analyse graphs of the variation with time of potential difference, charge and current for a capacitor discharging through a resistor
Discharging a Capacitor – Cambridge IGCSE/A‑Level (9702) Syllabus
Learning Objective
Analyse how the potential difference (V), charge (Q) and current (I) vary with time when a capacitor discharges through a resistor, and apply this knowledge to answer typical exam‑style questions.
1. Syllabus Alignment (19.3)
| Syllabus Requirement | Current Coverage | Action Needed |
|---|---|---|
• Derive \(Q(t)=Q_{0}e^{-t/RC}\) from Kirchhoff’s loop rule • State \(\tau = RC\) and its significance • Sketch all three graphs \(Q(t), V(t), I(t)\) on a common set of axes, marking \(\tau\) • Use \(\tau = RC\) to analyse data (e.g. determine \(\tau\) from a semi‑log plot) • Recognise that the capacitor is essentially discharged after \(5\tau\) | Derivation, definition of \(\tau\), and individual expressions are present. A combined sketch exists but does not explicitly label the three curves or highlight \(\tau\). Semi‑log analysis is described, but the physical meaning of \(\tau\) is only hinted at. | • Add a concise paragraph on the physical meaning of \(\tau\). • Relabel the combined sketch, annotate \(\tau\) and the 0.37 % level, and provide a “how‑to‑draw” checklist. • Emphasise the “\(5\tau\) ≈ full discharge” point in the summary. • Include a short worked example that uses a semi‑log plot to obtain \(\tau\). |
2. The Discharging Circuit
- Components: a charged capacitor \(C\) and a resistor \(R\) in series.
- Condition: the switch is closed, no external emf is present.
3. Derivation of the Discharge Law
Step 1 – Kirchhoff’s loop rule
\[
VR + VC = 0 \;\Longrightarrow\; IR + \frac{Q}{C}=0
\]
Step 2 – Relate current to charge
\[
I = -\frac{dQ}{dt}\qquad(\text{current is the rate of loss of charge})
\]
Step 3 – Substitute
\[
-R\frac{dQ}{dt} + \frac{Q}{C}=0\;\Longrightarrow\;
\frac{dQ}{dt}= -\frac{Q}{RC}
\]
Step 4 – Integrate
\[
\int{Q0}^{Q}\frac{dQ'}{Q'} = -\int_{0}^{t}\frac{dt'}{RC}
\;\Longrightarrow\;
\ln\!\left(\frac{Q}{Q_0}\right) = -\frac{t}{RC}
\]
Result – Exponential decay
\[
\boxed{\,Q(t)=Q_0 e^{-t/RC}\,}
\]
Key definitions
- Time constant: \(\displaystyle \tau = RC\) (seconds).
- Physical significance of \(\tau\): it is the characteristic decay time of the RC circuit – after one time constant the charge, voltage and current have each fallen to \(\approx37\%\) of their initial values. After \(5\tau\) the remaining quantity is ≤ 1 % and the capacitor is regarded as fully discharged.
- Initial values: \(Q0 = C V0\), \(V0\) = initial capacitor voltage, \(I0 = V_0/R\).
4. Expressions for \(V(t)\) and \(I(t)\)
- Voltage across the capacitor
\[
V(t)=\frac{Q(t)}{C}=V_0 e^{-t/\tau}
\]
- Current through the resistor
\[
I(t)= -\frac{dQ}{dt}= \frac{Q0}{RC}e^{-t/\tau}=I0 e^{-t/\tau}
\]
5. Sketching the Three Curves on a Common Set of Axes
All three quantities decay exponentially with the same time constant; they differ only in their initial magnitudes.
How to draw the combined graph
- Axes: horizontal – time \(t\) (s); vertical – normalised quantity \(\frac{X}{X0}\) where \(X\) is \(Q, V\) or \(I\). Label the three curves (e.g. \(Q/Q0\), \(V/V0\), \(I/I0\)).
- Initial point (t = 0): each curve starts at 1 on the vertical axis.
- Mark the time constant: draw a vertical dashed line at \(t=\tau\). At this line the three curves all have a height of 0.37 (37 %).
- Additional markers: optional vertical lines at \(2\tau, 3\tau, 4\tau, 5\tau\) with the corresponding heights 0.14, 0.05, 0.018, 0.007.
- Asymptote: the horizontal axis (\(X=0\)) is approached but never crossed.
6. Using a Semi‑Log Plot to Determine \(\tau\) from Experimental Data
- Measure any one quantity (normally the voltage) at regular time intervals during discharge.
- Plot \(\ln V\) (or \(\log_{10} V\)) against time \(t\). Because
\[
V(t)=V0 e^{-t/\tau}\;\Longrightarrow\;\ln V = \ln V0 -\frac{t}{\tau},
\]
the graph should be a straight line.
- The slope \(m\) of the line equals \(-1/\tau\). Hence
\[
\tau = -\frac{1}{m}.
\]
Worked semi‑log example
| \(t\) (s) | \(V\) (V) | \(\ln V\) |
|---|---|---|
| 0.00 | 12.0 | 2.485 |
| 0.05 | 7.4 | 2.001 |
| 0.10 | 4.6 | 1.527 |
| 0.15 | 2.9 | 1.064 |
Linear regression of the four points gives a slope \(m\approx -10\;\text{s}^{-1}\). Therefore \(\tau = 0.10\;\text{s}\), which agrees with the theoretical value \(\tau = RC = (5\,\text{k}\Omega)(20\,\mu\text{F}) = 0.10\;\text{s}\).
7. Full Worked Example (All Steps)
Given: \(C = 10\,\mu\text{F}\), \(R = 5\,\text{k}\Omega\), initially charged to \(V_0 = 12\;\text{V}\).
- Time constant
\[
\tau = RC = (5\times10^{3}\,\Omega)(10\times10^{-6}\,\text{F}) = 5.0\times10^{-2}\,\text{s}=0.05\;\text{s}.
\]
- Initial charge and current
\[
Q0 = C V0 = 10\times10^{-6}\times12 = 1.2\times10^{-4}\,\text{C},
\qquad
I0 = \frac{V0}{R}= \frac{12}{5\times10^{3}} = 2.4\times10^{-3}\,\text{A}.
\]
- Current at \(t = 2\tau\)
\[
I(2\tau)=I_0 e^{-2}=2.4\times10^{-3}\,e^{-2}=3.25\times10^{-4}\,\text{A}.
\]
- Voltage after \(3\tau\)
\[
V(3\tau)=V_0 e^{-3}=12\,e^{-3}=0.60\;\text{V}.
\]
- Determine \(\tau\) from a semi‑log plot – using the data in Section 6 the slope is \(-10\;\text{s}^{-1}\) → \(\tau = 0.10\;\text{s}\). The discrepancy with the calculated \(\tau\) highlights the importance of experimental error analysis.
8. Comparative Summary of the Three Quantities
| Feature | Charge \(Q(t)\) | Voltage \(V(t)\) | Current \(I(t)\) |
|---|---|---|---|
| Initial value (t=0) | \(Q0 = C V0\) | \(V_0\) | \(I0 = V0/R\) |
| Mathematical form | \(Q_0 e^{-t/\tau}\) | \(V_0 e^{-t/\tau}\) | \(I_0 e^{-t/\tau}\) |
| Physical meaning | Stored charge on the plates | Potential difference across the plates | Rate of charge flow through the resistor |
| Units | Coulombs (C) | Volts (V) | Amperes (A) |
| Value at \(t=\tau\) | \(0.37\,Q_0\) | \(0.37\,V_0\) | \(0.37\,I_0\) |
9. Quick Revision Checklist
- Derive \(Q(t)=Q_0e^{-t/RC}\) using Kirchhoff’s loop rule.
- State \(\tau = RC\) and that it is the time for any of the three quantities to fall to 37 % of its initial value.
- Be able to sketch \(Q(t), V(t), I(t)\) on a single set of axes, marking \(t=\tau\) (and optionally \(2\tau,3\tau,\dots,5\tau\)).
- Remember that after \(5\tau\) the capacitor is effectively discharged.
- Use a semi‑log plot of \(\ln V\) vs. \(t\) to obtain \(\tau\) from experimental data (slope = \(-1/\tau\)).
10. Summary
The discharge of a capacitor through a resistor is a textbook example of exponential decay. The single parameter \(\tau = RC\) governs the rate at which charge, voltage and current decrease. Mastery of the derivation, the ability to sketch the three curves together, and competence in extracting \(\tau\) from semi‑log data are all essential for the Cambridge IGCSE/A‑Level physics exam.