And as I previously posted, you don't need to know the values of R and C f you know the voltage to which the cap is charged. It's an either-or situation: If you know V, you don't need to know R and C. If you know R and C, you don't need to know V (except for voltage breakdown and safety considerations).Chippie said:Ahh now I see......I re read you post and saw the reference to R/C...
As I previously posted, we need to know the value of the cap and the resistor or load.......then I concurr with what you wrote 8)
It would have helped this discussion if Petersmith has explained his requirements more clearly. It always pays dividends to read and re-read a question before posting it.Ron H said:Kinjalgp, that's wrong. The units of the product of R and C is seconds. If RC=1, then that is a one second time constant. Petersmith says he wants to deliver energy for one time constant, not one second. Suppose R=1 kohm and C=1 uF. Then RC (one time constant)=1 millisecond. Does RC=1 in this case?
If we want to deliver energy for one time constant, then in the equation you posted, which I agree is correct, t=RC, or, as I said, t/RC=1.
I do want to clarify the meaning of VF. In this case, I have assumed that we are discharging the cap through a resistor whose other end is connected to ground (while the switch is closed). In this case then, VF=0, and the equation simplifies to the equation I posted: