The idea here is to bleed the voltage off the caps. This is normally done by placing what is commonly called a "bleeder resistor across a charged capacitor. The cap can be any value and charged to any potential. The LED and LED series resistor are across the cap (in parallel with the cap). The switch is in series with the resistor and when the switch is turned on it places the bleeder resistance across (parallel with) the cap. The cap discharges through the switched in resistor. The switch needs to be rated for the max voltage and current it will see. When the system is running the LED will be illuminated. When the system is shut down and the bleeder switch turned on the cap will discharge and when the voltage is low enough the LED will fade out. This is the simple and crude but effective way to discharge the cap(s). There can be more elaborate methods using more parts. That is why I mentioned "My best guess" above.Thank you for responding. My question is how that would be wired. There are 4 components. Cap, switch, LED and resistor.
Don't understand why the LED would be illuminated if the momentary switch isn't closed. If it's illuminated it is bleeding the caps which defeats the purpose of the momentary switch doesn't it? The goal is for the caps to stay charged until the switch is flipped illuminating the LED which bleeds the caps.When the system is running the LED will be illuminated. When the system is shut down and the bleeder switch turned on the cap will discharge and when the voltage is low enough the LED will fade out. Ron