Electronic – How to calculate instantaneous active power from sampled values of voltage and current.

I think that trying to use RMS values of voltage and current for this is not going to work. Imagine shifting the current waveform 4ms later; neither the RMS voltage nor the RMS current would change at all, but the power drawn would change by an order of magnitude.

The instantaneous power drawn by your circuit is V * I. In any small time dt, the energy consumed will be V * I * dt. The energy consumed in 1s, the power drawn, will be the integral of V * I * dt from T=0 to T=1s. You could compute this directly from the sample values in your excel spreadsheet. At each time sample, multiply the instantaneous voltage by the instantaneous current, and that gives the instantaneous power drawn. Multiply that by the sample interval, and that is the energy consumed in that sample interval. Add all those up over an AC cycle, and multiply by the number of cycles per second and that is the energy drawn per second, otherwise known as the power.

Looking at the scope traces, the current drawn by the circuit is usually 0. Once per AC half-cycle, the current increases to about 90mA very quickly, then drops linearly to 0 over about 820us. It's a 60Hz circuit so it does this every 8.3ms. When the circuit is drawing current, the voltage is more-or-less constant at 170V. That's an average current of 45mA over the 820us at 170V = 7.65 W, but it only takes this power 1/10 of the total time, so the final power consumption is 0.76 W.

In my experience, the probability of wiring up a current probe backwards is exactly 0.5!

Any help would be appreciated

You can measure phase-neutral voltage and if the supply and load are balanced then you can infer line voltage by: -

line volts = phase volts\$\times\sqrt3\$

This should be one of the easier measurements yet you say you can calculate "real power, reactive power and apparent with ease". This does make me think that you have used one of these measurements, and the RMS measurement of current and back-calculated phase/line voltage and that's when you are seeing a discrepency.

If this is so then I suspect your current measurement may be at fault either through incorrect use of a current transformer or some ratio being incorrect. Or, it could be the \$\sqrt3\$ thing mentioned above.

If you need any further help with this you should consider detailing how you make the other measurements and why you believe voltage to be incorrectly calculated/measured.