OxfordAQA International AS and A-level Physics: Capacitor charge and discharge through resistors (9630) – Revision Notes

Condensed notes for section 3.9.1 of the OxfordAQA International AS and A-level Physics (9630) specification, version 4.4. The specification marks section 3.9 as International A-level only. For full explanations, use the study guide; for questions, use the practice set. Course pages: physics hub and checklist. To find weak spots, try the free 10-minute diagnostics.

Key ideas in one place

• Discharge: capacitor loses charge through R. I = V/R = Q/(RC), so the rate of loss is proportional to the charge left. Result: exponential decay.
• Charge: capacitor gains charge from a supply of e.m.f. V₀ through R. V₀ = V_R + V_C at all times. The current is largest at the start and falls.
• Time constant = RC, in seconds when R is in Ω and C in F.
• Neither process finishes in a finite time. Each is exponential: equal time steps give equal fractional changes.

Equations

Quantity	Discharge	Charge
Charge	Q = Q₀e^(−t/RC), Q₀ = CV₀	Q = Q₀(1 − e^(−t/RC)), Q₀ = CV₀ is the final charge
Capacitor p.d.	V = V₀e^(−t/RC)	V = V₀(1 − e^(−t/RC))
Current	I = I₀e^(−t/RC), I₀ = V₀/R	I = I₀e^(−t/RC), I₀ = V₀/R
Resistor p.d.	equals the capacitor p.d.	V_R = V₀e^(−t/RC)
Time to halve	T½ = ln 2 × RC = 0.693 RC	the quantity that is falling halves in T½

Useful fractions:

Time	Fraction left (discharge)	Fraction of final value reached (charge)
RC	0.368 (about 37%)	0.632 (about 63%)
2RC	0.135	0.865
3RC	0.050	0.950
5RC	0.0067	0.993

Graphs: shapes, gradients, areas

Graph	Shape	Gradient	Area
Q–t (discharge)	Falling curve, steepest at t = 0	Equals −I	—
V–t (discharge)	Same shape as Q–t	Equals −I/C	—
I–t (discharge)	Same shape, starts at V₀/R	—	Charge transferred; total = Q₀
Q–t, V–t (charge)	Rising from zero to CV₀, V₀; steepest at t = 0	Q–t gradient = I	—
I–t (charge)	Falling curve from V₀/R to zero	—	Charge stored so far; total = CV₀

Tangent at t = 0 to a discharge curve meets the time axis at t = RC. For a charging curve the tangent at t = 0 reaches the final value at t = RC.

Method boxes

Find a value at time t (discharge).

1. Convert to F, Ω, s.
2. RC = R × C.
3. Quantity = start value × e^(−t/RC).

Find the time for a given value.

1. Divide: target/start.
2. t = −RC ln(target/start) for discharge.
3. For charge, first find the fraction left to go: (final − target)/final, then t = −RC ln(that fraction) = RC ln(final/(final − target)).

Find RC from data (log-linear).

1. Add a column ln V (or ln Q, ln I). For charging use ln(V₀ − V).
2. Plot against t, draw the best line, take the gradient with a large triangle.
3. RC = −1/gradient. The intercept is ln V₀ (discharge).
4. For C, divide RC by the known R.

Half-time check. RC = T½ / 0.693. Read T½ at two places on the curve: if the values are the same, the decay is exponential.

Worked reminder. RC = 20 s and V₀ = 15 V. Discharging, at t = 20 s: V = 15e⁻¹ = 5.5 V. Charging from zero, at t = 20 s: V = 15(1 − e⁻¹) = 9.5 V. At t = 40 s: discharge 15e⁻² = 2.0 V; charge 15(1 − e⁻²) = 13.0 V.

Worked reminder (half-time). A discharge takes 26 s for V to halve. RC = 26/0.693 = 37.5 s. With C = 750 µF, R = 37.5 / (750 × 10⁻⁶) = 50 kΩ. After 78 s (three half-times) one eighth of V₀ remains.

Worked reminder (sloping tangent). On a discharge Q–t graph, Q₀ = 6.0 mC and the tangent at t = 0 meets the time axis at 12 s. RC = 12 s. The initial current is the magnitude of the gradient: 6.0 mC / 12 s = 0.50 mA.

Describing and sketching graphs

A good description or sketch includes these features:

• The axes labelled with quantity and unit, and the starting value marked.
• The exponential shape: steepest at t = 0, with the gradient decreasing.
• The curve approaches the final value (zero for discharge; CV₀ or V₀ for charging) but does not reach it.
• For several curves on one axis, the one with the larger RC is drawn flatter, and curves that start at the same value do not cross.
• The same time axis used for Q, V and I so that the half-times can be compared.

Units and powers of ten

Prefix	Meaning	Example in RC work
m (milli)	× 10⁻³	3.0 mF = 3.0 × 10⁻³ F
µ (micro)	× 10⁻⁶	470 µA = 4.7 × 10⁻⁴ A
k (kilo)	× 10³	5.0 kΩ = 5.0 × 10³ Ω
M (mega)	× 10⁶	2.0 MΩ = 2.0 × 10⁶ Ω

A quick size check: a microfarad capacitor with a megohm resistor gives a time constant of 1 s, since 10⁻⁶ × 10⁶ = 1. A millifarad capacitor with a kilohm resistor also gives 1 s. Use these to see if an answer is the right size before you move on.

Required practical 6

Investigate charge and discharge. Typical set-up: capacitor, resistor, switch, supply, voltmeter or data logger.

1. Charge the capacitor fully. Switch to discharge. Start timing at the same moment.
2. Record V at regular times across several time constants.
3. Plot ln V against t. Gradient = −1/RC.
4. Check by changing R or C: RC changes in proportion.

Evaluation points: finite voltmeter resistance in parallel with R shortens the measured time constant; capacitor tolerance means the marked C is not exact; polarity of an electrolytic capacitor; read timings from one start point; the uncertainty in RC has the same percentage uncertainty as the gradient.

Must-know distinctions

• Discharge form Q₀e^(−t/RC) versus charging form Q₀(1 − e^(−t/RC)). In the second, Q₀ is the final charge.
• Gradient of the Q–t graph (current) versus area under the I–t graph (charge).
• Time constant (to fall to 37%) versus half-time (to fall to 50%). T½ is shorter than RC.
• The quantity never reaches zero in a finite time: after 5RC about 0.7% remains.
• Charging current falls even though Q rises.

Quick self-test

1. A 3.0 mF capacitor discharges through a 5.0 kΩ resistor. Find the time constant.
2. For the same circuit, find T½.
3. A 100 µF capacitor charged to 12 V discharges through 200 kΩ. Find Q₀, I₀ and V after 30 s.
4. What fraction of the initial charge remains after 3RC?
5. What fraction of the final p.d. has been reached after 2RC while charging?
6. A ln V against t graph for a discharge through 25 kΩ has gradient −0.040 s⁻¹. Find RC and C.
7. The half-time of a discharge is 8.0 s. How long does it take for the charge to fall to 1/16 of its start?
8. What should be plotted against t to get a straight line for a charging capacitor?
9. A discharge has I₀ = 50 µA and RC = 30 s. What charge was stored initially?
10. R is doubled with C and V₀ unchanged. What happens to Q₀, I₀ and RC?
11. A discharging capacitor has RC = 15 s. How long does V take to fall from 9.0 V to 3.0 V?

Answers

1. RC = 3.0 × 10⁻³ × 5.0 × 10³ = 15 s.
2. T½ = 0.693 × 15 = 10.4 s.
3. RC = 20 s; Q₀ = 100 × 10⁻⁶ × 12 = 1.2 × 10⁻³ C; I₀ = 12/200 000 = 6.0 × 10⁻⁵ A; V = 12e^(−30/20) = 2.7 V.
4. e⁻³ = 0.050 (5.0%).
5. 1 − e⁻² = 0.865 (about 86%).
6. RC = 1/0.040 = 25 s; C = 25/25 000 = 1.0 × 10⁻³ F.
7. 1/16 = four halvings, so 4 × 8.0 = 32 s.
8. ln(V₀ − V) against t, with gradient −1/RC. (ln I against t also works.)
9. Q₀ = I₀ × RC = 50 × 10⁻⁶ × 30 = 1.5 × 10⁻³ C.
10. Q₀ = CV₀ is unchanged; I₀ = V₀/R halves; RC doubles.
11. t = 15 ln(9.0/3.0) = 15 ln 3 = 16.5 s.

Where marks are usually lost

• Putting µF or kΩ into RC without converting, so the answer is wrong by a power of ten.
• Using the discharge equation for a charging question, or the reverse.
• Entering e^(−t/RC) without brackets round the whole exponent, so only part of it is divided. Write the exponent in brackets and check the answer is smaller than the starting value.
• Giving the gradient of a ln graph as +1/RC.
• Plotting ln V against t for charging instead of ln(V₀ − V).
• Saying the capacitor is empty after one time constant.
• Describing curves without saying “exponential”, “never reaches zero” or “steepest at the start”, so a description loses detail.
• No units, or a wrong unit, on RC, charge or current.
• Forgetting that the area under an I–t graph is charge, and quoting energy.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams from May/June 2020 onwards, published by OxfordAQA. These notes follow section 3.9.1 and Required practical 6.