capacitor & voltage source power transfer

Question

When I have a capacitor connected to an AC voltage source, we can get from i=C*dv/dt the current through the capacitor (therefore through the voltage source). Than we can multiply the voltage across it with the current and we get: Power_over_capacitor(t) = 0.5i_m*v_m*sin(2wt) During half cycly it gets power from the voltage source. I dont understand however what does the negative power mean? when the capacitor discharges - where does the power go to? The voltage source, as far as I understand, always provide power. So how can it get the power from the capacitor and where does it go to?

Answer 1

It may be a voltage source but it can still take reverse current and the "negative power" goes back into the voltage source because the current is reversed. In a real supply this may not be possible of course (think voltage regulator) but, in something that behaves like an ideal voltage source, then energy is returned.

Answer 2

+1 to Andy. The capacitor stores energy (e.g. \$\frac{1}{2}Cv^2)\$ during first \$\frac{1}{4}\$ cycle, then gives it up during the next \$\frac{1}{4}\$ cycle. This is just how capacitors do business.

In this plot below (from simple simulation of your circuit) the red trace is voltage across the cap, green is power,\$v(t)\times i(t)\$.

And here is a plot of the power (blue trace) with the energy (red trace). This is probably what @analogsystemsrf is thinking of with his answer (energy, not power).

Answer 3

If I recall rightly, sin(aT) * sin(aT) is

• 0.5 [ 1 + sin (2aT) ]

so there is no negative result.