Electronic – How to find True power dissipated and apparent power

Use these identities :-

\$z = R.e^{j\theta}\$
\$Re(z) = R\cos(\theta) = a\$
\$Im(z) = R\sin(\theta) = b\$
\$z = a + jb\$
\$R = |z| = \sqrt{a^2 + b^2}\$
\$\theta = Arg(z) = \arctan(\frac{b}{a})\$

For example:
\$56.e^{j40} = 56\cos(40) + 56j\sin(40) = 42.9 + 36.0j\$
\$75 - j22 = \sqrt{75^2 + 22^2}.e^\arctan(\frac{-22}{75}) = 78.16.e^{-16.3j}\$

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