There is a BIG difference between reactive, active and apparent power and it is why the terms were invented for AC circuits. One must be careful not mixing them with instantaneous power. In AC circuits, power is a function of time t given by
p(t)=e(t)i(t)
where e(t) is the instantaneous voltage and i(t) is the instantaneous current. Note that in DC, everything is constant, so instantaneous power is just power.
Now let us examine 3 basic cases for this instantaneous power:
The Resistor
Let us assume we have a resistor of linear resistance R connected to a voltage source e(t)=E\cos{\omega t} (i.e. a sinusoidal waveform of magnitude E and angular frequency \omega). Since the current in the resistor is in phase with the voltage, we get some current i(t)=I\cos{\omega t} in the resistor. From the definition of p(t) above, we get
p(t)=e(t)i(t)
=E\cos{\omega t}\cdot I\cos{\omega t}
=EI\cos^2{\omega t}
=\frac{1}{2}EI(1+\cos{2\omega t})
(the last step included the identity 2\cos^2{\omega t}=1+\cos{2\omega t})
Hence, we can express the instantaneous power of the resistor as 2 parts: a first part which is independent of time (mean value) and a second part which oscillates at twice the frequency. The active power P is defined as the mean value of the instantaneous power
P=\frac{1}{2}EI
Notice here the presence of the \frac{1}{2}, because E and I are the magnitudes of the cosines, not the RMS value. This is where the definition of RMS kicks in, since we know the power in DC (instantaneous or mean value, all the same) is
P=EI
and we know for sinusoidal waveforms the RMS values are E_{RMS}=\frac{E}{\sqrt{2}} and I_{RMS}=\frac{I}{\sqrt{2}}. Hence, we get
P=E_{RMS}I_{RMS}
for the sinusoidal waveforms of the resistor. We see the equivalence of the AC so-called active power with the power in DC. The RMS values in AC represent the values needed to perform the same amount of work in DC with a resistor. We can express this active power by using Ohm's law in other forms for the resistor
P=E_{RMS}I_{RMS}=RI_{RMS}^2=\frac{E_{RMS}^2}{R}
The Inductor
This case is a bit trickier. Let us assume we have the same voltage source e(t)=E\cos{\omega t} applied to an inductor of inductance L. The current of an inductor is lagging by 90 degrees its voltage, so the current is i(t)=I\sin{\omega t}
From the definition of p(t) above, we get
p(t)=e(t)i(t)
=E\cos{\omega t}\cdot I\sin{\omega t}
=\frac{1}{2}EI(\sin{2\omega t})
(the last step included the identity 2\cos{\omega t}\cdot \sin{\omega t}=\sin{2\omega t})
Here, we see a big difference from the case of the resistor. The mean value of instantaneous power is 0. Which means that energy is transferred back and forth within a cycle (actually twice within a period of e(t)).
Even though the mean instantaneous power is 0, that doesn't mean we don't have current in the source necessary for the inductance (in practice, this current has to flow in the power lines, cables, etc.). Hence, we need a measure of this fictitious necessary power that draws AC current from the source. In a similar fashion as in the resistor (p(t)=P+P\cos{2\omega t}), we can define this reactive power Q as p(t)=Q\sin{2\omega t}, where Q will be again given by
Q=\frac{1}{2}EI
or
Q=E_{RMS}I_{RMS}
These formulas are, of course, for the inductor. Then again, we could replace the last equation by using Ohm's law (in complex form)
Q=E_{RMS}I_{RMS}=L\omega I_{RMS}^2=\frac{E_{RMS}^2}{L\omega}
The impedance
Since the capacitor case is relatively simple, let us now look at the impedance case, where we get an arbitrary phase difference \theta between voltage and current waveforms. Again, let us assume we have the same voltage source e(t)=E\cos{\omega t}. The current of the impedance will be of the form i(t)=I\cos{(\omega t-\theta)}. So we will get for the instantaneous power
p(t)=e(t)i(t)
=E\cos{\omega t}\cdot I\cos{(\omega t-\theta)}
=EI\cos{\omega t}(\cos{\omega t}\cos{\theta}+\sin{\omega t}\sin{\theta})
=EI(\cos^2{\omega t}\cos{\theta}+\cos{\omega t}\sin{\omega t}\sin{\theta})
with the 2 identities mentioned earlier, we finally get
p(t)=\frac{1}{2}EI\cos{\theta}(1+\cos{2\omega t})+\frac{1}{2}EI\sin{\theta}(\sin{2\omega t})
=P(1+\cos{2\omega t})+Q(\sin{2\omega t})
We can see that in the general case (impedance), the active power in the last equation is now \frac{1}{2}EI\cos{\theta} and the reactive power is now \frac{1}{2}EI\sin{\theta}. Since both powers do affect the current that must be drawn from the AC source, there was also a name given for that power, which is the apparent power. Following the complex notation for the impedance, we can formulate this power as a complex power, which real part comes from active power P and the imaginary part comes from Q
\vec{S}=P+jQ
S=\frac{1}{2}EI=E_{RMS}I_{RMS}
**From an AC point of view, this is all very important. In order not to mix these powers, each one has a different unit (even though all equivalent to the SI unit W for power). Active power is in W, reactive power is in var (lower-case), which stands for volt-ampere reactive and apparent power is in VA (volt-ampere). The active power is the power that creates work (lift charges, heating, etc.). The reactive power is stored and restored power (in the form of electromagnetic field). The apparent power is the result of both powers, it is what you will measure in the end, by multiplying the RMS voltage and current. Hence, this is why you usually have ratings for transformers in VA, because depending on what the impedance (connected to it) is, you must not have more than a certain amount of current (RMS) in the conductors (in order for them not to overheat) and a certain amount of voltage (RMS) on the ends of the conductors (in order for the dielectric to withstand this voltage). Of course, there is always a margin for these (steady-state) values, but one has to understand that if there was a transformer rated 10W, that would mean I could connect 10W and whatever reactive power to it (such as 1Mvar). But the currents drawn from that load would melt the conductors right away... This is why the distinction in AC is important.
Sorry for the long post, perhaps only the last part is necessary.
M.
