Discharging Capacitors

Spec mapping: OCR H556 Module 6.1 — Capacitors (discharging of a capacitor through a resistor; the exponential decay laws $Q(t) = Q_0 e^{-t/RC}$Q(t)=Q0​e−t/RC, $V(t) = V_0 e^{-t/RC}$V(t)=V0​e−t/RC, $I(t) = I_0 e^{-t/RC}$I(t)=I0​e−t/RC; the time constant $\tau = RC$τ=RC and its physical interpretation; the use of logarithmic plots to determine the time constant; PAG 8 — investigating the discharge of a capacitor). Refer to the official OCR H556 specification document for exact wording.

A charged capacitor connected to a resistor is one of the purest examples of exponential decay in physics. Left alone, a capacitor on its own would stay charged indefinitely — the plates are insulated and have nowhere to lose their charge. But connect the plates through a resistor and the charge drains away with a beautifully simple mathematical form: an exponential. This lesson introduces the time constant $\tau = RC$τ=RC and the exponential decay laws for charge, voltage and current.

This lesson continues Module 6.1. It is also a PAG 8 anchor lesson — investigating the discharge of a capacitor is a required practical for OCR H556. Every numerical idea you meet here will reappear in lesson 5 (charging) with a subtle but important sign change, and in Module 6.5 (medical physics) when we look at MRI relaxation times. The mathematics of exponential decay is universal in physics, and the discharge of a capacitor is by far the cleanest classroom demonstration of it.

---

The Discharge Circuit

Consider a capacitor $C$C charged to some initial voltage $V_0$V0​, and then connected through a switch to a resistor $R$R:

graph LR
    Cap[Capacitor C] --> S{Switch}
    S --> R1[Resistor R]
    R1 --> Cap

When the switch closes, current flows from the positive plate, through the resistor, to the negative plate, neutralising the charge bit by bit.

At any instant during the discharge:

• The voltage across the capacitor is $v = q/C$v=q/C.
• The voltage across the resistor must also be $v$v (it is in parallel with the capacitor; Kirchhoff's voltage law around the loop gives $v_C = v_R$vC​=vR​).
• The current is $i = v/R = q/(RC)$i=v/R=q/(RC).

But the current is the rate at which charge leaves the capacitor: $i = -\mathrm{d}q/\mathrm{d}t$i=−dq/dt. Combining,

$\frac{\mathrm{d}q}{\mathrm{d}t} = -\frac{q}{RC}.$dtdq​=−RCq​.

This is the defining equation of exponential decay — the rate of change of $q$q is proportional to $q$q itself. The same mathematical form appears in radioactive decay (Module 6.4), Newton's law of cooling, atmospheric pressure with altitude, and the absorption of light through an attenuating medium. The physical content differs but the mathematics is identical.

---

Solving the Decay Equation

The solution to $\mathrm{d}q/\mathrm{d}t = -q/(RC)$dq/dt=−q/(RC) with initial charge $Q_0$Q0​ is

$\boxed{\,q(t) = Q_0\,e^{-t/RC}\,}$q(t)=Q0​e−t/RC​

The voltage is proportional to the charge, so

$v(t) = V_0\,e^{-t/RC}$v(t)=V0​e−t/RC

and the current is proportional to the voltage through Ohm's law:

$i(t) = I_0\,e^{-t/RC}, \quad \text{with } I_0 = \frac{V_0}{R}.$i(t)=I0​e−t/RC,with I0​=RV0​​.

Every quantity — charge, voltage, current — decays with the same exponential and the same time constant. This is one of the cleanest features of the topic: a single exponential $e^{-t/RC}$e−t/RC runs through every formula in lessons 4 and 5.

To verify that $q(t) = Q_0 e^{-t/RC}$q(t)=Q0​e−t/RC is the solution, differentiate:

$\frac{\mathrm{d}q}{\mathrm{d}t} = -\frac{1}{RC} Q_0 e^{-t/RC} = -\frac{q(t)}{RC}\ \checkmark$dtdq​=−RC1​Q0​e−t/RC=−RCq(t)​ ✓

and check the initial condition: $q(0) = Q_0 e^{0} = Q_0$q(0)=Q0​e0=Q0​. $\checkmark$✓ The exponential is the only function whose derivative is proportional to itself, which is why exponentials appear so universally in decay processes.

---

The Time Constant τ = RC

The quantity $RC$RC has units of seconds. You can prove this with dimensional analysis: $[R] =$[R]= V A$^{-1}$−1, $[C] =$[C]= A s V$^{-1}$−1, so $[RC] =$[RC]= V A$^{-1}$−1 $\times$× A s V$^{-1}$−1 $=$= s. It is given a special name, the time constant, symbol $\tau$τ (Greek tau):

$\boxed{\,\tau = RC\,}$τ=RC​

The time constant has a clear physical meaning: after one time constant, the charge has dropped to $e^{-1}$e−1 of its initial value — about 37%. After two time constants, it is $e^{-2} \approx 13.5\%$e−2≈13.5%. After five time constants, it is $e^{-5} \approx 0.67\%$e−5≈0.67% — the capacitor is essentially fully discharged.

Time	Fraction of initial charge remaining
0	100.0 %
τ	36.79 % (1/e)
2τ	13.53 %
3τ	4.98 %
5τ	0.67 %
10τ	0.0045 %

graph LR
    A[t=0] -->|"Q₀"| B[t=τ]
    B -->|"0.37 Q₀"| C[t=2τ]
    C -->|"0.14 Q₀"| D[t=5τ]
    D -->|"0.007 Q₀"| E[t=10τ]

The decay curve for charge or voltage looks like this (time on x-axis, fraction of initial value on y-axis):

The tangent to the curve at $t = 0$t=0 meets the time axis at exactly $t = \tau$t=τ. This is a graphical way to measure $\tau$τ from an experimental discharge trace — extend the initial slope and read off where it cuts the time axis.

---

Worked Example 1 — Discharge Through a Resistor

A $100\ \mu$100 μF capacitor is charged to $12$12 V and then discharged through a $47\ \text{k}\Omega$47 kΩ resistor. Calculate (a) the time constant, (b) the initial current, (c) the voltage after $10$10 s, and (d) the time for the voltage to fall to $1$1 V.

(a) Time constant.

$\tau = RC = (47 \times 10^3)(100 \times 10^{-6}) = 4.7\ \text{s}.$τ=RC=(47×103)(100×10−6)=4.7 s.

(b) Initial current.

$I_0 = \frac{V_0}{R} = \frac{12}{47\,000} \approx 2.55 \times 10^{-4}\ \text{A} \approx 0.26\ \text{mA}.$I0​=RV0​​=4700012​≈2.55×10−4 A≈0.26 mA.

(c) Voltage after $10$10 s.

$$\begin{aligned} V &= V_0\,e^{-t/\tau} \\ &= 12 \times e^{-10/4.7} \\ &= 12 \times e^{-2.128} \\ &= 12 \times 0.1191 \\ &\approx 1.43\ \text{V}. \end{aligned}$$V​=V0​e−t/τ=12×e−10/4.7=12×e−2.128=12×0.1191≈1.43 V.​

(d) Time to fall to $1$1 V.

$$\begin{aligned} 1 &= 12\,e^{-t/4.7} \\ \frac{1}{12} &= e^{-t/4.7} \\ -\frac{t}{4.7} &= \ln(1/12) = -2.485 \\ t &= 4.7 \times 2.485 \approx 11.7\ \text{s}. \end{aligned}$$1121​−4.7t​t​=12e−t/4.7=e−t/4.7=ln(1/12)=−2.485=4.7×2.485≈11.7 s.​

So in about $12$12 s the voltage has fallen from $12$12 V to $1$1 V — roughly $2.5$2.5 time constants. The same calculation runs identically for $Q(t)$Q(t) and $i(t)$i(t) because all three quantities share the same $e^{-t/RC}$e−t/RC factor.

---

Logarithmic Form

Taking the natural log of the decay equation gives a useful linear form:

$\ln q(t) = \ln Q_0 - \frac{t}{RC}.$lnq(t)=lnQ0​−RCt​.

So a graph of $\ln q$lnq against $t$t is a straight line with intercept $\ln Q_0$lnQ0​ and gradient $-1/RC$−1/RC. This is a standard OCR exam technique for finding the time constant from experimental data: rearrange to a straight line, fit, and read off the gradient.

The same works for $V$V or $I$I:

$\ln v(t) = \ln V_0 - \frac{t}{\tau}, \quad \ln i(t) = \ln I_0 - \frac{t}{\tau}.$lnv(t)=lnV0​−τt​,lni(t)=lnI0​−τt​.

The intercept differs (it depends on which initial value you start with), but the gradient $-1/\tau$−1/τ is the same in all three plots — a useful consistency check on experimental data.

Exam tip. If a question gives you a discharge curve and asks for $RC$RC, the easiest method is to plot $\ln V$lnV (or $\ln Q$lnQ) against $t$t, fit a straight line, and take the negative reciprocal of the gradient. This is more accurate than the "$37\%$37% method" because it uses every data point, not just two.

---

Half-Life of a Capacitor

Just as radioactive decay has a half-life $T_{1/2}$T1/2​, so does a capacitor discharge. Setting $q = Q_0/2$q=Q0​/2:

$$\begin{aligned} \tfrac{1}{2} &= e^{-T_{1/2}/RC} \\ \ln\tfrac{1}{2} &= -\frac{T_{1/2}}{RC} \\ T_{1/2} &= RC \ln 2 \approx 0.693\,RC = 0.693\,\tau. \end{aligned}$$21​ln21​T1/2​​=e−T1/2​/RC=−RCT1/2​​=RCln2≈0.693RC=0.693τ.​

Halving time and time constant are related by the same factor $\ln 2 \approx 0.693$ln2≈0.693 as in nuclear physics. This is the same mathematics all over again — exponential decay is exponential decay, whether it is charge on a capacitor, radioactive nuclei in a sample, or the temperature difference between a cup of coffee and the room.

---

PAG 8 Practical Notes — Investigating the Discharge of a Capacitor

A typical OCR Practical Activity Group 8 (PAG 8) experiment asks you to:

1. Charge a capacitor through a switch to a known voltage.
2. Discharge it through a fixed resistor, recording voltage against time with a data logger or voltmeter and stopwatch.
3. Plot $V$V against $t$t (exponential) and $\ln V$lnV against $t$t (straight line).
4. Calculate the time constant from either curve.
5. Compare with $RC$RC from the nominal component values.

Electrolytic capacitors have wide tolerances (often $\pm 20\%$±20%), so the experimental $\tau$τ and the nominal $RC$RC rarely agree better than about $10\%$10%. Always quote a realistic uncertainty. The dominant uncertainty in a school PAG 8 practical is usually the capacitor tolerance — the resistor is typically $1\%$1%, but a $100\ \mu$100 μF electrolytic might be anywhere from $80$80 to $120\ \mu$120 μF. A clean data-logger gradient-fit of $\ln V$lnV vs $t$t is the gold-standard analysis.

Common practical hazards:

• A high-voltage capacitor (above $\sim 30$∼30 V) can deliver a painful shock if you short it across your fingers. Discharge through a resistor before handling.
• Electrolytic capacitors have a polarity. Connecting them the wrong way can cause them to fail explosively, especially at high voltage.
• The voltmeter has its own internal resistance. If $R_\text{voltmeter}$Rvoltmeter​ is comparable to $R$R, the effective resistance in the circuit is $R$R in parallel with $R_\text{voltmeter}$Rvoltmeter​, and your measured $\tau$τ will be too short. Use a digital voltmeter with $R_\text{voltmeter} > 10\,\text{M}\Omega$Rvoltmeter​>10MΩ, or a data logger with high input impedance.

---

Worked Example 2 — Discharge with Energy Tracking

A $470\ \mu$470 μF capacitor is charged to $V_0 = 12$V0​=12 V and then discharged through a $2.2\ \text{k}\Omega$2.2 kΩ resistor. Calculate (a) the time constant, (b) the time for the voltage to fall to half its initial value, (c) the total energy initially stored, and (d) the energy dissipated as a function of time $E_R(t)$ER​(t).

(a) Time constant.

$\tau = RC = (2.2 \times 10^3)(470 \times 10^{-6}) = 1.034\ \text{s} \approx 1.03\ \text{s}.$τ=RC=(2.2×103)(470×10−6)=1.034 s≈1.03 s.

(b) Time to halve. From $\tfrac{1}{2} = e^{-t/\tau}$21​=e−t/τ:

$T_{1/2} = \tau\,\ln 2 = 1.034 \times 0.693 \approx 0.717\ \text{s}.$T1/2​=τln2=1.034×0.693≈0.717 s.

So in just under three-quarters of a second the capacitor voltage has dropped from $12$12 V to $6$6 V.

(c) Initial energy stored.

$E_0 = \tfrac{1}{2}CV_0^2 = \tfrac{1}{2}(470 \times 10^{-6})(12)^2 \approx 0.0338\ \text{J} = 33.8\ \text{mJ}.$E0​=21​CV02​=21​(470×10−6)(12)2≈0.0338 J=33.8 mJ.

(d) Energy dissipated as a function of time. The remaining stored energy at time $t$t is

$E_C(t) = \tfrac{1}{2}C\bigl[V_0 e^{-t/\tau}\bigr]^2 = \tfrac{1}{2}CV_0^2\,e^{-2t/RC}.$EC​(t)=21​C[V0​e−t/τ]2=21​CV02​e−2t/RC.

Energy conservation requires that whatever has left the capacitor must have been dissipated in $R$R, so the energy dissipated by time $t$t is

$\boxed{\,E_R(t) = \tfrac{1}{2}CV_0^2\bigl(1 - e^{-2t/RC}\bigr)\,}$ER​(t)=21​CV02​(1−e−2t/RC)​

Notice the factor of $2$2 in the exponent: energy decays twice as fast as voltage or charge, because energy goes like $V^2$V2. So while $V$V has time constant $\tau = RC$τ=RC, the energy has effective time constant $\tau/2 = RC/2$τ/2=RC/2.

Numerical check at $t = \tau = 1.034$t=τ=1.034 s:

$E_R(\tau) = 33.8 \times (1 - e^{-2}) = 33.8 \times 0.865 \approx 29.2\ \text{mJ}.$ER​(τ)=33.8×(1−e−2)=33.8×0.865≈29.2 mJ.

After just one time constant, $86.5\%$86.5% of the initial energy has already been dissipated — a striking contrast with the $63\%$63% figure for voltage. By $t = 2\tau$t=2τ the dissipated fraction is $1 - e^{-4} \approx 98.2\%$1−e−4≈98.2%; by $t = 5\tau$t=5τ it is $1 - e^{-10} \approx 99.995\%$1−e−10≈99.995%. Energy dies away dramatically faster than voltage does, a fact that surprises many students.

---

Half-Life and the Analogy with Radioactive Decay

The discharge half-life

$T_{1/2} = RC\,\ln 2 \approx 0.693\,\tau$T1/2​=RCln2≈0.693τ

is structurally identical to the radioactive half-life $T_{1/2} = \ln 2/\lambda$T1/2​=ln2/λ from Module 6.4. The correspondence is exact:

Capacitor discharge	Radioactive decay
Charge $Q(t) = Q_0 e^{-t/RC}$Q(t)=Q0​e−t/RC	Activity $N(t) = N_0 e^{-\lambda t}$N(t)=N0​e−λt
Decay constant $1/\tau = 1/RC$1/τ=1/RC	Decay constant $\lambda$λ
Half-life $T_{1/2} = \tau\,\ln 2$T1/2​=τln2	Half-life $T_{1/2} = \ln 2/\lambda$T1/2​=ln2/λ
Linear plot: $\ln Q$lnQ vs $t$t, gradient $-1/\tau$−1/τ	Linear plot: $\ln N$lnN vs $t$t, gradient $-\lambda$−λ
$5\tau$5τ "effectively discharged"	$5\,T_{1/2}/\ln 2$5T1/2​/ln2 "effectively decayed"

The mathematics is the same because both processes obey a rate equation in which the rate of loss is proportional to what is left. For a capacitor, $\mathrm{d}Q/\mathrm{d}t = -Q/(RC)$dQ/dt=−Q/(RC); for a sample of radionuclide, $\mathrm{d}N/\mathrm{d}t = -\lambda N$dN/dt=−λN. Same differential equation, same exponential solution, same half-life formula. OCR examiners deliberately draw out this parallel in synoptic questions — recognising it lets you reuse Module 6.1 mathematics in Module 6.4 and vice versa.

The physical content differs sharply: in the capacitor, charge has a well-defined deterministic route (through the resistor) and the "decay" is a smooth bulk current; in radioactivity, individual nuclei decay randomly, and the smooth exponential emerges only as a statistical average over many nuclei. But for exam-grade calculations, the formulae are interchangeable up to a symbol rename.

---

Practical Applications of RC Discharge