To charge a capacitor in a RC arrangement it takes 5xRC seconds, right, where:
RC = R x C (where R is resistors value in ohms, and C capacitance in faradays)
So if R = 10k, C=100uF --> RC = 10000 Ohm x 0,0001F = 1s
and it would take 5x1s = 5s to charge the circuit.
Again if R=1k, C=5000uF --> RC = 1000 Ohm x 0,001F = 1s
and it would take 5s to charge the circuit.
My question is:
- If I build a classic RC circuit that makes a LED fade out when a switch is opened is there any difference in the behavior of the fade if I use R=10k C=100uF or if I use R=1k C=1000uF?
According to my understanding in both configurations the fade takes 5s. But would exist a difference in the effect (like one configuration being smoother than the other one)?
Thanks in advance.
RC = R x C (where R is resistors value in ohms, and C capacitance in faradays)
So if R = 10k, C=100uF --> RC = 10000 Ohm x 0,0001F = 1s
and it would take 5x1s = 5s to charge the circuit.
Again if R=1k, C=5000uF --> RC = 1000 Ohm x 0,001F = 1s
and it would take 5s to charge the circuit.
My question is:
- If I build a classic RC circuit that makes a LED fade out when a switch is opened is there any difference in the behavior of the fade if I use R=10k C=100uF or if I use R=1k C=1000uF?
According to my understanding in both configurations the fade takes 5s. But would exist a difference in the effect (like one configuration being smoother than the other one)?
Thanks in advance.