Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Are there any such things as real and reactive energies just like real and reactive power? if so, how is reactive energy dissipated?

share|improve this question

Energy is just power integrated over time, so real and reactive energy exists or not just like real and reactive power, respectively.

As for power, real power exists and reactive power is a mathematical convenience to simplify expressing certain things. By using the mental shortcut of imagining imaginary power, we can simplify calculations and explain real observed parameters more easily than without.

Imaginary power doesn't exist, but the effects of it projected back to real power are real. Large scale power grid producers and consumers are often rated in both the instantaneous real and imaginary power they are producing or consuming. However, the same real observable characteristics can be explained other ways. Explaining them in terms of imaginary power is merely a mental and mathematical convenience.

share|improve this answer
So is the use of imaginary power a mathematical simplification that is hiding real physical phenomena behind it that could all be explained using real-number physics (ie things we could directly measure given the correct instruments)? – Gonzik Dec 26 '13 at 5:30
@Gonzik, AC (phasor) analysis "pretends" that the circuit is excited with sources of the form \$e^{j\omega t}\$, which is unphysical, so that we can just keep track of the complex voltage and current variables that carry the amplitude and phase information. However, it is easy to show that this pretense gives physical results if, at the end, we add back the time dependence and then take the real part of the solution. Now, we could do this in the time domain using real valued variables but, since we need to keep track of both amplitude and phase, it's far easier to do this with phasors. – Alfred Centauri Dec 26 '13 at 15:51
Thanks for that AlfredCentauri – Gonzik Dec 26 '13 at 22:02

First of all, remember that in the context of AC (phasor) analysis, real and reactive power, unlike voltage and current, are not phasors, i.e., they do not represent the amplitude and phase of a sinusoid in the time domain. Thus, we cannot "tack on" the time dependence and take the real and imaginary parts to calculate the associated energies in time domain.

Sometimes it is helpful to "go back to basics" to gain insight into a problem. This is such a case. Reactive power is a useful concept in AC analysis but what it represents physically is best seen in the time domain.

First, consider a sinusoidal voltage source \$v_s(t) = V\cos\omega t\$ driving a resistor R. The power delivered to the resistor is:

$$p_R = \dfrac{v^2_s(t)}{R} = \dfrac{V^2\cos^2\omega t}{R} = \dfrac{V^2}{2R}(1 + \cos2\omega t)$$

The key observation here is that the power is never negative, i.e., the flow of energy is from the source to the resistor always. Thus, the energy supplied by the source increases over time.

The energy supplied by the source over a period \$\dfrac{\pi}{\omega}\$ is:

$$W_R = \dfrac{\pi V^2}{2\omega R}$$

Now, replace the resistor with a capacitor. The power delivered to the capacitor is:

$$p_C = v_s(t) \cdot i_C = V\cos\omega t \cdot (-\omega C)V\sin\omega t = -\dfrac{CV^2}{2}\sin2\omega t$$

The energy supplied by the source over a period \$\dfrac{\pi}{\omega}\$ is:

$$W_C = 0 $$

The key observation here is that the power is alternately and equally positive and negative, i.e., the flow of energy is back and forth between the source and the capacitor. Thus, the energy supplied by the source over a period is zero.

But, as we know, the power associated with a capacitor is reactive power in phasor analysis, and now we can answer your question:

Are there any such things as real and reactive energies

We have shown that reactive power is associated with an alternating energy flow between the source and load that is zero over a period. In other words, it is associated with the energy that "sloshes" back and forth between the source and load without any dissipation.

share|improve this answer

how is reactive energy dissipated?

Reactive energy has no definition that I'm aware of but, it could be taken to mean the energy interactions between a power source and a reactive component. If that power source is a sinusoidal AC voltage and an inductor is connected to that source, energy flows into the inductor then out from the inductor as the AC waveform alternates. The average energy is zero i.e. the same flows "in" that flows "out" but, if a small resistor were inserted in series with the inductor both forward energy (into the inductor) and reverse energy (back from the inductor) would cause some of that energy to be dissipated as heat in the resistor.

share|improve this answer

Is there something more to Reactive Power than a mathematical convenience?

Reactive Power in the News (New York Times, Sept. 26, 2003):

“Experts now think that on Aug. 14, northern Ohio had a severe shortage of reactive power, which ultimately caused the power plant and transmission line failures that set the blackout in motion. Demand for reactive power was unusually high because of a large volume of long-distance transmissions streaming through Ohio to areas, including Canada, than needed to import power to meet local demand. But the supply of reactive power was low because some plants were out of service and, possibly, because other plants were not producing enough of it.”


share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.