# Modelling Capacitor Discharge ## Exponential Decay Graph for Capacitors - To verify if potential difference, $V$, or charge, $Q$, on a capacitor decreases exponentially, there are two methods: - **Constant ratio method**: Plot a $V$-$t$ graph and check the time constant is constant, or check if the time to halve from its initial value is constant - **Logarithmic graph method**: Plot a graph of $\ln V$ against $t$ and check if a straight line graph is obtained ## Constant Ratio Method - A general form of the exponential decay equation is given by $$x = x_0 e^{-At}$$ - Where $A$ is a constant - This equation shows that when $t = A^{-1}$ the value of $x$ will have decreased to approximately 37% of its original value, $x_0$: $$x = x_0 e^{-\frac{A}{A}} = x_0 e^{-1} = x_0 \left(\frac{1}{e}\right) \approx 0.37x_0$$ - Comparing this to the discharge equation for a capacitor: $$V = V_0 e^{-\frac{t}{\tau}}$$ - Therefore, for a discharging capacitor, when $t = \tau$ the potential difference on the capacitor will have decreased to approximately 37% of its original value

Equal intervals of time give equal fractional changes of 1/e in potential difference.

- Hence, to validate if potential difference across a capacitor decreases exponentially: - The time constant, or the time taken for the potential difference to decrease to 37% of its original value, will be constant - To find the time constant from a voltage-time graph, calculate $0.37V_0$ and determine the corresponding time for that value | Feature | Description | |---------|-------------| | Time at $0.37V_0$ | First time constant, $\tau$ | | Time between successive 37% drops | Also equals $\tau$ | | Pattern | Each time interval of $\tau$ produces the same fractional decrease | ### Visualizing Exponential Decay The graph below shows how potential difference decreases exponentially during capacitor discharge. Notice how at each time constant $\tau$, the voltage drops to 37% of its previous value:

- From the graph, if $\tau = 20$ s and $V_0 = 10$ V: - At $t = 20$ s (one time constant): $V \approx 3.7$ V - At $t = 40$ s (two time constants): $V \approx 1.4$ V (which is 37% of 3.7 V) - At $t = 60$ s (three time constants): $V \approx 0.5$ V (which is 37% of 1.4 V) ## Logarithmic Graph Method - The potential difference (p.d.) across the capacitance is defined by the equation: $$V = V_0 e^{-\frac{t}{RC}}$$ - Where: - $V$ = p.d. across the capacitor (V) - $V_0$ = initial p.d. across the capacitor (V) - $t$ = time (s) - $e$ = exponential function - $R$ = resistance of the resistor (Ω) - $C$ = capacitance of the capacitor (F) - Rearranging this equation for $\ln(V)$ by taking the natural log (ln) of both sides: $$\ln\left(\frac{V}{V_0}\right) = -\frac{t}{RC}$$ $$\ln(V) - \ln(V_0) = -\frac{t}{RC}$$ $$\ln(V) = -\frac{t}{RC} + \ln(V_0)$$ - Comparing this to the equation of a straight line: $y = mx + c$ - $y = \ln(V)$ - $x = t$ - gradient = $-1/RC$ - $c = \ln(V_0)$

A straight-line logarithmic graph of ln V against t confirms exponential decay.

### Analysis Steps 1. Plot a graph of $\ln(V)$ against $t$ and draw a line of best fit 2. Calculate the gradient (this should be negative) 3. The capacitance of the capacitor is equal to: $$C = -\frac{1}{R \times \text{gradient}}$$ ### Worked Example: Logarithmic Method **Scenario**: A student investigates the relationship between the potential difference and the time it takes to discharge a capacitor. The capacitor is labelled with a capacitance of 4200 µF and discharged through a 10 kΩ resistor. **Step 1: Complete the table** Add an extra column $\ln(V)$ and calculate this for each p.d. | Time $t$ / s | Potential Difference / V | $\ln(V)$ / V | |--------------|--------------------------|--------------| | 0.00 | 10.0 | 2.30 | | 10.00 | 8.95 | 2.19 | | 20.00 | 7.58 | 2.03 | | 30.00 | 6.13 | 1.81 | | 40.00 | 4.90 | 1.59 | | 50.00 | 4.12 | 1.42 | | 60.00 | 3.00 | 1.10 | | 70.00 | 2.03 | 0.71 | **Step 2: Plot the graph of $\ln(V)$ against time $t$** Make sure the axes are properly labelled and the line of best fit is drawn with a ruler

**Step 3: Calculate the gradient of the graph** The gradient is calculated by: $$\text{gradient} = \frac{0.9 - 2.2}{67.5 - 10} = -0.0226$$ **Step 4: Calculate the capacitance, $C$** $$C = -\frac{1}{R \times \text{gradient}}$$ $$C = -\frac{1}{10\,000 \times (-0.0226)} = 4.42 \times 10^{-3}\text{ F} = 4400\text{ µF}$$ ## Modelling the Discharge of a Capacitor - From electricity, the charge is defined as: $$\Delta Q = I\Delta t$$ - Where: - $I$ = current (A) - $\Delta Q$ = change in charge (C) - $\Delta t$ = change in time (s) - This means that the **area under a current-time graph** for a charging (or discharging) capacitor is the charge stored for a certain time interval - Rearranging for the current: $$I = \frac{\Delta Q}{\Delta t}$$

The gradient of the charge-time graph is the current at that time

- In the discharging graph, this is the discharging current at that time - In the charging graph, this is the charging current at that time - To calculate the gradient of a curve, draw a tangent at that point and calculate the gradient of that tangent ### Combining Relationships - As a capacitor charges or discharges, the current at any time can be found from Ohm's law: $$I = \frac{V}{R}$$ - From the definition of capacitance, the value of potential difference at any time is given by: $$V = \frac{Q}{C}$$ - Combining these equations gives: $$I = \frac{1}{R}\left(\frac{Q}{C}\right) = \frac{Q}{RC}$$ - For a discharging capacitor, the current decreases with time, hence: $$I = -\frac{\Delta Q}{\Delta t}$$ - This leads to an expression which can be used to solve for the time constant of a discharging capacitor: $$\frac{\Delta Q}{\Delta t} = -\frac{Q}{RC}$$

This equation is useful for modelling using spreadsheets

### Worked Example: Current from a Q-t Graph **Scenario**: A charge-time graph shows how the charge stored on a capacitor with capacitance $C$ varies with time as it discharges through a resistor. **Task**: Calculate the current through the circuit after 4 s. The graph below shows a typical charge-time relationship for a discharging capacitor:

**Answer**: **Step 1: Draw a tangent at $t = 4$** Draw a tangent line touching the curve at exactly $t = 4$ s using a ruler **Step 2: Calculate the gradient of the tangent to find the current $I$** From the graph, identify two clear points on the tangent line. For instance, if the tangent passes through $(0.4 \text{ s}, 0.17 \text{ µC})$ and $(7.6 \text{ s}, 0.04 \text{ µC})$: $$\text{gradient} = I = \frac{\Delta Q}{\Delta t} = \frac{(0.04 - 0.17) \times 10^{-6}}{(7.6 - 0.4)} = \frac{-0.13 \times 10^{-6}}{7.2}$$ $$I = -1.8 \times 10^{-8}\text{ A}$$ The magnitude of the current is $1.8 \times 10^{-8}$ A (the negative sign indicates discharge) ## Summary of Verification Methods | Method | What to Plot | What to Check | Result if Exponential | |--------|-------------|---------------|----------------------| | Constant Ratio | $V$ vs $t$ | Time constant $\tau$ | $\tau$ is constant at all points | | Constant Ratio | $V$ vs $t$ | Time to halve | Half-life is constant | | Logarithmic | $\ln(V)$ vs $t$ | Shape of graph | Produces a straight line | | Logarithmic | $\ln(V)$ vs $t$ | Gradient | Gradient = $-1/RC$ |