Electronic – Calculating instantaneous and average power

Real power is based on the difference in phase between the voltage and the current waveforms. You mainly see it in calculation of power factor. From wikipedia:

The ratio between real power and apparent power in a circuit is called the power factor. It's a practical measure of the efficiency of a power distribution system. For two systems transmitting the same amount of real power, the system with the lower power factor will have higher circulating currents due to energy that returns to the source from energy storage in the load. These higher currents produce higher losses and reduce overall transmission efficiency. A lower power factor circuit will have a higher apparent power and higher losses for the same amount of real power. The power factor is one when the voltage and current are in phase. It is zero when the current leads or lags the voltage by 90 degrees. Power factors are usually stated as "leading" or "lagging" to show the sign of the phase angle of current with respect to voltage.

In your application I believe they are using shiftedV as a way to see how much the phase has changed since the last value. With a purely resistive load, the power factor would be 1...so the real power=apparent power. So you are probably suppose to use a resistive load to determine your phasecal number(set it so real=apparent). Then you can find the power factor for various loads.

Edit: If you take the time average of the instantaneous product of voltage and current, you will surely have a real power measurement. As I (poorly) implied above, I believe the shiftedV is used to adjust for any reactance elements introduced by the measurement process. As a way to null out the effects of the electronics, so you get the real power of the load. Here good reading and pictures National Instruments

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!