recall and use τ = RC for the time constant for a capacitor discharging through a resistor

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 \(R\).

1. Apply Kirchhoff’s loop rule (taking the direction of current as positive): \[ V_{C}+V_{R}=0. \]
2. Express the voltages in terms of charge and current: \[ \frac{Q}{C}+IR=0. \]
3. Current is the rate of loss of charge: \[ I=-\frac{dQ}{dt}. \]
4. Substitute (3) into (2): \[ \frac{Q}{C}-R\frac{dQ}{dt}=0\;\Longrightarrow\; \frac{dQ}{dt}=-\frac{Q}{RC}. \]
5. Integrate (separating variables): \[ \int_{Q_{0}}^{Q}\frac{dQ'}{Q'}=-\int_{0}^{t}\frac{dt'}{RC} \;\Longrightarrow\; \ln\!\left(\frac{Q}{Q_{0}}\right)=-\frac{t}{RC}. \]
6. Exponentiate: \[ \boxed{Q(t)=Q_{0}\,e^{-t/RC}}. \]
7. Since \(V=Q/C\): \[ \boxed{V(t)=V_{0}\,e^{-t/RC}}. \]
8. 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.

   1. For a natural‑log plot the gradient is \(-1/\tau\). Hence \[ \tau = \frac{1}{25\;\text{s}^{-1}} = 0.040\;\text{s}. \]
   2. 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 \(3\;\text{s}\).

   1. τ = RC = \((2.0\times10^{3}\,\Omega)(10\times10^{-6}\,\text{F}) = 2.0\times10^{-2}\,\text{s}=0.020\;\text{s}\).
   2. 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 \(2\tau\).

   1. τ = RC = \((1.0\times10^{6}\,\Omega)(4.7\times10^{-6}\,\text{F}) = 4.7\;\text{s}\).
   2. 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)

   1. 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.)
   2. 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?
   3. 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.
   4. 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.
   5. 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.)