Discharging a Capacitor – Cambridge International AS & A Level Physics (9702) – Topic 19.3
Learning Objectives (AO1‑AO3)
- Define the time constant τ for an RC discharge and explain its physical meaning.
- Derive and use the exponential decay expressions for charge, voltage and current.
- State and justify why a capacitor is regarded as “effectively discharged” after 5τ (≈0.7 % of the initial value).
- Analyse RC‑discharge data on semi‑logarithmic paper and extract τ, times or voltage fractions.
- Calculate τ, V(t), I(t), Q(t) and the remaining energy U(t) for any given instant.
Key Concepts
- Capacitance (C) – stores charge \(Q\) with \(Q = C V\). Energy stored:
\[
U = \tfrac12 C V^{2}.
\]
- Resistance (R) – opposes the flow of charge; determines the rate at which the capacitor discharges.
- Time constant (τ) – the product of resistance and capacitance,
\[
\boxed{\tau = RC}\quad\text{(seconds)}.
\]
It is the time required for the charge **or** the voltage on the capacitor to fall to \(\frac1e\) (≈ 37 %) of its initial value.
- During discharge the charge, voltage and current all decay exponentially with the **same** τ.
Derivation of the Discharge Equations
Consider a capacitor initially charged to \(V_{0}\) (\(Q_{0}=CV_{0}\)) that is suddenly connected across a resistor \(R\).
- Apply Kirchhoff’s loop rule (taking the direction of current as positive):
\[
V_{C}+V_{R}=0.
\]
- Express the voltages in terms of charge and current:
\[
\frac{Q}{C}+IR=0.
\]
- Current is the rate of loss of charge:
\[
I=-\frac{dQ}{dt}.
\]
- Substitute (3) into (2):
\[
\frac{Q}{C}-R\frac{dQ}{dt}=0\;\Longrightarrow\;
\frac{dQ}{dt}=-\frac{Q}{RC}.
\]
- Integrate (separating variables):
\[
\int_{Q_{0}}^{Q}\frac{dQ'}{Q'}=-\int_{0}^{t}\frac{dt'}{RC}
\;\Longrightarrow\;
\ln\!\left(\frac{Q}{Q_{0}}\right)=-\frac{t}{RC}.
\]
- Exponentiate:
\[
\boxed{Q(t)=Q_{0}\,e^{-t/RC}}.
\]
- Since \(V=Q/C\):
\[
\boxed{V(t)=V_{0}\,e^{-t/RC}}.
\]
- From Ohm’s law \(I=V/R\):
\[
\boxed{I(t)=\frac{V_{0}}{R}\,e^{-t/RC}}.
\]
\endol>
Energy During Discharge
\[
U(t)=\tfrac12 C[V(t)]^{2}
=\tfrac12 C V_{0}^{2}\,e^{-2t/RC}
=U_{0}\,e^{-2t/\tau},
\]
where \(U_{0}= \tfrac12 C V_{0}^{2}\) is the initial energy.
Because of the factor 2, the energy falls twice as fast as the voltage (or charge).
Interpretation of the Time Constant τ
| Time | Voltage (or Charge, or Current) | Fraction of Initial Value |
|---|
| \(t = \tau\) | \(V = V_{0}e^{-1}\) | 0.368 ≈ 37 % |
| \(t = 2\tau\) | \(V = V_{0}e^{-2}\) | 0.135 ≈ 14 % |
| \(t = 5\tau\) | \(V = V_{0}e^{-5}\) | 0.007 ≈ 0.7 % |
The same percentages apply to the charge \(Q(t)\) and the current \(I(t)\) because they share the identical exponential factor.
Graphical Analysis on Semi‑Logarithmic Paper
- Plot \(\log_{10} V\) (or \(\log_{10} Q\), \(\log_{10} I\)) on the vertical axis against linear time on the horizontal axis.
- The discharge curve becomes a straight line:
\[
\log_{10} V = \log_{10} V_{0} -\frac{t}{\tau}\log_{10}e,
\]
whose gradient \(m = -\frac{1}{\tau}\log_{10}e\) (≈ \(-0.434/\tau\)).
- Finding τ from a graph – Measure the slope \(m\) of the straight‑line portion, then
\[
\tau = -\frac{1}{m}\,\log_{10}e \approx -\frac{0.434}{m}.
\]
- Time for a given voltage fraction – Rearrange the exponential relation:
\[
\frac{V}{V_{0}} = e^{-t/\tau}
\;\Longrightarrow\;
t = -\tau\ln\!\left(\frac{V}{V_{0}}\right).
\]
Example: for \(V = 0.20V_{0}\), \(t = -\tau\ln(0.20) \approx 1.61\tau\).
Worked Example A – Determining τ from a Semi‑Log Plot
Problem: A semi‑log plot of voltage versus time for a discharging capacitor shows a straight line with a gradient of \(-25\;\text{s}^{-1}\) when the vertical axis is natural‑log (ln) scale. Find τ and the time at which the voltage has fallen to 10 % of its initial value.
- For a natural‑log plot the gradient is \(-1/\tau\). Hence
\[
\tau = \frac{1}{25\;\text{s}^{-1}} = 0.040\;\text{s}.
\]
- Set \(V/V_{0}=0.10\) and solve for \(t\):
\[
t = -\tau\ln(0.10)=0.040\times2.303 = 0.092\;\text{s}\approx2.3\tau.
\]
Worked Example B – Voltage after a Given Time
Problem: A \(10\;\mu\text{F}\) capacitor is discharged through a \(2\;\text{k}\Omega\) resistor. The initial voltage is \(12\;\text{V}\). Find τ and the voltage after \(3\;\text{s}\).
- τ = RC = \((2.0\times10^{3}\,\Omega)(10\times10^{-6}\,\text{F}) = 2.0\times10^{-2}\,\text{s}=0.020\;\text{s}\).
- Use \(V(t)=V_{0}e^{-t/\tau}\):
\[
V(3\;\text{s}) = 12\,e^{-3/0.020}=12\,e^{-150}\approx 0\;\text{V}.
\]
After 3 s (≈150τ) the capacitor is essentially fully discharged.
Worked Example C – Current and Energy after 2τ
Problem: A \(4.7\;\mu\text{F}\) capacitor charged to \(5\;\text{V}\) discharges through a \(1\;\text{M}\Omega\) resistor. Find τ, then the voltage, current and remaining energy after \(2\tau\).
- τ = RC = \((1.0\times10^{6}\,\Omega)(4.7\times10^{-6}\,\text{F}) = 4.7\;\text{s}\).
- At \(t=2\tau=9.4\;\text{s}\):
- Voltage: \(V = 5e^{-2}=5\times0.1353=0.68\;\text{V}\).
- Current: \(I = V/R = 0.68/1.0\times10^{6}=6.8\times10^{-7}\;\text{A}\).
- Energy remaining:
\[
U = \tfrac12 C V^{2}= \tfrac12 (4.7\times10^{-6})(0.68)^{2}\approx1.1\times10^{-6}\;\text{J}.
\]
This is about 1.9 % of the initial energy \(U_{0}=5.9\times10^{-5}\;\text{J}\).
Summary Table of RC‑Discharge Quantities
| Quantity |
Symbol |
Formula / Definition |
Typical Example |
| Capacitance |
\(C\) |
Given or measured |
10 µF |
| Resistance |
\(R\) |
Given or measured |
2 kΩ |
| Time constant |
\(\tau\) |
\(\tau = RC\) |
0.020 s |
| Voltage (time t) |
\(V(t)\) |
\(V_{0}e^{-t/\tau}\) |
≈0 V after 3 s |
| Charge (time t) |
\(Q(t)\) |
\(Q_{0}e^{-t/\tau}\) |
0.68 µC after 2τ (example) |
| Current (time t) |
\(I(t)\) |
\(\dfrac{V_{0}}{R}e^{-t/\tau}\) |
6.8 × 10⁻⁷ A after 2τ |
| Energy (time t) |
\(U(t)\) |
\(\tfrac12 C V_{0}^{2}e^{-2t/\tau}\) |
≈1.1 µJ after 2τ |
Common Mistakes & How to Avoid Them
- Sign of the exponent – The exponent must be negative; a positive exponent would describe growth, not decay.
- Unit conversion – Convert µF → F, kΩ → Ω, etc., before calculating τ.
- Interpretation of τ – After one τ the capacitor still holds 37 % of its initial voltage; only after ≈5τ is it effectively empty (≈0.7 % remains).
- Omitting the factor 2 in energy – Energy decays as \(e^{-2t/\tau}\); forgetting the “2” over‑estimates the remaining energy.
- Reading semi‑log graphs – Remember the vertical axis is logarithmic; the slope gives \(-1/\tau\) (natural‑log) or \(-0.434/\tau\) (base‑10 log).
- Current expression – Always write \(I(t)=\dfrac{V_{0}}{R}e^{-t/\tau}\) when asked for discharge current.
Practice Questions (Cambridge Syllabus Alignment)
- τ and voltage after 2τ – A \(4.7\;\mu\text{F}\) capacitor discharges through a \(1\;\text{M}\Omega\) resistor. Find τ and the voltage after \(2\tau\) if \(V_{0}=5\;\text{V}\). (Also give the current and remaining energy.)
- Designing a time constant – What resistance is required to give a time constant of \(0.5\;\text{s}\) for a \(22\;\mu\text{F}\) capacitor?
- Graph analysis – A semi‑log plot of voltage versus time for a discharge shows a straight line with gradient \(-20\;\text{s}^{-1}\) (natural‑log scale). Determine τ and the time at which the voltage has fallen to 10 % of its initial value.
- Extracting time from a voltage fraction – In a circuit with τ = 0.15 s, the voltage is measured to be 0.25 \(V_{0}\). Calculate the elapsed time.
- Energy loss – A \(12\;\mu\text{F}\) capacitor initially at 10 V is discharged through a 500 Ω resistor. Compute the total energy dissipated in the resistor after the capacitor is effectively discharged (assume 5τ).
Suggested Diagram (for the learner)
A labelled schematic showing a charged capacitor \(C\) connected across a resistor \(R\) with the direction of discharge current indicated, together with a plot of \(V\) versus \(t\) that marks the points \(t=\tau\), \(2\tau\) and \(5\tau\). (Insert as a figure in your notes.)