Charging Capacitors

Spec mapping: OCR H556 Module 6.1 — Capacitors (charging of a capacitor through a resistor; the rising exponential laws $Q(t) = Q_0(1 - e^{-t/RC})$ and $V(t) = V_0(1 - e^{-t/RC})$ ; the decaying current $I(t) = I_0 e^{-t/RC}$ ; the same time constant $\tau = RC$ as in discharging; the use of logarithmic plots of $V_0 - V_C$ to determine $\tau$ ; PAG 8 — investigating the charging of a capacitor). Refer to the official OCR H556 specification document for exact wording.

The reverse of discharging is charging. Connect an uncharged capacitor to a battery through a resistor and the capacitor fills up — fast at first, then more slowly, then tailing off as it approaches the supply voltage. This lesson derives the charging equations, shows how they differ subtly from the discharge equations, and shows how to read charging curves in exam questions.

This lesson continues Module 6.1 and is the partner to lesson 4. It is also a PAG 8 anchor lesson — OCR allow either charging or discharging as the required-practical investigation, and most teachers do both. The mathematics is identical except for one sign change, but that sign change makes the charge and voltage curves " $1 - e^{-t/\tau}$ " rather than " $e^{-t/\tau}$ ". The current is still $e^{-t/\tau}$ — a subtle but heavily examined distinction.

The Charging Circuit

graph LR
    B((V₀ battery)) --- R1[Resistor R] --- C1[Capacitor C] --- B

When the switch closes at $t = 0$ , the capacitor is empty: its voltage is $0$ and the full e.m.f. $V_0$ appears across the resistor. The initial current is therefore the maximum possible,

$I_0 = \frac{V_0}{R}.$

As charge builds up on the capacitor, the voltage across it rises and the voltage across the resistor falls. The current drops because it is proportional to the resistor voltage (which is $V_0 - v_C$ ). Eventually the capacitor voltage equals $V_0$ , no more current flows, and the capacitor is fully charged.

This is a key conceptual point: at $t = 0$ the current is largest, not smallest, because the empty capacitor offers no back-voltage and the full battery voltage drives current through $R$ . The current then falls as the capacitor "fills up" and resists further charge.

Deriving the Charging Equations

By Kirchhoff's voltage law around the loop,

$V_0 = v_R + v_C = iR + \frac{q}{C}.$

With $i = \mathrm{d}q/\mathrm{d}t$ this becomes

$V_0 = R\,\frac{\mathrm{d}q}{\mathrm{d}t} + \frac{q}{C}.$

Rearranging,

$\frac{\mathrm{d}q}{\mathrm{d}t} = \frac{CV_0 - q}{RC}.$

Let $Q_0 = CV_0$ be the final charge on the capacitor. Substituting $u = Q_0 - q$ gives $\mathrm{d}u/\mathrm{d}t = -u/(RC)$ , the familiar exponential decay equation of lesson 4, whose solution is $u = Q_0 e^{-t/RC}$ . Therefore

$\boxed{\,q(t) = Q_0\bigl(1 - e^{-t/RC}\bigr)\,}$

and correspondingly

$v_C(t) = V_0\bigl(1 - e^{-t/RC}\bigr).$

The current, however, decays from $I_0$ just as in a discharge circuit:

$i(t) = I_0\,e^{-t/RC}, \quad I_0 = \frac{V_0}{R}.$

This sign-asymmetry is the central subtlety of the topic:

Charge and voltage across the capacitor: $1 - e^{-t/RC}$ — rising from $0$ to the final value.
Current: $e^{-t/RC}$ — falling from $I_0$ to $0$ .

At any instant, $V_0 = v_C + iR$ , which you can check by adding the two expressions:

$v_C + iR = V_0(1 - e^{-t/\tau}) + (V_0/R)\,e^{-t/\tau}\,R = V_0 - V_0 e^{-t/\tau} + V_0 e^{-t/\tau} = V_0.\ \checkmark$

Time	Capacitor voltage $v_C/V_0$	Current $i/I_0$
$0$	$0$	$1$
$\tau$	$1 - 1/e \approx 0.632$	$1/e \approx 0.368$
$2\tau$	$0.865$	$0.135$
$3\tau$	$0.950$	$0.050$
$5\tau$	$0.993$	$0.007$

The capacitor is essentially fully charged after about $5$ time constants — the same " $5\tau$ rule" as for discharging.

Charging Curves

Notice the two curves are mirror images about the horizontal line at $V_0/2$ : as the capacitor voltage rises from $0$ to $V_0$ , the current falls from $I_0$ to $0$ , and at all times

$V_0 = v_C + v_R = v_C + iR,$

confirming Kirchhoff's voltage law (and, behind it, energy conservation).

Worked Example 1 — Charging Through a Resistor

A $220\ \mu$ F capacitor is charged from $0$ V through a $10\ \text{k}\Omega$ resistor by a $9.0$ V battery. Calculate (a) the initial current, (b) the time constant, (c) the voltage after $3.0$ s, (d) the current at the same moment, (e) the time for the voltage to reach $8.0$ V.

(a) Initial current.

$I_0 = \frac{V_0}{R} = \frac{9.0}{10\,000} = 9.0 \times 10^{-4}\ \text{A} = 0.90\ \text{mA}.$

(b) Time constant.

$\tau = RC = (10 \times 10^3)(220 \times 10^{-6}) = 2.2\ \text{s}.$

(c) Voltage after $3.0$ s.

\begin{aligned} V_C &= V_0(1 - e^{-t/\tau}) \\ &= 9.0 \times (1 - e^{-3/2.2}) \\ &= 9.0 \times (1 - e^{-1.3636}) \\ &= 9.0 \times (1 - 0.2557) \\ &= 9.0 \times 0.7443 \\ &\approx 6.7\ \text{V}. \end{aligned}

(d) Current after $3.0$ s. At the same instant the voltage across the resistor is $V_0 - V_C = 9.0 - 6.7 = 2.3$ V, so

$i = \frac{2.3}{10\,000} \approx 2.3 \times 10^{-4}\ \text{A} = 0.23\ \text{mA}.$

Equivalently, $i = I_0\,e^{-t/\tau} = 0.90 \times 0.2557 \approx 0.23$ mA. The two methods agree.

(e) Time to reach $8.0$ V.

\begin{aligned} 8.0 &= 9.0\,(1 - e^{-t/2.2}) \\ 1 - \frac{8.0}{9.0} &= e^{-t/2.2} \\ \frac{1}{9} &= e^{-t/2.2} \\ \frac{t}{2.2} &= \ln 9 = 2.197 \\ t &\approx 4.83\ \text{s}. \end{aligned}

About $4.8$ s — roughly $2.2$ time constants. This is reasonable: at $t = \tau$ the voltage is $\approx 63\%$ of $V_0 = 5.7$ V; at $t = 2\tau$ it is $\approx 86\%$ $= 7.8$ V; at $t = 3\tau$ it is $\approx 95\%$ $= 8.55$ V. So " $8.0$ V" should land between $2\tau$ and $3\tau$ , which it does.

Graphical Analysis

A common OCR question gives you a charging graph of $V_C$ against $t$ and asks you to extract the time constant. Two techniques:

Method 1 — reading at $63\%$ . Draw a horizontal line at $V_C = 0.63\,V_0$ (because $1 - 1/e \approx 0.632$ ). The time at which the curve crosses this line is one time constant. This is the charging analogue of the " $37\%$ " reading used in the discharge case.

Method 2 — linearisation. Compute $\ln(V_0 - V_C)$ and plot against $t$ . From

\begin{aligned} V_C &= V_0(1 - e^{-t/RC}) \\ V_0 - V_C &= V_0\,e^{-t/RC} \\ \ln(V_0 - V_C) &= \ln V_0 - \frac{t}{RC} \end{aligned}

you get a straight line with gradient $-1/RC$ and intercept $\ln V_0$ . Note the quantity you linearise is $V_0 - V_C$ , not $V_C$ itself, for the charging case. This is the subtle but heavily examined trap.

Exam tip. For a discharge plot, linearise $V_C$ directly ( $\ln V_C$ vs $t$ ). For a charge plot, linearise $V_0 - V_C$ ( $\ln(V_0 - V_C)$ vs $t$ ). Getting these two the wrong way round is a classic mistake — losing the linearity by trying to take $\ln$ of a $(1 - e^{-t/\tau})$ shape gives a curve, not a line.

Energy During Charging — A Reminder

As we saw in lesson 3, the battery does $CV_0^2$ of work during the entire charging process, but only $\tfrac{1}{2}CV_0^2$ ends up stored in the capacitor. The other $\tfrac{1}{2}CV_0^2$ is dissipated in the charging resistor $R$ . You can verify this by integrating the instantaneous power $i^2 R = (V_0/R)^2 e^{-2t/RC}\,R$ from $0$ to $\infty$ :

$\int_0^\infty \frac{V_0^2}{R}e^{-2t/RC}\,\mathrm{d}t = \frac{V_0^2}{R} \cdot \frac{RC}{2} = \tfrac{1}{2}CV_0^2.\ \checkmark$

The integral is exactly $\tfrac{1}{2}CV_0^2$ , independent of $R$ — the 50%-of-battery-work-is-dissipated result of lesson 3.

Worked Example 2 — Tabulating the Rise

A $100\ \mu$ F capacitor charges through a $10\ \text{k}\Omega$ resistor from a $9.0$ V supply, starting uncharged at $t = 0$ . Tabulate $V_C(t)$ and $i(t)$ at $t = 0,\ \tau,\ 2\tau,\ 3\tau,\ 5\tau$ , and confirm that $V_C$ reaches $63.2\%$ of $V_0$ at one time constant.

Time constant. $\tau = RC = (10 \times 10^3)(100 \times 10^{-6}) = 1.0$ s.

Initial current. $I_0 = V_0/R = 9.0/10\,000 = 9.0 \times 10^{-4}$ A $= 0.90$ mA.

Using $V_C(t) = V_0(1 - e^{-t/\tau})$ and $i(t) = I_0\,e^{-t/\tau}$ :

$t$	$t/\tau$	$1 - e^{-t/\tau}$	$V_C$ (V)	$V_C/V_0$	$i$ (mA)
$0$	$0$	$0$	$0.00$	$0\%$	$0.900$
$\tau = 1.0$ s	$1$	$0.632$	$5.69$	$63.2\%$	$0.331$
$2\tau = 2.0$ s	$2$	$0.865$	$7.78$	$86.5\%$	$0.122$
$3\tau = 3.0$ s	$3$	$0.950$	$8.55$	$95.0\%$	$0.045$
$5\tau = 5.0$ s	$5$	$0.993$	$8.94$	$99.3\%$	$0.006$

At $t = \tau$ , $V_C/V_0 = 1 - 1/e = 1 - 0.3679 = 0.6321 \approx 63.2\%$ — the canonical OCR figure. Notice also how the current has fallen to $I_0/e \approx 37\%$ of its initial value, mirroring the discharge case exactly. At every instant the sum $V_C + iR$ equals $V_0$ : at $t = \tau$ , $5.69 + (0.331 \times 10^{-3})(10\,000) = 5.69 + 3.31 = 9.0$ V. $\checkmark$

What the Resistor "Feels" During Charging

The current through the resistor is largest at $t = 0$ and decays exponentially to zero. The instantaneous power dissipated in $R$ is therefore

$P_R(t) = i^2 R = \left(\frac{V_0}{R}\right)^2 e^{-2t/RC}\,R = \frac{V_0^2}{R}\,e^{-2t/RC}.$

At $t = 0$ the power is $V_0^2/R$ — the maximum the circuit ever delivers to the resistor. As time passes, the current falls and the power drops sharply, with effective time constant $RC/2$ (twice as fast as the current itself, because power goes like $i^2$ ).

The total energy dissipated in $R$ over the full charge-up is found by integrating from $t = 0$ to $\infty$ :

$E_R = \int_0^\infty \frac{V_0^2}{R}\,e^{-2t/RC}\,\mathrm{d}t = \frac{V_0^2}{R} \cdot \frac{RC}{2} = \tfrac{1}{2}CV_0^2.$

This is exactly equal to the final energy stored in the capacitor — the same $\tfrac{1}{2}CV_0^2$ that we calculated at the start of Module 6.1. The remarkable result is:

Charging always dissipates as much energy in the resistor as it stores in the capacitor — exactly half each, regardless of the value of $R$ .

Make $R$ small and the current is large but brief; make $R$ large and the current is small but long-lasting. Either way the integral gives $\tfrac{1}{2}CV_0^2$ . The 50-50 split is a topological feature of the RC circuit, not a property of any particular component value. This is also why "fast charging" of any RC-modelled load (batteries, supercapacitors) is intrinsically inefficient: half the battery's work is always wasted as heat in the series resistance, no matter how quickly or slowly you charge.

Comparing Charge-Up and Discharge Curves

The rising charge-up curve $1 - e^{-t/\tau}$ and the falling discharge curve $e^{-t/\tau}$ are mirror images of each other about the horizontal line $0.5\,V_0$ . Their numerical values at the standard checkpoints:

$t/\tau$	Charge-up $1 - e^{-t/\tau}$	Discharge $e^{-t/\tau}$
$0$	$0.000$	$1.000$
$1$	$0.632$	$0.368$
$2$	$0.865$	$0.135$
$3$	$0.950$	$0.050$
$5$	$0.993$	$0.007$

Add the two columns row by row and you always get exactly $1.000$ : the charge-up and discharge fractions are complementary, summing to unity at every instant. This is the " $5\tau$ rule" working both ways: after five time constants, $99.3\%$ of the way up for charging is the same statement as $0.7\%$ left for discharging.

Mathematically the symmetry is trivial — $1 - e^{-t/\tau}$ and $e^{-t/\tau}$ are constructed to add to $1$ . Physically it is significant: the fraction of the final state achieved during charging equals the fraction lost during discharge in the same time. The two processes share not only the same time constant $\tau$ but the same numerical "shape" inverted.

Linearisation for Graphical Determination of τ — The PAG 8 Method

Plotting $V_C$ against $t$ for a charging trace gives a curve, not a straight line, so the time constant cannot be read off a gradient directly. The fix is to linearise $V_0 - V_C$ (the resistor voltage) instead:

$V_0 - V_C = V_0\,e^{-t/RC} \quad \Longrightarrow \quad \ln(V_0 - V_C) = \ln V_0 - \frac{t}{RC}.$

A plot of $\ln(V_0 - V_C)$ against $t$ is therefore a straight line with:

gradient $-1/RC = -1/\tau$ (negative, slope downwards),
$y$ -intercept $\ln V_0$ (the natural log of the supply voltage).

This is the standard analytical method for the OCR PAG 8 capacitor-charging practical. The experimental procedure is:

Charge the capacitor through the chosen $R$ from a known $V_0$ , with the switch closed at $t = 0$ .
Log $V_C(t)$ every second (or every fraction of a second, depending on $\tau$ ) for at least $5\tau$ .
Compute $V_0 - V_C$ at each time point.
Plot $\ln(V_0 - V_C)$ against $t$ .
Fit a straight line. Take $\tau = -1/\text{gradient}$ and $V_0^\text{exp} = e^\text{intercept}$ .

Charging Capacitors

Charging Capacitors

The Charging Circuit

Deriving the Charging Equations

Charging Curves

Worked Example 1 — Charging Through a Resistor

Graphical Analysis

Energy During Charging — A Reminder

Worked Example 2 — Tabulating the Rise

What the Resistor "Feels" During Charging

Comparing Charge-Up and Discharge Curves

Linearisation for Graphical Determination of τ — The PAG 8 Method

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