OCR A A-level Physics · H556 · 6.1

Capacitors: charging, discharging and time constants

Understand where charge moves, choose the right exponential equation and turn capacitor graphs into calculations.

Complete 6.1 notes7 worked examplesInteractive capacitor lab38 marks of practice

One RC means 63.2% charged from empty or 36.8% remaining during discharge. These are fractions of different reference values.

Start with a two-minute check

A capacitor starts to discharge through a resistor. After one time constant, what fraction of its initial charge remains?

Your prediction

Use Q/Q_i = e^−t/RC to check your prediction.

By the end, you should be able to

Explain charging and discharging in terms of electron flow, and calculate charge, capacitance and energy.
Reduce series and parallel networks and analyse a circuit in each switch position.
Use RC, exponential equations, graph gradients and areas to solve problems.
Plan a capacitor investigation and build a finite-step spreadsheet discharge model.

Need the circuit foundations first? Revisit charge and current and electrical circuits.

What is capacitance?

A capacitor consists of two conductors separated by an insulator. Charging separates charge: one plate has an excess of electrons and the other has an equal electron deficit. In the ideal model, no electrons pass through the insulating gap.

C = Q/V and Q = CV

Capacitance C is the magnitude of charge Q on one plate per unit potential difference V between the plates. It is measured in farads: 1 F = 1 C V⁻¹. The two plates have +Q and −Q, so an initially neutral capacitor still has zero net charge overall.

The supply transfers electrons through the external circuit onto the negative plate and away from the positive plate. During discharge, electrons flow through the external resistor from the negative plate towards the positive plate. Conventional current is in the opposite direction to electron flow.

Read the symbol in context. C can mean the quantity capacitance or the unit coulomb. Here Q is charge in coulombs and C is capacitance in farads. Convert μF × 10⁻⁶, nF × 10⁻⁹, mC × 10⁻³ and kΩ × 10³ before substituting SI values.

Worked example 1 · charge on one plate

A 240 μF capacitor is charged to 9.0 V. What charge magnitude is stored?

C = 240 × 10⁻⁶ F, so Q = CV = 240 × 10⁻⁶ × 9.0 = 2.16 × 10⁻³ C = 2.16 mC (2.2 mC to two significant figures). Each plate has this magnitude, with opposite signs. Do not add them and report 4.32 mC.

Capacitors in series and parallel

Parallel: same p.d., charges add

Each capacitor connects across the same two nodes, so each has the same p.d. Total charge supplied is Q_total = Q₁ + Q₂.

C_eq = C₁ + C₂ + …

The equivalent capacitance is larger than any individual capacitance.

Series: same charge magnitude, p.d.s add

For an initially uncharged series chain, the intermediate conductor remains neutral overall. Equal and opposite induced charges give each capacitor the same charge magnitude. The p.d.s sum to the supply p.d.

1/C_eq = 1/C₁ + 1/C₂ + …

For two capacitors, C_eq = C₁C₂/(C₁ + C₂). It is smaller than either individual capacitance. The smaller capacitor has the larger p.d. because V = Q/C.

Identify branches by their endpoints. Components drawn next to one another are not necessarily in series.

Worked example 2 · a mixed network

Reduce the upper series branch: C_s = (10 × 15)/(10 + 15) = 6.0 μF.
Add the parallel branch: C_eq = 6.0 + 4.0 = 10 μF.
The series branch is across 12 V, so each series capacitor has charge magnitude Q_s = 6.0 × 12 = 72 μC.
Its p.d.s are 72/10 = 7.2 V and 72/15 = 4.8 V; they sum to 12 V.
The lower capacitor has Q = 4.0 × 12 = 48 μC. Total charge supplied is 72 + 48 = 120 μC, consistent with C_eqV.

Check the direction of the result. Capacitor combination rules are the reverse of resistor combination rules. Use the size of C_eq as a check before calculating individual p.d.s.

Energy stored in a capacitor

As charge accumulates, the capacitor p.d. increases. Moving each extra amount of charge therefore requires more work. Stored energy is the area under a graph with V on the vertical axis and Q on the horizontal axis: W = ∫V dQ.

W = ½QV = ½CV² = Q²/(2C)

For a constant capacitance, V = Q/C is a straight line. The average p.d. during charging from empty is V/2.

Read the axes before the gradient

Q vertically against V: gradient = C.
V vertically against Q: gradient = 1/C.
For these straight-line capacitor graphs, the triangular area is numerically ½QV in either orientation. The work integral is specifically ∫V dQ; do not assume every area or gradient in a physics graph gives energy.

At fixed C, doubling V doubles Q and quadruples W. At fixed Q, increasing C reduces W = Q²/(2C).

Why store energy in a capacitor?

A capacitor can release energy quickly in a camera flash or a pulsed electrical device. In a smoothing circuit, it charges when the supply p.d. is high and supplies energy to the load between peaks. Stored energy falls during discharge as energy is transferred to the resistor or load.

Worked example 3 · energy loss and average power

A 680 μF capacitor falls from 12 V to 7.0 V in 4.0 s.

W_i = ½ × 680 × 10⁻⁶ × 12² = 0.04896 J.
W_f = ½ × 680 × 10⁻⁶ × 7.0² = 0.01666 J.
ΔW = W_i − W_f = 0.03230 J.
P_average = ΔW/Δt = 0.03230/4.0 = 8.075 mW ≈ 8.1 mW.

Using ½C(12 − 7)² would give the wrong energy loss. Instantaneous power I²R changes during discharge, so it is different from this interval average.

Draw a complete charging and discharging circuit

A series switch that simply opens the charging circuit leaves the capacitor isolated. It does not create a discharge path. Use a changeover switch to connect the resistor either to the supply or to the return wire.

Switch at A: supply → R → C → supply. Switch at B: C → R → return → C, with the supply positive terminal disconnected. The voltmeter measures capacitor p.d.

Charging from empty

Initially V_C = 0, so all supply p.d. is across R and I is largest. As charge builds, V_C rises, leaving less p.d. across R. Current decreases.

V_s = V_R + V_C
I = (V_s − V_C)/R

At the ideal long-time limit, V_C = V_s and I = 0.

Discharging

The charged capacitor provides the p.d. that drives current through R. As charge falls, both p.d. and current magnitude fall.

|I| = V_C/R

If charging current is defined as positive, discharge current is negative. Current can reverse immediately when the switch moves; capacitor p.d. is continuous in this finite-resistance model.

Reset the initial condition after a switch. If a capacitor begins discharge at 5.2 V, that is its initial p.d. for the new process, even if the charging supply was 9.0 V. Measure time from the switch event.

Choose the correct capacitor equation

Assume a constant capacitance, a fixed total series resistance, an ideal constant-voltage supply for charging and negligible leakage or meter loading. V_s is the charging supply p.d.; V_i is the p.d. at the start of discharge. The symbols have different meanings.

Charging from empty compared with discharge from an initial charge
Quantity	Charging from empty	Discharging
Capacitor p.d.	V_C = V_s(1 − e^−t/RC)	V_C = V_ie^−t/RC
Charge magnitude	Q = CV_s(1 − e^−t/RC)	Q = Q_ie^−t/RC, where Q_i = CV_i
Current	I = (V_s/R)e^−t/RC	\|I\| = (V_i/R)e^−t/RC
Resistor p.d. magnitude	V_R = V_se^−t/RC	\|V_R\| = V_ie^−t/RC

In charging, capacitor Q and V increase, while resistor p.d. and current decrease. Identify the measured component before choosing the equation.

What if charging starts with a non-zero capacitor p.d.?

For initial capacitor p.d. V_i, the more general charging equation is V_C = V_s + (V_i − V_s)e^−t/RC. The familiar charging-from-empty equation follows when V_i = 0.

Worked example 4 · a discharge calculation

A 680 μF capacitor starts at 9.0 V and discharges through 15 kΩ. Find V, Q and |I| after 8.0 s.

τ = 15 × 10³ × 680 × 10⁻⁶ = 10.2 s.
V = 9.0e^−8.0/10.2 = 4.108 V.
Q = CV = 680 × 10⁻⁶ × V = 2.793 mC (using unrounded V).
|I| = V/R = 0.274 mA.

The ratio 8.0/10.2 is dimensionless. Keep unrounded values until the final step, then report about 4.1 V, 2.8 mC and 0.27 mA.

Worked example 5 · time to charge to a target

An empty 250 μF capacitor charges through 30 kΩ from a 10 V supply. When does its p.d. reach 8.0 V?

RC = 30000 × 250 × 10⁻⁶ = 7.5 s.
8.0 = 10(1 − e^−t/7.5).
e^−t/7.5 = 1 − 8.0/10 = 0.20.
−t/7.5 = ln(0.20).
t = −7.5 ln(0.20) = 12.07 s ≈ 12 s.

Check: at t = 0 the selected equation must give V_C = 0. The discharge equation would give 10 V and cannot describe this start.

What does the time constant RC mean?

τ = RC T_½ = RC ln 2 ≈ 0.693RC

The time constant is the time for charge or p.d. during discharge to fall to 1/e of its initial value. This means 36.8% remains and 63.2% has been transferred. It is not the time to halve or the time to finish.

Time	Discharge fraction remaining	Charge fraction reached from empty
RC	36.8%	63.2%
2RC	13.5%	86.5%
3RC	4.98%	95.0%
5RC	0.674%	99.3%

For a fixed interval Δt, Q(t + Δt)/Q(t) = e^−Δt/RC. Equal intervals remove the same fraction of the remaining charge, rather than the same amount of charge.

Doubling R doubles τ and halves the initial current, with C and supply p.d. fixed.
Doubling C doubles τ and the final charge, with R and supply p.d. fixed; the initial current is unchanged.
Changing supply p.d. changes Q, V, I and stored energy, but does not change RC in the ideal linear model.

Capacitor graphs: gradients, areas and logarithms

Gradient of Q against t

For Q defined as the positive charge magnitude on the positive plate, dQ/dt is positive while charging and negative while discharging. Its magnitude is current magnitude. A discharge curve is steepest initially, when current is largest.

The initial tangent to an ideal exponential discharge curve meets the time axis at t = RC. Read the intercept from the tangent, not where the curved graph appears to reach zero.

Area under I against t

Current is charge-transfer rate, so ∫I dt is signed charge transferred. An area under |I| against t gives its magnitude. Over a complete discharge this magnitude is Q_i.

A straight line from discharge

ln(V/V_ref) = ln(V_i/V_ref) − t/(RC)

Choose one fixed reference, such as V_ref = 1 V. The vertical axis is dimensionless. The gradient of ln(V/1 V) against t is m = −1/(RC), with units s⁻¹ when time is in seconds.

RC = −1/m and C = −1/(mR)

Use a large triangle on a best-fit line. If time is plotted in milliseconds, convert the gradient to s⁻¹. Do not use ln(0), and avoid readings close to the sensor's noise floor.

Worked example 6 · extracting capacitance from data

Discharge measurements through a 20 kΩ resistor give this idealised data:

t / s	V / V	ln(V/1 V)
0	8.0000	2.079
2	4.8522	1.579
4	2.9430	1.079
6	1.7850	0.579
8	1.0827	0.079

The straight-line gradient is (0.079 − 2.079)/(8 − 0) = −0.250 s⁻¹. Hence RC = 4.00 s and C = 4.00/(20 × 10³) = 2.00 × 10⁻⁴ F = 200 μF.

For experimental data, calculate the gradient from the fitted line; two raw readings may not represent the whole trend. Use the shallowest and steepest acceptable lines through uncertainty ranges if estimating a gradient uncertainty.

Explore the charging and discharging capacitor lab

Use the component controls, play/pause button and individual Q–t, V_C–t and I–t graph views. The lab displays RC and elapsed time in multiples of RC. Each graph starts again from the latest switch event.

1 · Predict the checkpoint

Use Reset empty, then play with the battery connected. Pause near t/RC = 1. Expect about 63.2% of the final capacitor p.d. and charge, but about 36.8% of the initial current. Open a graph to relate the number to its shape.

2 · Change R, then C

Record RC and the initial current. Double R with C and supply p.d. fixed, then reset and compare. Restore R and double C. Predict the changes using the three rules above before reading the display. Component changes reset the lab.

3 · Switch before full charge

Pause during charging and record the actual capacitor p.d. Select Use resistor. Check that p.d. is initially unchanged while current reverses. Use the recorded p.d. as V_i, then resume to observe the decay.

4 · Compare charge and energy

During discharge, pause when p.d. is approximately half its starting value. Q is also half, but W should be about one quarter. Explain this using W = ½CV². The playback-speed control changes viewing speed, not RC.

Open the capacitor simulator in a full page

Investigating capacitance and RC in the laboratory

Charging and discharge measurements

Build the changeover-switch circuit above. Measure capacitor p.d. with a high-input-resistance voltmeter or voltage sensor.
Charge to a measured initial p.d., then switch to discharge through a known resistor.
Record V at known times. For slow decay, meters and a timer may be suitable; for rapid decay use a triggered data logger or oscilloscope.
Use a sampling interval much shorter than RC, so several readings resolve the early part of the decay.
Plot V against t, then ln(V/1 V) against t. Find RC and calculate C using the measured R.
Repeat from the same initial p.d. Compare with component values and their tolerances.

Testing series and parallel combinations

Use the same method on each capacitor and then on its combination with a fixed known R. For each arrangement determine C from the logarithmic gradient and compare with the predicted equivalent capacitance.

During charging, an ammeter is placed in series with the charging path; a voltmeter is placed across the capacitor or network. Since current changes, use the area under an I–t record to obtain charge, rather than multiplying an instantaneous current by total time. Then C = Q/V.

Limitation	Effect	Useful response
Human reaction time	Large timing uncertainty for a fast decay	Use automated, triggered voltage recording. If the question fixes R and C, keep those components.
Finite meter input resistance R_meter	The meter adds a parallel discharge path; effective R = R ∥ R_meter can be smaller	Choose a sensor input resistance much larger than the intended discharge resistance.
Component tolerance	Nominal R and C may differ from actual values	Measure R; compare the fitted C with the manufacturer's tolerance rather than expecting exact agreement.
Low p.d. readings	A fixed voltage uncertainty becomes a large fractional uncertainty; logs amplify this problem	Use suitable range/resolution and exclude readings near zero or the noise floor with a stated reason.
Inconsistent start	Different initial p.d. or timing delay changes the recorded curve	Charge to a measured starting value and trigger acquisition with the switch event.

Evaluation language: repeats reveal spread and can reduce random uncertainty in a mean. They do not remove a systematic error such as extra discharge through the meter. Explain which measured quantity each improvement affects.

Use an appropriate low-voltage school circuit, observe capacitor voltage and polarity ratings, and discharge through a resistor before changing connections.

Spreadsheet modelling: calculate the next row

OCR 6.1.3(d) includes modelling with the rate equation. Approximate the changing current as constant over a short interval Δt, then update the charge.

ΔQ/Δt ≈ −Q/(RC)
ΔQ ≈ −QΔt/(RC)
Q_next = Q + ΔQ = Q(1 − Δt/RC)

ΔQ is a signed negative change. If a table instead records the positive amount of charge lost, subtract that amount. Recalculate it from the new charge each time.

Worked example 7 · build and check the model

Take C = 80 μF, R = 50 kΩ and initial p.d. 12 V. Then RC = 4.0 s and Q_i = 960 μC. With Δt = 1.0 s, each step leaves 1 − 1/4 = 0.75 of the previous charge. The first rows are 960, 720, 540, 405 and 303.75 μC.

The exact value at 4.0 s is 960e^−4/4 = 353.16 μC. The one-second step model gives 303.75 μC, about 14.0% too low. This is an approximation error, not random experimental error.

Try a smaller time step

Spreadsheet interval Δt

At 4.00 s: step model 303.75 μC; exact exponential 353.16 μC; the model is 14.0% low.

Every calculated row for Δt = 1.00 s
t / s	Q / μC	Signed ΔQ for next step / μC
0.00	960.00	−240.00
1.00	720.00	−180.00
2.00	540.00	−135.00
3.00	405.00	−101.25
4.00	303.75	—

As Δt decreases, the step model approaches the exponential. For physical, steadily decreasing positive charges in this simple update, use Δt much smaller than RC; a large interval can even give negative charges.

Spreadsheet cell formulae to try

Use column A for time in seconds, B for charge in μC and C for signed change in μC. Set F1 = 4.0 (RC in s) and F2 = 1.0 (Δt in s). Start with A2 = 0 and B2 = 960.

C2: =-B2*$F$2/$F$1
A3: =A2+$F$2
B3: =B2+C2
C3: =-B3*$F$2/$F$1
Copy the row-3 formulae down. Optional exact comparison in D2: =960*EXP(-A2/$F$1), then copy down.

The fixed references keep RC and Δt constant while the charge reference advances row by row. Label charge as μC throughout this sheet; a plot using these values is not in coulombs unless you convert.

Common capacitor mistakes to avoid

Using the simpler discharge formula for charging. Test the initial value and identify whether capacitor p.d. rises or falls.
Assuming every voltage in a charging circuit rises. Capacitor p.d. rises; resistor p.d. falls because V_R = V_s − V_C.
Saying charge has “fallen by 37%” after RC. It has fallen to about 37%; about 63% has been transferred.
Using resistor rules for capacitor combinations. Parallel capacitances add; series reciprocal capacitances add.
Calling the gradient energy. Read both axes: Q–V gradient is C, V–Q gradient is 1/C. Stored energy comes from ½QV.
Squaring the voltage difference to find energy loss. Subtract the squared-voltage energies instead.
Disconnecting the supply without providing a closed discharge loop. Trace the current path in each switch position.
Subtracting the same charge in every spreadsheet row. An exponential loses the same fraction in equal intervals; update the charge used for the next change.
Giving “human error” as the whole practical evaluation. Identify reaction time, sampling interval, meter loading or component tolerance and explain its effect.

Original capacitor exam practice · 38 marks

Attempt the questions before opening the mark points. Show the equation, SI conversions and substitution, and keep intermediate values unrounded. Numerical answers below include extra digits for checking; round your final answers to a sensible precision.