Denis Allard said:

A 1-farad cap connected to a 1-volt supply will store 1 coulomb and 1

coulomb/second = 1amp and 1amp passed through 1ohm/second = 1 joule and 1

joule/second = 1watt

Question: if the input is 24volts and is pulsed at 60hz., how will this

affect the standard formula: W=1/2xQxU

or W=1/2xCxU2 where QorC is the charge in one of the plates in coulombs or

capacitance,W is the stored energy in joules and U is the voltage between

plates. Urgent need

In addition to what else has been posted, it is not true that one

farad charged to one volt holds one joule of energy. (Perhaps you

didn't exactly say that in the first paragraph, but many might

interpret it that way...) Remember, as the capacitor discharges, the

voltage drops, and the energy is thus 1/2 joule.

One thing to realize about energy coming into and going out of a

capacitor is that if you are not very careful, much of the energy will

be dissipated as heat. If you connect a 1 farad capacitor, charged to

1 volt, to an identical capacitance with no charge, charge must be

conserved, so 1 coulomb is shared by the two capacitors: each is

charged to half a volt, and the energy stored is 2 capacitors times

1/8 joule per capacitor (half a volt, half a coulomb). You've lost

half the energy to radiation (heat and/or RF radiated energy). You

can answer pretty much any question about an ideal capacitor knowing

that i=C*dv/dt (or more accurately, i=C*dv/dt+v*dC/dt). That, coupled

with some basic formulas about charge and energy, should cover the

ground you need.

Cheers,

Tom