Capacitor Discharge Time: How can I calculate it?

Capacitor

Voltage: 400 Volts
uF: 14

If I want to run a device with that capacitor... How can I calculate the time?
(Device: 250 Watt)
I know the time is in MiliSecs, but I don't know how to calculate it.

Second question:
if I have 3 capacitors like the first one (connected in serie)
A) What is tht total voltage?
B) Total uF?

The time constant (tau) capacitor circuit is resistance * capacitance. This is the amount of time it takes the voltage to fall to 1/e (37%) of the original value.

in your case ..

P = V^2 / R
R = 400^2 / 250 = 640 Ohms

tau = 640 X 14 e-6 = 9 ms = .009 seconds

so after 9 miliseconds the voltage will be 400/e = 147 V
after 18 miliseconds .. 147/ e = 54 ms

Question 2 :

The total voltage will still be whatever voltage you apply to the circuit. The voltage across each capacitor will be 1/3 of this.

The total capacitance will be1/( 1/C1 + 1/C2 + 1/C3)

= 1/(1/14 + 1/14 + 1/14) = 4.7 uF

there will be a time constant if the load is resistive. if it is a constant current load, then just use dv/dt=i/C

if you want to use more caps to get a longer time period of operation, put them in parallel.

Sorry, I meant parallel instead of serie.

How affect the parallel caps connection, to the voltage and the uF?

total capacitance of parallel capacitors is the sum of the capacitances.

i have no idea what you mean by affect on the voltage - since the caps are in parallel, they will have the same voltage across them.

OutToLunch said:

total capacitance of parallel capacitors is the sum of the capacitances.

i have no idea what you mean by affect on the voltage - since the caps are in parallel, they will have the same voltage across them.

Click to expand...

Example:
parallel connections of 4 capacitors:
Each one: 400 Volts 16 uF

I think the total is: 400 Volts 64 uF, right?

Looks right to me.