Time to Halve & Quantitative Discharge Revision Notes | A Level AQA Physics

1. Why Capacitor Discharge is Never Instant

Picture a capacitor fully charged to voltage $V_0$, then connected across a resistor. You might expect the charge to drain away at a steady rate — but it does not. The discharge slows down continuously, because the current driving charge off the plates depends on the voltage still remaining, and that voltage is itself falling.

This self-limiting behaviour produces an exponential decay: the same fraction of charge is lost in each equal time interval, rather than the same amount. Understanding this quantitatively is the core of this lesson — and it unlocks two powerful tools: the exponential decay equations and the half-life formula.

2. The Exponential Decay Equations

The starting point is charge. At any time $t$ after discharge begins, the charge remaining on the capacitor is given by the equation $Q = Q_0 e^{-t/RC}$. Here $Q_0$ is the charge at $t = 0$, $R$ is the resistance in ohms, $C$ is the capacitance in farads, and the product $RC$ has units of seconds.

Because voltage across the capacitor is $V = Q/C$ and the initial voltage is $V_0 = Q_0/C$, dividing both sides of the charge equation by $C$ gives the corresponding voltage equation. Similarly, because the discharge current at any instant is $I = V/R$, dividing the voltage equation by $R$ gives the current equation. All three quantities decay with exactly the same exponential shape.

The exponent $-t/RC$ is dimensionless, which is a useful self-check: $t$ is in seconds and $RC$ is in $\Omega \cdot \text{F} = \text{s}$, so the ratio is unitless. If you ever find yourself putting a value with units into the exponent without cancellation, something has gone wrong.

A larger $RC$ means a longer time constant, which means a slower discharge. Doubling $R$ doubles $\tau$; doubling $C$ also doubles $\tau$. This makes physical sense: a larger resistor restricts the current that can flow, and a larger capacitor holds more charge to begin with, so both slow the process down.

3. The Half-Life of Discharge

The time constant tells you when charge reaches 37% of its starting value. But there is a second, equally useful quantity: the time for charge to halve. This is the half-life of discharge, $T_{\frac{1}{2}}$, and it is directly linked to $RC$.

To derive it, set $Q = Q_0/2$ in the discharge equation and solve for $t$. The derivation is short and worth knowing, because examiners sometimes ask you to justify the formula rather than just quote it.

Worked example: Deriving the Half-Life Formula $T_{\frac{1}{2}} = 0.69RC$

Show that the time for the charge on a discharging capacitor to halve is $T_{\frac{1}{2}} = 0.69RC$.

1. $Q = Q_0 \, e^{-t/RC}$ — This is the standard AQA equation for charge during discharge.
2. $\frac{Q_0}{2} = Q_0 \, e^{-T_{\frac{1}{2}}/RC}$ — We want the time at which the charge has fallen to exactly half its initial value.
3. $\frac{1}{2} = e^{-T_{\frac{1}{2}}/RC}$ — $Q_0$ cancels — the half-life is independent of the initial charge.
4. $\ln\left(\frac{1}{2}\right) = -\frac{T_{\frac{1}{2}}}{RC}$ — $\ln(e^x) = x$, so the right side simplifies to $-T_{\frac{1}{2}}/RC$.
5. $-\ln 2 = -\frac{T_{\frac{1}{2}}}{RC}$ $\Rightarrow$ $T_{\frac{1}{2}} = RC \ln 2 \approx 0.693\,RC$ — $\ln(1/2) = -\ln 2 \approx -0.693$. The negatives cancel, giving the result. AQA rounds this to $0.69RC$.

Answer: $T_{\frac{1}{2}} = 0.69RC$ (to 2 significant figures, using $\ln 2 \approx 0.693$)

Notice that $T_{\frac{1}{2}}$ does not depend on $Q_0$ at all. The charge halves in the same time interval regardless of where you start measuring — exactly as with radioactive decay. This is the hallmark of exponential behaviour.

4. Using the Equations: Worked Calculations

The equations look clean on paper, but exam questions require careful substitution and confident use of the natural logarithm. The two worked examples below cover the two most common question types: finding a quantity at a given time, and finding the time at which a quantity reaches a given value.

Worked example: Finding Voltage at a Given Time

A $470\,\mu\text{F}$ capacitor is charged to $9.0\,\text{V}$ and then discharged through a $15\,\text{k}\Omega$ resistor. Calculate the voltage across the capacitor $12\,\text{s}$ after discharge begins. Give your answer to 2 significant figures.

1. $V_0 = 9.0\,\text{V}$, $R = 15 \times 10^3\,\Omega$, $C = 470 \times 10^{-6}\,\text{F}$, $t = 12\,\text{s}$ — Convert all values to SI base units before substituting.
2. $RC = 15 \times 10^3 \times 470 \times 10^{-6} = 7.05\,\text{s}$ — $15000 \times 0.000470 = 7.05\,\text{s}$. Keep this unrounded for the next step.
3. $-\frac{t}{RC} = -\frac{12}{7.05} = -1.702...$ — Keep at least 3 significant figures here to avoid rounding errors in the final answer.
4. $e^{-1.702} \approx 0.1824$ — Use the $e^x$ function on your calculator. A quick check: $e^{-1} \approx 0.368$, so $e^{-1.7}$ should be smaller — 0.18 is plausible.
5. $V = 9.0 \times 0.1824 = 1.642\,\text{V}$ — $9.0 \times 0.1824 = 1.6416\,\text{V}$
6. $V \approx 1.6\,\text{V}$ — Match the significant figures to the data given in the question. Always state the unit.

Answer: $V \approx 1.6\,\text{V}$

Worked example: Finding the Time for Charge to Reach a Given Value

A capacitor has initial charge $Q_0 = 8.0 \times 10^{-3}\,\text{C}$ and discharges through a $25\,\text{k}\Omega$ resistor. The capacitance is $200\,\mu\text{F}$. Find the time at which the charge has fallen to $2.0 \times 10^{-3}\,\text{C}$.

1. $Q = Q_0 \, e^{-t/RC}$ — State the equation first — this earns the method mark [M1].
2. $2.0 \times 10^{-3} = 8.0 \times 10^{-3} \times e^{-t/RC}$ — Substituting $Q = 2.0 \times 10^{-3}\,\text{C}$ and $Q_0 = 8.0 \times 10^{-3}\,\text{C}$ [M1].
3. $\frac{2.0 \times 10^{-3}}{8.0 \times 10^{-3}} = e^{-t/RC}$ $\Rightarrow$ $0.25 = e^{-t/RC}$ — The ratio $Q/Q_0 = 0.25$. Notice this is $1/4$, which is two halvings — a useful cross-check.
4. $\ln(0.25) = -\frac{t}{RC}$ — $\ln(0.25) = \ln(1/4) = -\ln 4 \approx -1.386$
5. $RC = 25 \times 10^3 \times 200 \times 10^{-6} = 5.0\,\text{s}$ — $25000 \times 0.000200 = 5.0\,\text{s}$
6. $t = -RC \times \ln(0.25) = -5.0 \times (-1.386) = 6.93\,\text{s}$ — The two negatives cancel. $5.0 \times 1.386 = 6.93\,\text{s}$ [A1]. Cross-check: two half-lives would be $2 \times 0.69 \times 5.0 = 6.9\,\text{s}$ — consistent.

Answer: $t \approx 6.9\,\text{s}$ (to 2 significant figures)

5. Reading Discharge Graphs

Exam questions frequently present a graph of $V$ against $t$ (or $Q$ against $t$, or $I$ against $t$) and ask you to extract $RC$ or $T_{\frac{1}{2}}$. The shape is always the same: a curve that starts steeply and flattens asymptotically towards zero.

To find $T_{\frac{1}{2}}$ from a graph, read off the initial value, halve it, and find the corresponding time on the $x$-axis. For example, if $V_0 = 10\,\text{V}$, find where the curve crosses $5\,\text{V}$ and read off that time. Repeat from $5\,\text{V}$ to $2.5\,\text{V}$ to confirm the half-life is constant — this is a key check that the decay is genuinely exponential.

Once you have $T_{\frac{1}{2}}$, you can find $RC$ by rearranging: $RC = T_{\frac{1}{2}} / 0.69$. Alternatively, read off the time at which the voltage has fallen to 37% of its initial value — that time is exactly one time constant $\tau = RC$.

7. Exam Technique and Common Mistakes

Capacitor discharge questions appear in multiple formats: numerical calculation, graph interpretation, and extended explanation. The mark scheme rewards precise scientific language and correct equation use, so the guidance below targets the specific points where students most often drop marks.

Every spec requirement in this lesson — the half-life formula $T_{\frac{1}{2}} = 0.69RC$, the exponential equations for $Q$, $V$, and $I$, and their quantitative application — can appear in calculation, graph, or extended-writing format. Practise all three formats on past papers to build fluency with the full range of question styles.