Let's assume you have a non dissipative circuit represented by a box: it involves only capacitance and inductance for example.

You drive this box via a sinusoidal voltage generator.

I call $U(t)$ and $I(t)$ the voltage and total current in receptor convention.

At some instant of the evolution, you might have: $P(t)=U(t)I(t)>0$ (the box receives some power) and at some other: $P(t)=U(t)I(t)<0$: (the box releases some power).

For example, we can consider a capacitance only in this box and I drive it via $U(t)=U_0 \cos(\omega_0 t)=Q/C$

$$P(t)=U(t)I(t)=C U(t) \dot{U}(t)=- C U_0^2 \omega_0 \sin(\omega_0 t) \cos(\omega_0 t) = -\frac{ C U_0^2 \omega_0}{2} \sin(2 \omega_0 t) $$

We see that sometimes the power is positive, sometimes it is negative, $0$ in average.

My question: When the box releases some power, does this power really goes back in the generator and then to the electric network ?

Or it could be the case conceptually but in practice it won't be true: i.e the power that goes back in the generator is not recycled with "standard" generator used in experiments. So it is dissipated somehow.


1 Answer 1


This oscillating reactive power is "converted" to some extent into real power due to resistors in transformer coils and power lines. This decreases the capacity to transfer real power. So the goal is to compensate this effect as close as possible near the source of this reactive power, called power factor correction.

  • $\begingroup$ So in the end, would you agree that without specific "recycling system" of the power that gets back in the generator, it will always get lost because of some resistor inside of it. So when you provide power to a reactive circuit, the power it gives you back is, excepted for some "specifically designed" generator always lost. $\endgroup$
    – StarBucK
    Feb 2, 2020 at 13:07
  • $\begingroup$ Yes, the loss is due to the oscillating current between ps and LC circuit, if that current flows through real resistors. Also power could be lost due to electromagnetic antenna effects. $\endgroup$
    – xeeka
    Feb 2, 2020 at 13:24

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