I've been googling it for hours and frustratingly all I could find are vague hand-waving explanations. I get that \$\textbf{S}=\textbf{VI}^\textbf{*}\$ is useful because \$Re(\textbf{S})= P_{avg}\$, which is the average power used by the load. But how exactly does \$Im(\textbf{S})\$ relate to the power that goes back and forth?
Mathematically, how does \$Im(\textbf{S})\$ relate to \$P=V_{rms} I_{rms} cos(2\omega t +\theta_V +\theta_I)\$?
Best Answer
I quess some images can be useful. The wanted formula is at the end.
V1 is 100V peak sinusoidal voltage f=50Hz. There's 100 Ohm resistor R1 and 0.5H inductor L1 as load. The load current is about 530mA peak and its phase lags from V1 as it should be with inductive load.
In the next image the red curve is the momentary power from the source to the load i.e. the voltage of node 1 multiplied by the current of R1:
We see that the power is flowing in 50% shorter cycles and it is a part of the time negative. Negative power means that the load sends energy back to the source. That energy was stored in the magnetic field but it was not dissipated.
Most of us can surely believe that reactive power is the power which goes from source to the inductor and back. But without a formula that's only qualitative definition. It's wrong to say it's the negative part because there's all the time some dissipation in R1, too. The red curve is only the net flow from the source, one cannot easily see from it the reactive power. The real power can be seen easily. It must be the average of the curve.
Real power and reactive power can be calculated with the phasors but they can be also drawn in the simulator:
The upper curve is the dissipated power in R1, it's the voltage over R1 multiplied by the current. The peak dissipated power is 28.5 Watts.
The lower curve is the power into L1. It's half of the time negative because L1 returns all it has received. The peak value is in this case about 22.5 watts.
But how the red curve is related with the usual average reactive power which can be calculated with RMS phasors as Im(S)? The result is surely somehow surprising. The peak of the red is equal with the reactive power.
The lack of any numeric multiplier can be proven with trigonometric formulas. The red curve is the product of the current and the voltage of L1. Those quantities have 90 degrees phase difference. If we discard the common part of their phase angles we can see that the red curve is (Up)(Ip)cos(at)sin(at) where Up and Ip are the peak voltage and peak current and a is 2Pi*frequency.
cos(at)sin(at) is equal with sin(2at)/2 so the red curve has peak value UpIp/2. But that's the product of RMS values of inductor's voltage and current and that's the reactive power.
As a conclusion we can formulate the next rule: The absolute value of the reactive power Q=Im(S) (sign omitted) is physically the peak value of the total energy flow (=watts) from the source to the fields of the reactive components of the load when flow to inductances is considered to be positive and flow to capacitors is considered to be negative.
One should note that the load is handled as one. The input reactive power contains zero info of what happens inside the load. There can for ex. be resonant circuits where substantial energies oscillate between inductors and capacitors, but the total input reactive power can still be zero.