Discharging a capacitor
Leaking away
- Connect a charged capacitor to a resistor and it discharges.
- The current is large at first, then fades — fast at the start, slow at the end.
- This is exponential decay.
Setting up the decay
- Round the loop, $\dfrac{Q}{C} = -R\dfrac{dQ}{dt}$ (the capacitor's p.d. drives the current).
- This equation is solved by an exponential decay.
Practice
During discharge through a resistor, the charge on a capacitor:
The bigger the charge, the bigger the current — giving an exponential decay.
The discharge equations
- Charge, p.d. and current all decay together: $Q = Q_0 e^{-t/RC}$ (and $V$, $I$ the same way).
- Each falls by the same fraction in equal time steps.
Practice
The charge during discharge is given by:
Discharge falls toward zero: $Q = Q_0 e^{-t/RC}$ (the $1 - e^{-t/RC}$ form is for charging).
The time constant
- $\tau = RC$ (in seconds) is the time constant.
- After one $\tau$ the charge falls to $\dfrac{1}{e} \approx 37\%$; after $5\tau$, below $1\%$.
Practice
A capacitor of $500\ \mu\text{F}$ discharges through a $2000\ \Omega$ resistor. What is the time constant?
$\tau = RC = 2000 \times 500 \times 10^{-6} = 1.0\ \text{s}$.
Practice
After one time constant, the charge falls to about ____ % of its starting value.
It falls to $\dfrac{1}{e} \approx 0.37$, i.e. about 37%.
Practice
After 5 time constants, less than 1% of the charge remains.
$e^{-5} \approx 0.0067$, so under 1% is left — effectively fully discharged.
You've got it
Key idea
- discharge is exponential: $Q = Q_0 e^{-t/RC}$ (also $V$ and $I$)
- time constant $\tau = RC$ — time to fall to $\dfrac{1}{e} \approx 37\%$
- a $\ln$ plot is linear with gradient $-\dfrac{1}{RC}$