Petersmith
New Member
Let we have a capacitor charged about 300v. we want to deliver energy for one time constant. that means we want to truncate it. does any body has idea how to do so.
Can you be a bit more clear? What does truncating energy over here mean?means we want to truncate it
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:
Vc=VI*e^(-t/RC)
Ron