltage waveform. The accuracy will degrade if you also rectify that output to power your project. If the last 1% accuracy is important, then why not buy two, one for the reference, and one to power the project, or a DC output one to power the project without further messing about. Your current measurement is already transformer coupled, why not have the voltage measurement transformer coupled as well?

To confirm real and apparent power calculations ...

for real power, repeatedly compute the instantaneous power as the product of the instantaneous voltage and the instantaneous current many times a cycle, then average the power

for apparent power, compute the mean voltage by averaging, and seperately the mean current by averaging, then take the product of these

... and here an average means either a) average over a long time, the more cycles the more accurate or b) sum synchronously over an integer number of whole cycles