JimB said:Nigel Goodwin said:Do the maths! - "dissipation = voltage across capacitor x current through capacitor", capacitors aren't some 'magic' component where standard rules don't apply :lol:
Sorry Nigel but you are way off on this one.
Dont forget that in a capacitor the voltage and current are at 90 degrees to each other, a phase angle of 90deg.
In an AC circuit Power = V x I x Cos(phi)
Here phi is 90 deg so Cos(phi) = 0
So no power dissipated in the capacitor.
Nigel Goodwin said:Capacitors used to pass high currents DO get hot, and their failure is a common problem.
Ok, above I was referring to perfect capacitors, in practice there is a VERY SMALL Equivalent Series Resistance (ESR) to account for losses in the capacitor but for most capacitors it is very small.
However, with capacitors like big electrolytics the ESR is a bit higher, if they are passing large currents they warm up and start to age, the ESR will increase so they get hotter and eventually just give up or explode.
JimB
How does power factor relate to this issue i.e. the efficiency of the circuit in question (imagine the led reaaaaaally big!)
Also to Ron H how did you arrive at 680nf I tried using formula 1\2 pi f Xc = C but only got 265nF????