Discharging a capacitor
| English | Chinese | Pinyin |
|---|---|---|
| resistor | 电阻器 | diàn zǔ qì |
| discharge | 放电 | fàng diàn |
| exponential decay | 指数衰减 | zhǐ shù shuāi jiǎn |
| time constant | 时间常数 | shí jiān cháng shù |
Leaking away
- Connect a charged capacitor to a resistor 电阻器 and it discharges 放电.
- The current is large at first, then fades — fast at the start, slow at the end.
- This is exponential decay 指数衰减.
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.
Charge / discharge curve
The voltage rises (or decays) exponentially with time constant τ = RC.
Discharging a capacitor
Q = Q₀·bᵗ
Charge decays exponentially through the resistor.
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.
Charge decays exponentially during discharge, falling to Q0/e after one time constant 时间常数 RC
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 $1\%$.
A capacitor charges through switch A, then discharges through the resistor via switch B
A capacitor of $500\ \mu\text{F}$ discharges through a $2000\ \Omega$ resistor. What is the time constant?
$\tau = RC = 2000 \times 500 \times 10^{-6} = 1.0\ \text{s}$.
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%.
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
- 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 $-\dfrac{1}{RC}$