I am new to the world of electronics and I had done a simple circuit to verify the charging and discharging of a capacitor.
The circuit is as shown below:
My aim was to keep the lamp, L1 switched on for a further 5 seconds after the main switch has been opened. This is how I got the value 2.32F for the capacitor:
Hence I found that I needed a 2.32 F capacitor to have a discharge duration of 5 seconds.
(1) The charging is taking way more than the calculated value ( > 63.661 seconds )
(2) The capacitor is not charging upto 12V. Tha maximum voltage attained is 7.098 V.
Please help me understand these variations from the calculated value :-(
Solved! Go to Solution.
Simple Ohm's Law. The final value of the capacitor voltage in a DC circuit will be the voltage at that point as if the capacitor had been removed. Resistor R2 and lamp L1 create voltage divider so 7 V appears to be correct.