OCR A Level Physics A: Charging, discharging and exponential decay in RC circuits (H556) – Revision Notes

These notes condense section 6.1.3, Charging and discharging capacitors, of the OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017. They cover statements 6.1.3(a) to 6.1.3(e). For full explanations and longer worked examples, read the study guide, then test yourself with the practice questions. Course pages: OCR A Level Physics hub and checklist. Not sure where to start? Take the free 10-minute diagnostics.

Key ideas in one place

• Charging: cell p.d. V₀ = V_R + V_C at every instant. Current starts at V₀/R and decays. V_C and Q rise from zero.
• Discharging: the capacitor is the source. Q, V_C and I all decay, and I flows the opposite way to the charging current.
• Time constant: τ = CR, measured in seconds.
• Exponential decay: the rate of change is proportional to the amount remaining, so equal time intervals give equal ratios.

Formulas

Situation	Equation	Notes
Time constant	τ = CR	C in farads, R in ohms
Discharge, any of Q, V, I	x = x₀ e^(−t/CR)	x₀ is the value at t = 0
Charge or p.d. while charging	x = x₀(1 − e^(−t/CR))	x₀ is the final value (Q₀ = CV₀)
Current while charging	I = I₀ e^(−t/CR)	I₀ = V₀/R, always decays
Time to reach x	t = CR ln(x₀/x)	for decay
Half-time	t½ = CR ln 2 = 0.693 CR	same anywhere on the curve
Straight-line form	ln x = ln x₀ − t/CR	gradient −1/CR
Rate model	ΔQ/Δt = −Q/CR	step: ΔQ = −(Q/CR) Δt

Fractions at multiples of CR

Time	Left (discharge)	Reached (charge)
1 CR	36.8%	63.2%
2 CR	13.5%	86.5%
3 CR	5.0%	95.0%
5 CR	0.67%	99.3%

6.1.3(a) Circuit behaviour and the practical

Graph shapes (describe them in words).

• Discharge, Q or V or I against t: a curve from a maximum at t = 0, falling ever more gently towards zero.
• Charge, Q or V_C against t: a curve from zero, rising ever more gently towards a final value.
• Charge, I or V_R against t: the same shape as a discharge curve.

Investigation (practical activity group 9).

1. Series circuit: capacitor, resistor, supply, two-way switch. Voltmeter across the capacitor.
2. Charge fully, switch over, start timing.
3. Record V (or I) at regular intervals with meters and a stopwatch, or let a data-logger sample automatically.
4. Plot ln V against t. Gradient = −1/CR.

	Meters	Data-logger
Timing	Reaction time in each reading	Logger clock
Sample rate	Slow	Fast, many points
Best for	Slow decays (large CR)	Any CR, including short ones

Reduce error: voltmeter resistance far larger than R, electrolytic capacitor the right way round, repeats averaged, full discharge between runs, large CR for hand timing.

6.1.3(b) and (c) Method boxes

Method: discharge value at a given time.

1. Convert units and find τ = CR.
2. Find x₀ (for V and Q: Q₀ = CV₀; for I: I₀ = V₀/R).
3. Substitute into x = x₀ e^(−t/τ).

Method: time to reach a value.

1. Write x/x₀ as a fraction.
2. t = τ ln(x₀/x).
3. Check the sign: t must be positive for a decay.

Method: charging values.

1. Find V_C = V₀(1 − e^(−t/τ)) or Q = CV₀(1 − e^(−t/τ)).
2. Find V_R = V₀ − V_C.
3. Find I = V_R/R. Check against I₀ e^(−t/τ).

Worked reminder. A 47 µF capacitor and 82 kΩ resistor are connected to a 5.0 V supply, with the capacitor uncharged. τ = 47 × 10⁻⁶ × 82 × 10³ = 3.854 s. After 5.0 s: V_C = 5.0(1 − e^(−1.297)) = 5.0 × 0.7268 = 3.63 V, so V_R = 1.37 V and I = 1.37/82 000 = 1.7 × 10⁻⁵ A.

Worked reminder (ln graph). A graph of ln V against t has gradient −0.056 s⁻¹ and passes through ln V = 1.79 at t = 0. CR = 1/0.056 = 17.9 s. V₀ = e^1.79 = 6.0 V.

6.1.3(c) ln graphs: what to remember

• The line is straight for any exponential decay, so a straight ln x graph confirms the model.
• Gradient = −1/CR, intercept on the ln x axis = ln x₀.
• Base 10 logs: gradient = −0.434/CR.
• For charging, plot ln of a decaying quantity (I, V_R, or V₀ − V_C).
• Use a large gradient triangle on the best-fit line.

6.1.3(d) Modelling steps

1. Pick Δt much smaller than CR.
2. ΔQ = −(Q/CR) Δt.
3. Q_new = Q_old + ΔQ.
4. Copy down a spreadsheet column, plot Q against t.

The model always lies slightly below the exact curve because it uses the rate at the start of each step. A smaller Δt reduces the gap. Each step multiplies Q by (1 − Δt/CR).

Worked reminder. CR = 10 s, Q₀ = 5.0 mC, Δt = 2.0 s. Factor 0.80 per step: 5.0, 4.0, 3.2, 2.56, 2.05 mC at 0, 2, 4, 6, 8 s. The exact value at 8 s is 5.0 e^(−0.8) = 2.25 mC.

Graphically, the tangent at t = 0 meets the t-axis at t = CR.

6.1.3(e) Exponential decay and constant ratio

Feature	Exponential decay	Linear decrease
Equal time intervals	Same ratio	Same amount subtracted
Reaches zero?	Never	Yes, at a finite time
Gradient	Proportional to the value	Constant
ln graph	Straight line	Curve

The time for the value to halve is the same wherever you start. After n half-times the fraction left is (1/2)ⁿ.

Must-know distinctions

• Q, V_C against I, V_R when charging: the first pair rises, the second pair falls.
• x₀ when discharging against x₀ when charging: start value against final value.
• CR against t½: t½ = 0.693 CR.
• Gradient of ln x against CR: gradient = −1/CR, not CR.
• ΔQ negative: charge on the capacitor falls, so the model has a minus sign.

Quick self-test

1. State the unit of CR in terms of a single SI unit.
2. Find τ for 2.2 µF and 4.7 MΩ.
3. What percentage of the initial charge is left after 2 CR when discharging?
4. A 100 µF capacitor charged to 10 V discharges through 22 kΩ. Find the p.d. after 3.0 s.
5. A capacitor charges from a 6.0 V cell with τ = 4.0 s. Find V_C after 6.0 s.
6. A ln I against t graph has gradient −0.080 s⁻¹. Find CR.
7. Find the half-time for CR = 20 s.
8. A 12 V supply is connected through 4.7 kΩ to an uncharged capacitor. Find the initial current.
9. A discharge current starts at 3.0 mA. Find the current after one time constant.
10. A p.d. falls from 12 V to 8.0 V in 5.0 s. Predict the p.d. after another 5.0 s.
11. In a spreadsheet model with CR = 10 s, Q₀ = 5.0 mC and Δt = 2.0 s, find Q after the first step.
12. Why does a ln V against t graph for a discharge give a straight line?

Answers

1. The second (s), since CR = Q/I.
2. τ = 2.2 × 10⁻⁶ × 4.7 × 10⁶ = 10.3 s.
3. e⁻² = 0.135, so 13.5%.
4. τ = 2.2 s. V = 10 e^(−3.0/2.2) = 2.56 V.
5. V_C = 6.0(1 − e^(−1.5)) = 4.66 V.
6. CR = 1/0.080 = 12.5 s.
7. t½ = 0.693 × 20 = 13.9 s.
8. I₀ = 12 / 4700 = 2.55 mA.
9. I = 3.0 e⁻¹ = 1.10 mA.
10. Ratio 2/3, so 8.0 × 2/3 = 5.33 V.
11. ΔQ = −(5.0/10) × 2.0 = −1.0 mC, so Q = 4.0 mC.
12. Taking logs of V = V₀ e^(−t/CR) gives ln V = ln V₀ − t/CR, which has the form y = mx + c.

Where marks are usually lost

• Leaving C in µF or R in kΩ when evaluating CR.
• Using the charging equation for the current.
• Reading the gradient of a ln graph as CR instead of −1/gradient.
• Writing the gradient as positive.
• Taking a gradient from two data points off the best-fit line.
• Saying the capacitor is “fully discharged” after one time constant.
• Forgetting that x₀ is the final value in the charging equation.
• Describing a decay graph as reaching zero, or a curve as “linear then flat”.
• In an explanation of a data-logger’s advantage, writing “more accurate” without saying it removes reaction-time error or gives more readings.
• In a model, forgetting that a larger step gives a larger error.

Official syllabus

This page is aligned to the OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), first assessment 2017, published by OCR, section 6.1.3.