Why do capacitor equations have q=Q(1-e^-t/RC) and q=Qe^-t/RC?

[arthur01]

hey guys, have a quick question

just wanted to know what is the difference between these two equations, i couldn't find anything on google

q=Q(1-e^-t/RC)

and

q=Qe^-t/RC


why does one have 1 subtracting the rest of the equation, and the other doesn't

thanks!

 

[arthur01]

I understand that in the first equation the capacitor is uncharged before the switch is closed, and when the switch is closed at t=0, charge (q) will increase.

I just don't understand what the "1-" has to do with the problem

 

[arthur01]

wait i think i got it, in the second equation, the capacitor is charged, and it is discharging.

in the first equation, it is uncharged. is this correct?

 

[sandy.bridge]

The second equation is for a series connection of a voltage source, resistor, and a capacitor. Applying KVL around the loop and integrating, that is the result. The case is indeed for a capacitor discharging; hence the exponential decay.

 

[CWatters]

Try calculating Q for t=0 and t=∞...

1) q=Q(1-e^-t/RC)

t=0, q=0
t=∞, q=Q
= charging

2) q=Qe^-t/RC

t=0, q=Q
t=∞, q=0
= discharging