Discharging a capacitor
Introduced| English | Meaning |
|---|---|
| discharge/dɪsˈtʃɑːdʒ/ | discharge |
| time constant/taɪm ˈkɒnstənt/ | The characteristic time for an exponential process to change by a factor of e; in decay, the quantity falls to one over e of its starting value. |
| exponential decay/ˌekspəˈnenʃl dɪˈkeɪ/ | A decrease in which a quantity is multiplied by the same factor, between zero and one, over equal intervals. |
The decay that never quite finishes
- A discharging capacitor loses 63% of its charge in the first time constant, then 63% of what remains in the next, and so on for ever.
- After five time constants less than 1% is left, but the curve never reaches zero. It cannot, because the rate of loss is always proportional to how much is left.
- Every quantity that behaves that way, radioactive decay included, gives the same curve, so learning it once serves twice.
- This lesson is the discharge 放电 equations, the time constant 时间常数, and the log plot that turns the curve into a straight line.
Where the exponential comes from
- Close the switch and the capacitor's p.d. drives a current through the resistor. Kirchhoff's second law gives $V_C = V_R$, and with $V_C = Q/C$, $V_R = IR$ and $I = -dQ/dt$:
- The rate of loss is proportional to the charge remaining. That is the defining property of exponential decay 指数衰减, and it is why the curve flattens as it falls.
During discharge through a resistor, the charge on a capacitor:
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The bigger the charge, the bigger the current — giving an exponential decay.
Why is a capacitor's discharge exponential?
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Q/C = -R dQ/dt says exactly that. Every quantity with this property, radioactive decay included, gives the same curve.
The three equations
- Charge, p.d. and current all decay together, with the same time constant:
- The current is largest at the instant the switch closes, $I_0 = V_0/R$, because the full initial p.d. is then across the resistor. It falls as the p.d. falls.
- The area under the current-time graph is the charge that has flowed. All three curves have the same shape, so measuring $\tau$ from any one of them gives $RC$.
Steep at first, then ever flatter, and never quite zero
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.
The charge during discharge is given by:
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Discharge falls toward zero: $Q = Q_0 e^{-t/RC}$ (the $1 - e^{-t/RC}$ form is for charging).
The time constant
- The time constant is:
- Its unit is the second, since an ohm times a farad is a second.
- It is the time for the decaying quantity to fall to $\dfrac{1}{e} \approx 37\%$ of its starting value. After $2\tau$ about 13.5% remains, and after $5\tau$ less than 1%.
- A large $R$ or a large $C$ means a slow discharge: a big store emptying through a narrow pipe.
A capacitor of $500\ \mu\text{F}$ discharges through a $2000\ \Omega$ resistor. What is the time constant?
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$\tau = RC = 2000 \times 500 \times 10^{-6} = 1.0\ \text{s}$.
After one time constant, the charge falls to about ____ % of its starting value.
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It falls to $\dfrac{1}{e} \approx 0.37$, i.e. about 37%.
After 5 time constants, less than 1% of the charge remains.
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$e^{-5} \approx 0.0067$, so under 1% is left — effectively fully discharged.
A capacitor charged to 24 V discharges through a 5.6 kilo-ohm resistor. What is the initial current, in mA?
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At the instant the switch closes the whole p.d. is across the resistor, so I0 = V0/R = 24/5600 = 4.3 mA. It then decays with the same time constant as the charge.
Worked example: a full discharge question
- A $470\ \mu\text{F}$ capacitor charged to $24\ \text{V}$ discharges through a $5.6\ \text{k}\Omega$ resistor. Find the time constant, the initial current, the p.d. after $4.0\ \text{s}$, and the time for the p.d. to fall to $6.0\ \text{V}$.
- $\tau = RC = (5.6 \times 10^3)(470 \times 10^{-6}) = 2.6\ \text{s}$, and $I_0 = V_0/R = 24/5600 = 4.3\ \text{mA}$.
- After $4.0\ \text{s}$: $V = V_0 e^{-t/RC} = 24\,e^{-4.0/2.63} = 5.3\ \text{V}$.
- For the time, take logarithms: $t = -RC\ln(V/V_0) = -2.63\ln(6.0/24) = 3.6\ \text{s}$.
- Keep $\tau$ unrounded inside the calculation and round only at the end. And note that $\ln(V/V_0)$ is negative, so the minus sign makes $t$ positive: that rearrangement is the most often asked and the most often botched.
After one time constant, about 37% of the original charge remains on the capacitor.
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It falls to $\dfrac{1}{e}$ of its starting value, about $37\%$. After two it is at about $13.5\%$, and after five below $1\%$, though it never quite reaches zero.
Worked example: reading $\tau$ off a graph
- The p.d. across a $2200\ \mu\text{F}$ capacitor falls from $6.0\ \text{V}$ to $2.2\ \text{V}$ in $4.5\ \text{s}$. Find the resistance.
- Check the ratio first: $2.2/6.0 = 0.367$, which is $1/e$. So $4.5\ \text{s}$ is one time constant.
- Then $R = \tau/C = 4.5/(2200 \times 10^{-6}) = 2.0 \times 10^3\ \Omega$.
- If the fall is not to $1/e$, use $RC = -t/\ln(V/V_0)$ instead. The answer is the same either way, but spotting the $1/e$ saves a step.
A capacitor with time constant 2.63 s is charged to 24 V. How long, in seconds, until the p.d. falls to 6.0 V?
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t = -RC ln(V/V0) = -2.63 ln(0.25) = 3.6 s. ln(0.25) is negative, so the minus sign makes t positive; that rearrangement is the most often botched.
The log plot
- Take logarithms of $V = V_0 e^{-t/RC}$:
- So a graph of $\ln V$ against $t$ is a straight line with gradient $-1/RC$ and intercept $\ln V_0$.
- This is the standard experimental method: it turns a curve into a line, so $RC$ comes from a gradient rather than from a single reading, and any departure from a straight line shows the decay is not exponential.
The p.d. across a $2200\ \mu\text{F}$ capacitor falls from $6.0\ \text{V}$ to $2.2\ \text{V}$ in $4.5\ \text{s}$. Put the steps of finding the resistance in order.
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If the fall is not to $\dfrac{1}{e}$, use $RC = -\dfrac{t}{\ln(V/V_{0})}$ instead. The answer is the same, but spotting the $\dfrac{1}{e}$ saves a step.
Marks that slip away
- $\tau = RC$ is a time, in seconds, not a fraction. After one $\tau$ the quantity is at 37%, not 63% gone and stopped.
- The current is largest at the start, when the whole p.d. is across the resistor, and it decays with the same time constant as the charge.
- To find a time, take logs: $t = -RC\ln(V/V_0)$. Keep the time constant unrounded through the working.
- On a $\ln V$ against $t$ plot the gradient is $-1/RC$, so $RC$ is minus the reciprocal of the gradient.
A graph of ln V against t for a discharging capacitor is a straight line. What is its gradient?
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Taking logs of V = V0 exp(-t/RC) gives ln V = ln V0 - t/RC. So RC is minus the reciprocal of the gradient, and the straight line itself confirms the decay is exponential.
You've got it
- the rate of loss is proportional to the charge left, so the decay is exponential: $Q = Q_0e^{-t/RC}$, and $V$ and $I$ decay identically
- the current starts at $I_0 = V_0/R$ and the area under the current-time graph is the charge that has flowed
- the time constant $\tau = RC$ is the time to fall to $1/e \approx 37\%$; after $5\tau$ less than 1% remains
- take logs to find a time, $t = -RC\ln(V/V_0)$, and a plot of $\ln V$ against $t$ is a straight line of gradient $-1/RC$