An RC Circuit: Discharging
What happens if the capacitor is now fully charged and is then discharged through the resistor? Now the potential difference across the resistor is the capacitor voltage, but that decreases (as does the current) as time goes by.
Applying Kirchoff's loop rule:
| -IR |
- |
| Q
|  |
| C
|
|
= |
0 |
| I |
= |
| dQ
| |
| dt
|
|
, so the equation can be written: |
| R |
| dQ
| |
| dt
|
|
= |
| -Q
| |
| C
|
|
This is a differential equation that can be solved for Q as a function of time. The solution is:
Q(t) = Qo e-t/t
where Qo is the initial charge on the capacitor and the time constant t = RC.
Differentiating this expression to get the current as a function of time gives:
| I(t) |
= |
| -Qo
| |
| RC
|
|
e-t/t |
= |
-Io e-t/t |
| where | Io |
= |
| Qo
| |
| RC
|
|
Note that, except for the minus sign, this is the same expression for current we had when the capacitor was charging. The minus sign simply indicates that the charge flows in the opposite direction.
Here the time constant t = RC determines how quickly the capacitor discharges. If RC is small the capacitor discharges quickly; if RC is large the capacitor discharges more slowly.