[SOLVED] Discharge big capacitor

[Cecemel]

What type of resistor do i need to use for discharging a 10 000 uf capacitor(charged with little over 30 volts) without sparks. How long does it takes then to completely discharge it?

Regards, Cecemel!:razz:

 

[KlausST]

Hi,

With any resistor you can calculate a time constant by: tc = R x C
This tc says how many seconds it takes to discharge the capacitor to about 37% of its initial value.

Example:
R = 1000 ohms.
Tc = 1000 x 10000 uF = 10000000us = 10s
30V x 37% = 12V

With this resistor it takes 10 seconds to discharge to 12 V, another 10 s to discharge to 12 x 37% = 4.4V and so on....

Try it with another resistor value...

Good luck

 

[Cecemel]

Hi,

With any resistor you can calculate a time constant by: tc = R x C
This tc says how many seconds it takes to discharge the capacitor to about 37% of its initial value.

Example:
R = 1000 ohms.
Tc = 1000 x 10000 uF = 10000000us = 10s
30V x 37% = 12V

With this resistor it takes 10 seconds to discharge to 12 V, another 10 s to discharge to 12 x 37% = 4.4V and so on....

Try it with another resistor value...

Good luck

Thanks!
Very helpful formula that i'm gonna use a lot in the future!
But... Is that 1000 ohm resistor gonna give me a spark or not because i don't want to damage my cap:wink:

 

[WimRFP]

You will not get a spark as the initial current is low:

I = U/R = 30/1000 = 30 mA.

When using 100 Ohms, the current will be 300 mA.

If you use a 100 Ohms resistor, use 1 or 2 W resistor. You may damage a 0.25 W resistor. As the peak power dissipation in the resistor is:

P = U^2/R = 30^2/100 = 9 W.

You can also put 10, small 1 kOhms resistors in parallel to make 100 Ohms. This will discharge your capacitor to below 1V in 4 seconds.

Calculation (exponentially decaying function versus time):

V(t) = Vstart*e^(-t/(RC) ) = 30*e^(-4/(100*10000e-6)) = 0.55V.

OT: I like the chocolate drink CeCemel (in the Netherlands Chocomel)

 

[betwixt]

Most caps of that kind of size are safe to a few Amps and almost certainly wouldn't be damaged by being directly shorted out. The current is calculated by the voltage across the capacitor terminals (NOT it's rated voltage) divided by the resistor value in Ohms. Bear in mind that the voltage will drop as soon as the capacitor starts discharging so the maximum rating only applies at the instant you connect the resistor, after that the current will be lower.

Brian.

 

[Cecemel]

Thx everyone for replying