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If you want to store power in one circuit, while it is connected parallel to a different circuit that you don’t want to have the stored power be connected to, then you can use a diode to prevent the stored power from powering the other circuit when the main power is removed.
- Closing the switch powers both LEDs and charges the capacitor to about 0.7V less than the supply voltage thanks having to pass current through the forward biased to the silicon based rectifier diode. The LED that is powered by current going through the diode, also loses about 0.7V, and therefore will have a little less current flowing through it than the LED that doesn’t have current going through the diode. It will be a little less bright if the LEDs are the same brightness at the same current.
- Releasing the switch will cut power from the power supply to the circuits. The charged capacitor cannot power the LED on the other side of the diode, because when the capacitor is the only power source, then the diode is reverse biased. The capacitor will discharge through the LED and protective resistor that does not have a diode between them. The LEDs fades off due to the dropping voltage of the capacitor as it discharges through the LED/resistor. The LED stops current from flowing at about 1.5V if it is red. So the LED will probably be off completely, and the capacitor holding approx. 1.5V when the capacitor is completely discharged through the LED.
Demonstration video:
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