> > I've also seen C-R series voltage dropping circuits, here & there.
> Correct me if I'm wrong, but doesn't the series cap dissipate power just as > it would, were it a series resistor? I mean, if the LED is passing 20mA, > the cap is also doing 20mA - and at whatever the Vdrop is. > > Right? If not, why? Yes, the capacitor passes 20mA. But unlike a resistor the voltage drop across a capacitor is not in phase with the current. For a pure capacitor, it is out of phase by 90 degrees. For a pure inductor, BTW, it is 90 degrees out in the oposite direction. Now, if you consider the power at any instant, it is, indeed the product of the voltage and current at that instant. For a resistor, this is alwas positive, so if you add up all the instantaneous powers over a full cycle (mathematically, this is integration, of cource), the total power consumed is positive. But for a capacitor, sometimes the instantaneous power is negative, in effect the capacitor is supplying energy back to the circuit. And if you integrate that over a full cycle, you end up with zero. Intuitively : With a resistor, voltage and current are either both positive or both negative, so the product is positive. With a capacitor, due to that phase shift, there are 4 regions to consider : V +ve., I +ve; V +ve, I -ve; V -ve, I -ve; V -ve, I+ve. The first and third of those give a +ve power, the second and fourth a -ve power and they exactly cancel out. It turns out that the power in an AC circuit can be calcualted as the product of the RMS voltage, RMS current and the cosine of the phase angle between them. The RMS values are the ones normally quotes (115V mains had an RMS value of 115V). The cosine(phase) term is known as the 'power factor', and is probably the thing to look up in a book on AC electric circuits. -tony
