Discharging a capacitor

1. Overview

When a charged capacitor is connected in a closed loop with a resistor, it undergoes discharging. This is the process by which the stored electric potential energy in the capacitor's electric field is transferred to the resistor and dissipated as thermal energy (Joule heating).

The fundamental physics governing this process is the relationship between the charge $Q$ on the capacitor and the potential difference $V$ across it ($Q = CV$), combined with Ohm’s Law ($V = IR$). Because the current $I$ is the rate of flow of charge ($I = -dQ/dt$), the rate at which the capacitor loses charge is directly proportional to the amount of charge remaining. This leads to a mathematical pattern known as exponential decay, where the values of $Q$, $V$, and $I$ decrease by the same percentage over equal time intervals.

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Key Definitions

• Time Constant ($\tau$): The time taken for the charge, potential difference, or current of a discharging capacitor to fall to $1/e$ (approximately 37%) of its initial value. It is defined by the equation $\tau = RC$.
• Exponential Decay: A mathematical process where a quantity decreases at a rate proportional to its current value. In capacitor discharge, this means $\frac{dQ}{dt} \propto -Q$.
• Capacitance ($C$): The charge stored per unit potential difference across the plates of a capacitor. Unit: Farad (F), where $1 \text{ F} = 1 \text{ C V}^{-1}$.
• Initial Current ($I_0$): The maximum current in the circuit, occurring at the instant the discharge circuit is closed ($t = 0$). It is calculated using $I_0 = V_0 / R$.
• Initial Charge ($Q_0$): The maximum charge held by the capacitor at $t = 0$, calculated using $Q_0 = CV_0$.

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Content

3.1 The Physics of the Discharge Cycle

To understand why the discharge is exponential rather than linear, consider the sequence of events when a switch is closed in an $RC$ circuit:

1. Initial State ($t = 0$): The capacitor acts as a source of e.m.f. equal to $V_0$. The moment the circuit is completed, this $V_0$ drives an initial current $I_0 = V_0/R$ through the resistor.
2. Charge Depletion: As electrons flow from the negative plate to the positive plate through the external circuit, the charge $Q$ on the plates decreases.
3. Potential Drop: Since $V = Q/C$, as $Q$ decreases, the potential difference $V$ across the capacitor plates must also decrease.
4. Current Reduction: According to Ohm’s Law ($I = V/R$), the current $I$ is determined by the potential difference. As $V$ drops, the current $I$ must also drop.
5. Feedback Loop: Because the current is the rate of discharge ($I = -dQ/dt$), a smaller current means the charge $Q$ now decreases more slowly than it did initially. This results in a curve that levels off as it approaches zero, never theoretically reaching it.

3.2 Mathematical Derivation

The Cambridge 9702 syllabus requires you to use the exponential equations. Understanding the derivation aids in solving complex problems involving rates of change.

Apply Kirchhoff’s Second Law to the loop: The sum of e.m.f.s is equal to the sum of p.d.s. $$V_{capacitor} - V_{resistor} = 0 \implies V_C = V_R$$

Substitute the definitions for $V_C$ and $V_R$: $$\frac{Q}{C} = IR$$

Since the capacitor is losing charge, the current $I$ is the negative rate of change of charge: $I = -\frac{dQ}{dt}$. $$\frac{Q}{C} = -R \frac{dQ}{dt}$$

Rearrange to separate variables $Q$ and $t$: $$\frac{1}{Q} dQ = -\frac{1}{RC} dt$$

Integrate both sides from the initial state ($t=0, Q=Q_0$) to a later state ($t, Q$): $$\int_{Q_0}^{Q} \frac{1}{Q} dQ = -\frac{1}{RC} \int_{0}^{t} dt$$ $$[\ln Q]{Q_0}^{Q} = -\frac{1}{RC} [t]{0}^{t}$$ $$\ln Q - \ln Q_0 = -\frac{t}{RC}$$ $$\ln \left( \frac{Q}{Q_0} \right) = -\frac{t}{RC}$$

Taking the exponential ($e^x$) of both sides: $$\frac{Q}{Q_0} = e^{-\frac{t}{RC}}$$ $Q = Q_0 e^{-\frac{t}{RC}}$

Since $V \propto Q$ and $I \propto V$, the same derivation applies to potential difference and current: $V = V_0 e^{-\frac{t}{RC}}$ $I = I_0 e^{-\frac{t}{RC}}$

3.3 Graphical Analysis

Students must be able to interpret and sketch graphs of $Q$, $V$, and $I$ against time $t$.

• Common Features: All three graphs start at a maximum value ($Q_0, V_0, I_0$) and decay asymptotically toward the x-axis.
• The $Q-t$ Graph:
  • Gradient: The gradient at any point $t$ is the instantaneous current $I$ at that time.
  • Initial Gradient: The gradient at $t=0$ is the initial current $I_0$.
• The $I-t$ Graph:
  • Area: The total area under the $I-t$ graph from $t=0$ to $t=\infty$ represents the total initial charge $Q_0$ stored on the capacitor.
• The "Tangent at the Origin" Property:
  • If you draw a tangent to the curve at $t=0$, the point where this tangent intercepts the time axis (x-axis) is exactly $t = \tau$ (the time constant).

3.4 The Time Constant ($\tau = RC$)

The time constant is the characteristic scaling factor for the decay.

• Units: Resistance ($\Omega$) $\times$ Capacitance (F) = $(\text{V A}^{-1}) \times (\text{C V}^{-1}) = \text{C A}^{-1} = \text{C (C s}^{-1})^{-1} = \text{s}$.
• Physical Meaning:
  • After $1\tau$: $x = x_0 e^{-1} \approx 0.37 x_0$ (37% remains).
  • After $2\tau$: $x = x_0 e^{-2} \approx 0.135 x_0$ (13.5% remains).
  • After $5\tau$: $x = x_0 e^{-5} \approx 0.0067 x_0$ (Less than 1% remains; effectively fully discharged).
• Half-life ($t_{1/2}$): The time taken for the value to fall to half ($0.5 x_0$). $$0.5 x_0 = x_0 e^{-\frac{t_{1/2}}{RC}} \implies \ln(0.5) = -\frac{t_{1/2}}{RC}$$ $t_{1/2} = RC \ln 2 \approx 0.693 RC$

3.5 Linearization of Exponential Data

In practical exams or data analysis questions, we transform the exponential curve into a straight line to find $R$ or $C$ accurately.

Starting with $V = V_0 e^{-\frac{t}{RC}}$, take the natural logarithm ($\ln$) of both sides: $$\ln V = \ln(V_0 e^{-\frac{t}{RC}})$$ $$\ln V = \ln V_0 + \ln(e^{-\frac{t}{RC}})$$ $\ln V = -\left(\frac{1}{RC}\right)t + \ln V_0$

This matches the linear form $y = mx + c$:

• y-variable: $\ln V$ (or $\ln I$ or $\ln Q$)
• x-variable: $t$
• Gradient ($m$): $-\frac{1}{RC}$
• y-intercept ($c$): $\ln V_0$

3.6 Worked Examples

Worked example 1 — Calculating Time and Charge

A $2200 \mu\text{F}$ capacitor is charged to a potential difference of $9.0 \text{ V}$. It is then discharged through a $47 \text{ k}\Omega$ resistor. Calculate the time taken for the potential difference to fall to $2.0 \text{ V}$.

Step 1: Identify known values and convert to SI units. $C = 2200 \times 10^{-6} \text{ F}$ $R = 47 \times 10^3 \Omega$ $V_0 = 9.0 \text{ V}$ $V = 2.0 \text{ V}$

Step 2: Calculate the time constant $\tau$. $\tau = RC = (47 \times 10^3) \times (2200 \times 10^{-6}) = 103.4 \text{ s}$

Step 3: Use the discharge equation. $V = V_0 e^{-t/RC}$ $2.0 = 9.0 e^{-t/103.4}$

Step 4: Solve for $t$ using logarithms. $\frac{2.0}{9.0} = e^{-t/103.4}$ $\ln(0.2222) = -t/103.4$ $-1.504 = -t/103.4$ $t = 1.504 \times 103.4 = 155.5 \text{ s}$

Answer: $160 \text{ s}$ (to 2 s.f., matching the input data).

Worked example 2 — Determining Resistance from a Log-Graph

A student discharges a $100 \mu\text{F}$ capacitor through an unknown resistor $R$. A graph of $\ln(I/\mu\text{A})$ against $t/\text{s}$ is plotted. The line passes through $(0, 4.6)$ and $(20, 2.1)$. Determine the resistance $R$.

Step 1: Calculate the gradient of the line. $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2.1 - 4.6}{20 - 0} = \frac{-2.5}{20} = -0.125 \text{ s}^{-1}$

Step 2: Relate gradient to the time constant. $m = -\frac{1}{RC}$ $-0.125 = -\frac{1}{R \times (100 \times 10^{-6})}$

Step 3: Solve for $R$. $R = \frac{1}{0.125 \times 100 \times 10^{-6}}$ $R = \frac{1}{1.25 \times 10^{-5}} = 80,000 \Omega$

Answer: $80 \text{ k}\Omega$

Worked example 3 — Energy Loss during Discharge

A $500 \mu\text{F}$ capacitor is charged to $10 \text{ V}$ and then discharged through a resistor. Calculate the percentage of the initial energy remaining in the capacitor after a time equal to one time constant ($t = \tau$).

Step 1: Recall the energy equation. $E = \frac{1}{2}CV^2$

Step 2: Determine the voltage at $t = \tau$. $V = V_0 e^{-t/RC} = V_0 e^{-1}$

Step 3: Substitute $V$ into the energy equation for time $t$. $E_t = \frac{1}{2}C(V_0 e^{-1})^2 = \frac{1}{2}CV_0^2 e^{-2}$

Step 4: Compare to initial energy $E_0$. $E_0 = \frac{1}{2}CV_0^2$ $E_t = E_0 e^{-2}$

Step 5: Calculate the percentage. $\text{Percentage} = \frac{E_t}{E_0} \times 100 = e^{-2} \times 100 \approx 0.1353 \times 100 = 13.5%$

Answer: $13.5%$

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Key Equations

Equation	Description	Status
$\tau = RC$	Time constant for an $RC$ circuit	Data Sheet
$x = x_0 e^{-\frac{t}{RC}}$	Exponential decay for $x = Q, V, \text{ or } I$	Data Sheet
$I_0 = \frac{V_0}{R}$	Initial current at start of discharge	Memorise
$Q_0 = CV_0$	Initial charge stored	Memorise
$\ln x = -\frac{t}{RC} + \ln x_0$	Linearized form for graphing	Memorise
$t_{1/2} = RC \ln 2$	Time taken for value to halve	Memorise

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Common Mistakes to Avoid

• ❌ Wrong: Using $V = V_0(1 - e^{-t/RC})$ for discharging. ✓ Right: This is the charging equation for voltage. For discharging, all three variables ($Q, V, I$) use the simple $e^{-t/RC}$ form because they all decrease.
• ❌ Wrong: Forgetting to convert $\mu\text{F}$ or $\text{k}\Omega$ to base units. ✓ Right: Always use $10^{-6}$ for micro and $10^3$ for kilo. A single power-of-ten error will invalidate the entire calculation.
• ❌ Wrong: Thinking the capacitor is "empty" at $t = \tau$. ✓ Right: At $t = \tau$, the capacitor still has 37% of its charge. It is only "practically" empty after about $5\tau$.
• ❌ Wrong: Using $\log$ (base 10) instead of $\ln$ (natural log) on the calculator. ✓ Right: The exponential $e$ is the base of natural logarithms. Always use the $\ln$ key.
• ❌ Wrong: Assuming the area under a $V-t$ graph is charge. ✓ Right: Only the area under the $I-t$ graph represents charge ($Q = \int I dt$).

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Exam Tips

1. The Constant Ratio Test: If you are given a table of values and asked to show the decay is exponential, calculate the ratio $x_{t+\Delta t} / x_t$ for several equal time intervals $\Delta t$. If the ratio is constant, the decay is exponential.
2. Two-Point Ratio Method: If you don't know the initial value $x_0$, you can use the ratio of two points: $\frac{x_1}{x_2} = e^{\frac{t_2 - t_1}{RC}}$. This is very useful for finding $RC$ when the graph doesn't show the y-intercept.
3. Unit Check for Exponents: The term $\frac{t}{RC}$ must be dimensionless. If $t$ is in milliseconds, $RC$ must also be converted to milliseconds (or both to seconds) before dividing.
4. Significant Figures: Cambridge 9702 is strict on sig figs. If the resistance is $47 \text{ k}\Omega$ (2 s.f.) and capacitance is $2200 \mu\text{F}$ (2 s.f.), your final time or voltage should be given to 2 s.f.
5. Sketching Graphs: When asked to sketch a discharge graph, always label the y-intercept ($Q_0, V_0, \text{ or } I_0$) and at least one point on the curve (e.g., at $t = \tau$, $y = 0.37 y_0$) to show the scale.