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As above.

Digikey provides a calculator but it didn't really explain the pros/cons of higher resistance (slower discharge) vs lower resistance (faster discharge).

I guess there's the power loss as well but in my calculation the lower loss is not a lot.

Example: (30 V / 1 kΩ) x (30 V) = 0.9 W power loss

Are there factors that needs to be considered? (Like maybe capacitors last longer when it's discharged slowly?)

Thanks

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    \$\begingroup\$ The biggest factor to be considered is why you want to charge and then discharge any capacitor. What's up? \$\endgroup\$
    – jonk
    Commented Dec 26, 2022 at 7:05
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    \$\begingroup\$ You need to first ask yourself how fast does your application require the capacitors to be discharged and why. Edit in your answer in your question above. \$\endgroup\$
    – winny
    Commented Dec 26, 2022 at 9:06
  • \$\begingroup\$ By the way that's not how you would calculate energy lost discharging a capacitor. If it has 30V at the begining, it will drop voltage to 0 as it's being discharged. Proper formula would be CV²/2 which is basically all the energy stored in the capacitor. Note that it does not depend on the resistor value. \$\endgroup\$
    – floppydisk
    Commented Dec 26, 2022 at 9:13
  • \$\begingroup\$ I agree. The first calculation I do is not power loss but time. How fast do you want to discharge the capacitor? Then you look at energy loss and power dissipation, etc. I would not discount 0.9 W. Remember any time the voltage is at 30 V that 0.9 W dissipation is happening. \$\endgroup\$
    – user57037
    Commented Dec 26, 2022 at 9:19
  • \$\begingroup\$ @floppydisk: He is probably talking about a resistor permanently connected in parallel with the capacitor. The question is very low quality though... \$\endgroup\$
    – johnfound
    Commented Dec 26, 2022 at 9:20

2 Answers 2

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I think you are in this type of situation

enter image description here

Then you have to understand first if the resistor is needed, if the capacitor is feeding a load may you don’t really need this solution.

If the capacitor is needed you are introducing a constant power consumption into you circuit. This can be considered by the constant power consumption formula

P = R•V^2

Then, faster discharging time constant is, higher is the constant power consumption. You have to deal with the power dissipated and the specification your project gives.

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I guess there's the power loss as well but in my calculation the lower loss is not a lot.

You are always consuming power to charge and discharge a capacitor, this power is dissipated by the resistor. Comparing the energy stored by the capacitor this is the same as the energy dissipated on the resistance, sum of them gives the total energy supplied by the generator

Are there factors that needs to be considered? (Like maybe capacitors last longer when it's discharged slowly?)

The only one factor that changes in choosing with resistors is the time constant for the charging process, no differences are founded on power consumption.

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