Capacitors, charging and discharging: worked solutions

Answer: C = 47 µF; stored energy = 1.5 × 10⁻³ J. Electrons move through the external circuit, leaving opposite charges on the plates.

1. Capacitance is charge stored on one plate per unit potential difference: C = Q/V. For this ideal capacitor it is the same at every non-zero row.
2. At 8.0 V, C = (376 × 10⁻⁶)/8.0 = 47 × 10⁻⁶ F = 47 µF. On this V–Q plot the slope is V/Q = 1/C, not C.
3. The area is W = ½ × 376 × 10⁻⁶ × 8.0 = 1.504 × 10⁻³ J ≈ 1.5 mJ. As more charge is separated, the p.d. rises; that is why QV would overestimate the energy by a factor of two.
4. The battery removes electrons from the plate connected to its positive terminal and supplies electrons to the other plate through the wires. Equal and opposite plate charges develop. Electrons do not cross the ideal insulating gap.

Check: A few tens of microfarads storing a few volts should store millijoules, not thousands of joules. V × C gives the stated 376 µC.

Mark checklist · 5 marks

• 1 mark — Defines capacitance as charge on one plate per unit p.d., or gives C = Q/V with quantities identified.
• 1 mark — Calculates 47 µF (4.7 × 10⁻⁵ F), including the unit.
• 1 mark — Uses triangular area ½QV with charge in coulombs. The equivalent ½CV² form earns this method mark when linked to the triangular area using Q = CV.
• 1 mark — Obtains 1.5 × 10⁻³ J, or 1.504 × 10⁻³ J before rounding. Correct-method error carried forward from a calculated C is allowed.
• 1 mark — Explains removal of electrons from one plate and supply to the other through the external circuit, giving opposite charges.

Common mistake: Taking the V–Q gradient as capacitance reverses the ratio. Use the axis labels: this slope has units V/C, so its reciprocal has units farads.

Answer: Parallel: 550 µF, 4.95 mC and 22 mJ. Series: 5.4 V across 220 µF and 3.6 V across 330 µF.

1. Parallel capacitors have the same p.d. and their plate charges add: C_total = 220 + 330 = 550 µF = 550 × 10⁻⁶ F.
2. The total charge magnitude on the plates connected to one supply terminal is Q_total = 550 × 10⁻⁶ × 9.0 = 4.95 × 10⁻³ C. This does not mean a net charge of 4.95 mC has been added to the whole capacitor pair: the opposite plates carry −4.95 mC.
3. The stored energy is W = ½ × 550 × 10⁻⁶ × 9.0² = 0.022275 J ≈ 22 mJ. This is the energy available in the capacitors; the flash circuit's useful output may be smaller.
4. For series, C_total = (220 × 330)/(220 + 330) = 132 µF. Hence the common charge magnitude is Q = 132 × 10⁻⁶ × 9.0 = 1.188 mC.
5. Divide the common charge by each capacitance: V₂₂₀ = 1.188 × 10⁻³/(220 × 10⁻⁶) = 5.4 V;V₃₃₀ = 3.6 V. The smaller capacitor has the larger p.d.

Check: Series capacitance 132 µF is smaller than either component. The series voltages add to the supply p.d., and their inverse-capacitance ratio is 330:220.

Mark checklist · 6 marks

• 1 mark — Adds parallel capacitances to obtain 550 µF.
• 1 mark — Obtains 4.95 mC, or 5.0 mC to two significant figures, using Q = CV.
• 1 mark — Uses W = ½CV² or sums ½CiV² for the two capacitors.
• 1 mark — Obtains 0.022 J (22 mJ); 0.022275 J is accepted before rounding. Correct-method carry-forward from C is allowed.
• 1 mark — Uses equal series charge to obtain 5.4 V across 220 µF; a correct inverse-capacitance voltage ratio is an accepted method.
• 1 mark — Obtains 3.6 V across 330 µF; the two p.d.s sum to 9.0 V. Carry-forward by subtracting the candidate's first p.d. from 9.0 V is allowed if the series model is correct.

Common mistake: Using resistor combination rules gives the wrong capacitor result. Start from what the circuit shares: parallel gives equal V, while initially uncharged ideal series capacitors acquire equal |Q|.

Exam technique: OCR's June 2024 H556/02 report says that most candidates answered Q4 correctly; the most common distractor came from combining capacitors as if they were resistors. State the shared quantity before choosing the combination rule.

Answer: Before switching: 5.93 V and +60.7 µA. Immediately after: −126 µA in the discharge sense. After 5.00 s: 3.66 V. Charge and capacitor p.d. are continuous in this finite-resistance model.

1. Convert the prefixes and obtain τ_charge = 100 × 10³ × 220 × 10⁻⁶ = 22.0 s.
2. At 15.0 s, V_i = 12.0(1 − e^−15.0/22.0) = 5.9316… V. Keep extra digits for the following calculations.
3. The resistor has the remaining supply p.d.: I_before = (12.0 − 5.9316…)/(100 × 10³) = +60.7 µA. Dividing Vc by the charging resistor would use the wrong component p.d.
4. Charge cannot change instantaneously through a finite resistor, so Q and Vc = Q/C are continuous. The new discharge time constant is τ_discharge = 47.0 × 10³ × 220 × 10⁻⁶ = 10.34 s.
5. The discharge current is opposite to the chosen charging sense: I_after = −V_i/R_d = −5.9316…/(47.0 × 10³) = −126 µA. A signed circuit sketch or an explicit discharge direction is equally useful.
6. After 5.00 s of discharge, V_C = 5.9316… e^{−5.00/10.34} = 3.6574… V ≈ 3.66 V. The elapsed discharge time is 5.00 s, not 20.0 s.

Check: The smaller discharge resistor allows a larger initial current magnitude, even though the capacitor was only partly charged. The p.d. then falls below 5.93 V.

Mark checklist · 6 marks

• 1 mark — Finds charging RC = 22.0 s with prefixes converted correctly.
• 1 mark — Uses the charging exponential to obtain Vc = 5.93 V.
• 1 mark — Uses resistor p.d. 12.0 − Vc to obtain charging current +60.7 µA. Correct-method carry-forward from Vc is allowed.
• 1 mark — States charge or capacitor p.d. is continuous at switching and uses the new discharge RC = 10.34 s.
• 1 mark — Obtains magnitude 126 µA with an explicit discharge direction or negative sign under the stated convention. Carry-forward from Vi is allowed.
• 1 mark — Uses Vi exp(−5.00/10.34) to obtain 3.66 V. Correct-method carry-forward from Vi or discharge RC is allowed.

Common mistake: A switch does not automatically reset capacitor voltage to zero or 12 V. Carry the actual stored charge into the next circuit stage, then reset the elapsed-time origin.

Exam technique: The capacitor-versus-resistor p.d. distinction is also discussed in AQA's June 2024 Paper 2 report, Q02.3. This is transferable circuit reasoning; it does not imply an official OCR allocation for this original question.

Answer: Initial gradient = −0.75 V s⁻¹; C = 100 µF; time constant = 12.0 s, from logger time 30.0 − 18.0 s.

1. At the discharge start the resistor p.d. is 9.0 V. Its current magnitude is |I_i| = 9.0/(120 × 10³) = 75 µA. A later tangent would belong to a smaller current and voltage.
2. The initial tangent gives dV_C/dt ≈ (0 − 9.0)/(30.0 − 18.0) = −0.75 V s⁻¹. Use a large triangle along the tangent rather than a chord between two points on the curve.
3. For constant C, differentiating Q = CV gives dQ/dt = C dV/dt. Therefore C = |I_i|/|dV/dt| = (75 × 10⁻⁶)/0.75 = 100 µF. The initial gradient is required because the current value used is the initial current.
4. At one time constant, Vc = Vi/e = 9.0/e = 3.31 V. The data place this at logger time 30.0 s. Hence τ = 30.0 − 18.0 = 12.0 s. As a separate check, RC = 120 × 10³ × 100 × 10⁻⁶ = 12.0 s.

Check: Multiplying R by the graph-derived C reproduces 12 seconds. An answer of 30 seconds includes the 18 seconds before discharge began.

Mark checklist · 6 marks

• 1 mark — Selects the initial tangent and explains it must match the initial current 9.0/R.
• 1 mark — Calculates the tangent gradient −0.75 V s⁻¹, or magnitude 0.75 V s⁻¹ with discharge identified.
• 1 mark — Links I to charge rate and Q = CV to justify |I| = C|dV/dt|, or gives the equivalent discharge derivative.
• 1 mark — Obtains C = 100 µF (1.0 × 10⁻⁴ F). Correct-method carry-forward from a measured initial gradient is allowed.
• 1 mark — Uses 9.0/e = 3.31 V and reads the corresponding logger time 30.0 s from the data; an equivalent ratio/log method using data is accepted.
• 1 mark — Subtracts the start time to obtain elapsed time constant 12.0 s. A calculation using RC alone does not earn the data-reading mark.

Common mistake: The tangent intercept is not the time when the capacitor is empty. The exponential curve remains above zero; the intercept can help identify τ only because it is the initial tangent.

Exam technique: AQA's June 2024 Paper 2 report, Q02.1–02.2, stresses the initial gradient and elapsed time from the start of discharge. Here those ideas are applied to a new graph and different values.

Answer: Charge safely, isolate the supply, log Vc during discharge through measured R, and plot ln(Vc/1 V) against elapsed t. With gradient m, C = −1/(Rm), after accounting for any significant logger loading.

1. Measure R with the supply disconnected. Connect the voltage logger across the capacitor. Use a switch arrangement that charges the capacitor from a low-voltage supply and then disconnects the supply while completing a discharge loop through R. The resistor is not bypassed during discharge.
2. Respect capacitor voltage and polarity ratings; limit charging current, and discharge through a resistor before handling or reconnecting the component. Select R so RC is long enough to resolve many readings. A short direct discharge could damage the capacitor or switch.
3. Charge to a reproducible starting p.d. Vi. Begin logging and take the switching event as elapsed t = 0. Sample often enough to describe the early decay; record readings across several time constants while they remain above the logger's resolution/noise floor. Repeat the discharge.
4. Plot y = ln(Vc/1 V) against elapsed t in seconds. From Vc = Vi e^−t/RC, y = ln(Vi/1 V) − t/(RC). The best-fit straight-line gradient m has units s⁻¹ and is negative.
5. Use C = −1/(Rm). Compare repeat gradients and quote sensible precision. A worst-fit gradient range can estimate uncertainty in C alongside uncertainty in R.
6. The logger's finite input resistance is in parallel with the discharge resistor. If it is not much larger than R, use R_eff = (1/R + 1/R_in)⁻¹, or choose a higher-input-resistance logger. Ignoring that extra path makes the measured decay faster and underestimates C when R is used.

Check: A negative gradient must produce a positive capacitance. For example, R = 100 kΩ and m = −0.050 s⁻¹ would give C = 200 µF.

Mark checklist · 6 marks

• 1 mark — Describes a valid charge/discharge switch circuit with logger across C and supply disconnected during discharge through R.
• 1 mark — Measures R and specifies safe capacitor polarity/voltage and a current-limited charge/discharge procedure.
• 1 mark — Records multiple Vc–time readings with switching as the time origin and an interval appropriate to RC; repeated runs improve reliability.
• 1 mark — Plots ln(Vc/1 V) against time and fits a straight line, linking its gradient to −1/(RC). Standard ln V shorthand is accepted when the voltage unit is fixed.
• 1 mark — Uses C = −1/(Rm), or an equivalent clearly explained fit to determine C.
• 1 mark — Gives a justified contextual improvement/error, such as logger loading with its effect, avoiding low-voltage noise, or gradient/repeat uncertainty. A generic 'repeat to be accurate' is insufficient.

Common mistake: Plotting V against t and expecting a straight line loses the exponential relationship. Transform the voltage, or fit an exponential directly and explain what its fitted time constant means.

Answer: Time = 78 s (78.29 s before rounding). Doubling R doubles this time to 157 s to three significant figures (156.58 s before final rounding; 1.6 × 10² s to two significant figures) and halves the initial current magnitude from 88 µA to 44 µA.

1. The time constant is τ = 68 × 10³ × 1000 × 10⁻⁶ = 68 s. The initial energy is ½ × 0.001 × 6.0² = 0.018 J, although this value is not required for the ratio method.
2. Because C stays fixed, W/Wi = (V/Vi)² = e^−2t/RC. At 10% energy, V/Vi = √0.10 ≈ 0.316, not 0.10.
3. Take logs: ln(0.10) = −2t/68;t = −(68/2) ln(0.10) = 78.2879… s ≈ 78 s.
4. Doubling R doubles RC. The same fractional energy change therefore takes twice as long: 156.576… s. With the same starting voltage, |Ii| = Vi/R. It falls from 6.0/(68 × 10³) ≈ 88 µA to about 44 µA. The initial stored energy is unchanged because C and Vi are unchanged.

Check: At 78.29 seconds, voltage is about 1.90 V. Squaring 1.90/6.0 gives about 0.10. Treating 10% energy as 10% voltage would give a time twice as large.

Mark checklist · 6 marks

• 1 mark — Calculates RC = 68 s with the capacitance prefix converted correctly.
• 1 mark — Uses W/Wi = (V/Vi)² or the equivalent energy exponential, so the target voltage fraction is √0.10.
• 1 mark — Correctly rearranges the exponential using a logarithm: t = −RC ln(0.10)/2, or an equivalent voltage-ratio route.
• 1 mark — Obtains 78 s; 78.29 s is accepted before rounding. Correct-method carry-forward from RC is allowed.
• 1 mark — States time doubles and justifies it by RC doubling, with C unchanged.
• 1 mark — States initial current magnitude halves and justifies it by Vi/R with Vi unchanged.

Common mistake: Voltage, charge and current have the same exponential time constant here, but energy depends on their square. Always connect the requested quantity to the measured variable before using a percentage.