# Homework Help: What's the physical meaning of the reactive power signs?

1. Jan 12, 2012

### maCrobo

I wonder what really happens to powers in capacitors and inductors when they are connected in a circuit either in series or in parallel.
I know that inductors have a positive reactive power while capacitors a negative one, so I imagine inductors release energy while capacitors absorb it, but it sounds strange to me that they always do that without reversing their behavior so to have the inductors that absorb and viceversa. It's very improbable to me that a capacitor absorb energy continuously, I mean, at a certain point it should be full.
Anyway, I don't know...

2. Jan 12, 2012

### Staff: Mentor

Reactive power is associated with AC circuits, so voltage, current and power are in a state of constant change. The capacitor does absorb energy, but only for part of the cycle, then it releases it. The inductor does likewise, except it is 180 degrees behind the capacitor.

In a parallel resonant circuit, the energy 'sloshes' backwards and forwards between the capacitor and the inductor. The amplitude of the currents in these elements can be many times greater than the current drawn from the source.

3. Jan 12, 2012

### maCrobo

Did you mean power instead of current?

If not: I don't get this point, if I follow Kirchoff's Law the current that goes out from a source can only be divided among the components of a circuit in case they are in parallel. I mean it can't be that the current that get to an inductor is greater than the current provided by the source.

For the rest it's clear: charges go back and forth (sinusoidal sources) so the magnetic field inside the inductor increases and decrease so to increase and decrease the energy stored U=1/2μ B^2 and the same happens for the capacitor where the charge on the plates change so to change U=1/2 (q/C^2)
Thanks!

4. Jan 12, 2012

### Staff: Mentor

No, I did mean current.
That's why I spelled it out, in case you didn't already appreciate it.

The current drawn from the source equals the vector sum of the currents to the parallel elements. So if the capacitor is drawing +10.1A and the inductor -10.0A, the sum of the two is roughly 0.1A

So there is 100 times more current circulating in the L and C than is being drawn from the source.

Hope that clears things up.

5. Jan 12, 2012

### maCrobo

Ok :P

Last question: in the time domain I have a certain function for the power $p\left( t \right)=\frac{1}{2}VI\cos \left( 2wt+ϕ_{v}+ϕ_{I} \right)+\frac{1}{2}VI\cos \left( ϕ_{v}-ϕ_{I} \right)$ from where we define the active power Pa to be $P_{a}=\mbox{Re}\left[ \frac{1}{2}V\; I^{\mbox{conj}ugate} \right]$. Then we also define the reactive power to be the imaginary part of the last equation. I wonder why it has been chosen this definition, I got it under a practical point of view, but not mathematically. My doubt arose from the fact that the Power of a source is defined by $P_{}=\frac{1}{2}V\; I^{\mbox{conj}ugate}$ but the sine part of this equation is not present in the previous and general p(t).
So what?

Last edited: Jan 12, 2012
6. Jan 12, 2012

### technician

An important physics thing to realise is that capacitors and inductors do not dissipate energy, they store electrical energy as an electric field or as a magnetic field and this energy can be recovered as electrical energy.
Resistors dissipate electrical energy as heat which cannot be readily be recovered as electrical energy.

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