ExaminerPrep › Question bank › IAL Physics › 4.4 Capacitors
Edexcel International A Level Physics · Unit 4 · Further Mechanics, Fields and Particles
40 free, high-quality past-paper-style questions on Capacitors for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
Which of the following is equivalent to the farad?
| Answer | Mark | Notes |
|---|---|---|
| C = Q/V, so 1 F = 1 coulomb per volt = 1 C V−1. C s−1 is the ampere and J C−1 is the volt. | ✓ 1 | Answer B |
Answer: B
Exam tip: Derive units from the defining equation: capacitance is charge per unit p.d., so its unit is the coulomb per volt.
A 330 µF capacitor is charged until the potential difference across it is 12 V.
What is the charge stored?
| Answer | Mark | Notes |
|---|---|---|
| Q = CV = 330 × 10−6 × 12 = 3.96 × 10−3 C = 4.0 mC. | ✓ 1 | Answer C |
Answer: C
Common error: Using the energy equation by mistake. Charge is CV; ½CV has no physical meaning here.
A 2200 µF capacitor is charged to a potential difference of 6.0 V.
How much energy does it store?
| Answer | Mark | Notes |
|---|---|---|
| W = ½CV2 = ½ × 2200 × 10−6 × 6.02 = 0.0396 J = 0.040 J. | ✓ 1 | Answer C |
Answer: C
Exam tip: Choose the energy form that matches the data: with C and V given, use ½CV2 directly.
A 330 µF capacitor discharges through a 68 kΩ resistor.
What is the time constant of the circuit?
| Answer | Mark | Notes |
|---|---|---|
| RC = 68 × 103 × 330 × 10−6 = 22.4 s = 22 s. | ✓ 1 | Answer D |
Answer: D
Common error: Leaving kΩ or µF without the power of ten. Write 68 kΩ = 68 × 103 Ω and 330 µF = 330 × 10−6 F.
A 680 µF capacitor in the power supply of an audio amplifier is charged to a potential difference of 25 V.
Define capacitance.
Calculate the charge stored on the capacitor.
Calculate the energy stored in the capacitor.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the charge stored (on one plate) per unit potential difference across the capacitor | ✓ 1 | |
| Part (b) | ||
| Q = CV = 680 × 10−6 × 25 | ✓ 1 | |
| = 0.017 C (1.7 × 10−2 C) | ✓ 1 | |
| Part (c) | ||
| W = ½CV2 = ½ × 680 × 10−6 × 252 | ✓ 1 | |
| = 0.21 J (0.2125 J) (allow error carried forward from ½QV) | ✓ 1 | |
Exam tip: Convert µF to F before substituting: 680 µF = 680 × 10−6 F.
A 100 µF capacitor is charged to 12 V and then discharged through a 56 kΩ resistor.
Calculate the time constant of the circuit.
Calculate the current at the start of the discharge.
Calculate the potential difference across the capacitor 10 s after the discharge starts.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| RC = 56 × 103 × 100 × 10−6 = 5.6 s | ✓ 1 | |
| Part (b) | ||
| I0 = V0/R = 12 ÷ 56 × 103 = 2.1 × 10−4 A | ✓ 1 | |
| Part (c) | ||
| V = 12 × e−10/5.6 | ✓ 1 | |
| = 2.0 V (2.01 V) (allow error carried forward) | ✓ 1 | |
Exam tip: At the start of a discharge the whole capacitor p.d. is across the resistor, so the initial current is simply V0/R.
The graph shows how the charge Q stored varies with the potential difference V for two capacitors, X and Y.
Both capacitors are charged to a p.d. of 4.0 V. What is the ratio
charge stored on X ÷ charge stored on Y?
| Answer | Mark | Notes |
|---|---|---|
| Capacitance is the gradient of the Q–V line. CX = 2.0 mC ÷ 4.0 V = 0.50 mF and CY = 1.5 mC ÷ 8.0 V = 0.1875 mF. At the same p.d. Q ∝ C, so the ratio = 0.50 ÷ 0.1875 = 2.7. (Directly: at 4.0 V, QX = 2.0 mC and QY = 0.75 mC.) | ✓ 1 | Answer C |
Answer: C
Exam tip: On a Q–V graph the steeper line is the larger capacitance. Compare charges at the same p.d., not at the ends of the two lines.
In the circuit shown, the switch has been at X for a long time, so the capacitor is fully charged. The switch is then moved to Y.
How long does it take for the potential difference across the capacitor to fall to half its initial value?
| Answer | Mark | Notes |
|---|---|---|
| At Y the capacitor discharges through R2 only. RC = 47 × 103 × 470 × 10−6 = 22.1 s. Time to halve = RC ln 2 = 22.1 × 0.693 = 15 s. | ✓ 1 | Answer B |
Answer: B
Exam tip: Trace the discharge path on the diagram before calculating. Only the resistance in the loop that contains the capacitor after switching sets the time constant.
The graph shows how the potential difference V across a capacitor varies with time t as it discharges through a resistor.
What is the time constant of the circuit?
| Answer | Mark | Notes |
|---|---|---|
| From the graph, V falls from 10 V to 5.0 V in 7.0 s, so the time to halve is 7.0 s. Time to halve = RC ln 2, so RC = 7.0 ÷ 0.693 = 10 s. (Check: at 10 s the graph reads about 3.7 V, which is 37% of 10 V.) | ✓ 1 | Answer D |
Answer: D
Exam tip: The time constant is always longer than the time to halve: RC = t½ ÷ ln 2 ≈ 1.44 t½.
An uncharged capacitor is connected in series with a resistor, a switch and a d.c. supply. The switch is closed at time t = 0.
Which graph shows how the potential difference across the resistor varies with time?
| Answer | Mark | Notes |
|---|---|---|
| At t = 0 the capacitor p.d. is zero, so the whole supply p.d. is across the resistor. As the capacitor charges, VC rises and VR = Vsupply − VC falls, quickly at first and then more slowly, decaying exponentially towards zero: graph A. | ✓ 1 | Answer A |
Answer: A
Exam tip: During charging, the resistor p.d. (and the current) decays exponentially; the capacitor p.d. is the supply p.d. minus this, so it rises towards the supply value.
A camera flash unit stores energy in a capacitor. The graph shows how the potential difference V across the capacitor varies with the charge Q on it as it is charged to 300 V.
Use the graph to determine the capacitance.
State how the graph can be used to find the energy stored, and calculate the energy stored at 300 V.
The capacitor discharges completely through the flash tube in 1.5 ms. Calculate the mean power delivered.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| at 300 V the charge is 45 mC (read from the graph) | ✓ 1 | |
| C = Q/V = 45 × 10−3 ÷ 300 = 1.5 × 10−4 F (150 µF) | ✓ 1 | |
| Part (b) | ||
| energy stored = area under the V–Q graph | ✓ 1 | |
| W = ½ × 45 × 10−3 × 300 = 6.8 J (6.75 J) | ✓ 1 | |
| Part (c) | ||
| P = 6.75 ÷ 1.5 × 10−3 = 4.5 × 103 W (allow error carried forward) | ✓ 1 | |
Exam tip: On a V–Q graph the area is energy and the gradient is 1/C. Read the graph at a point where the line crosses grid lines exactly.
The figure shows a circuit used to charge an uncharged capacitor. The battery has negligible internal resistance and the voltmeters have very high resistance. Switch S is closed at time t = 0. While the capacitor charges, the p.d. across the resistor is VR = V0e−t/RC, where V0 is the battery p.d.
Calculate the time constant of the circuit.
Calculate the current immediately after S is closed.
Calculate the readings on V2 and V1 at t = 5.0 s.
Calculate the charge on the capacitor when it is fully charged.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| RC = 15 × 103 × 220 × 10−6 = 3.3 s | ✓ 1 | |
| Part (b) | ||
| I0 = 9.0 ÷ 15 × 103 = 6.0 × 10−4 A | ✓ 1 | |
| Part (c) | ||
| V2: VR = 9.0 e−5.0/3.3 = 2.0 V (1.98 V) | ✓ 1 | |
| V1 = 9.0 − 1.98 = 7.0 V (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| Q = CV = 220 × 10−6 × 9.0 = 2.0 × 10−3 C (1.98 mC) | ✓ 1 | |
Exam tip: The two voltmeter readings always add up to the battery p.d. while charging, so find the one that decays exponentially first and subtract.
The charge stored on a capacitor is increased by 50%.
By what factor does the energy stored increase?
| Answer | Mark | Notes |
|---|---|---|
| W = ½Q2/C, so for a fixed capacitor W ∝ Q2. Factor = 1.52 = 2.25. | ✓ 1 | Answer C |
Answer: C
Exam tip: For a fixed capacitor V rises in proportion to Q, so ½QV grows as the square of either.
A graph of potential difference V against charge Q for a capacitor is a straight line through the origin and the point (5.0 mC, 20 V).
How much energy is transferred to the capacitor as its charge increases from 2.5 mC to 5.0 mC?
| Answer | Mark | Notes |
|---|---|---|
| The energy is the area under the graph. At 5.0 mC: ½ × 5.0 × 10−3 × 20 = 50 mJ. At 2.5 mC (10 V): ½ × 2.5 × 10−3 × 10 = 12.5 mJ. Increase = 50 − 12.5 = 37.5 mJ = 38 mJ. | ✓ 1 | Answer B |
Answer: B
Common error: Finding only the triangle above the starting value. The area between 2.5 mC and 5.0 mC is a trapezium, not a triangle.
A capacitor charged to 12 V discharges through a resistor. The time constant of the circuit is RC.
What is the potential difference across the capacitor after a time 2RC?
| Answer | Mark | Notes |
|---|---|---|
| V = V0e−t/RC = 12 × e−2 = 12 × 0.135 = 1.6 V. | ✓ 1 | Answer A |
Answer: A
Exam tip: In each time constant the p.d. falls to 37% of its value at the start of that interval. Two time constants: 0.37 × 0.37 ≈ 0.14.
A 6.8 µF capacitor discharges through a 1.5 MΩ resistor.
How long does it take for the charge on the capacitor to halve?
| Answer | Mark | Notes |
|---|---|---|
| RC = 1.5 × 106 × 6.8 × 10−6 = 10.2 s. Time to halve = RC ln 2 = 10.2 × 0.693 = 7.1 s. | ✓ 1 | Answer B |
Answer: B
Exam tip: Put Q = ½Q0 into Q = Q0e−t/RC to get t = RC ln 2. The halving time is shorter than the time constant.
A toy car is powered by a 10 F supercapacitor. Before a run, the supercapacitor is charged to 2.5 V. The motor stops working when the potential difference across the supercapacitor has fallen to 1.0 V.
Calculate the energy stored when the supercapacitor is charged to 2.5 V.
The energy stored is ½QV, not QV. Explain why, with reference to a graph of potential difference against charge.
Calculate the energy transferred from the supercapacitor during a run.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| W = ½CV2 = ½ × 10 × 2.52 | ✓ 1 | |
| = 31 J (31.25 J) | ✓ 1 | |
| Part (b) | ||
| the graph of V against Q is a straight line through the origin / V ∝ Q, and the energy stored is the area under it | ✓ 1 | |
| the area is a triangle, ½QV / charge was added at an average p.d. of ½V, not at V | ✓ 1 | |
| Part (c) | ||
| energy left at 1.0 V = ½ × 10 × 1.02 = 5.0 J | ✓ 1 | |
| energy transferred = 31.25 − 5.0 = 26 J (26.25 J) | ✓ 1 | |
Model sketch
Common error: Using ½C(ΔV)2 = ½ × 10 × 1.52. Energy depends on V2, so find the energy at each p.d. and subtract.
A bicycle rear light keeps shining for a while after the bicycle stops. While the bicycle moves, a 1.0 F capacitor is charged to 4.2 V. When the bicycle stops, the capacitor discharges through the light circuit, which behaves as a constant resistance of 220 Ω. The light is useful until the potential difference falls to 2.6 V.
Calculate the time constant of the discharge.
Show that the light is useful for about 110 s after the bicycle stops.
Suggest one change to the capacitor that would make the light useful for longer.
Calculate the energy transferred from the capacitor to the light circuit while the light is useful.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| RC = 220 × 1.0 = 220 s | ✓ 1 | |
| Part (b) | ||
| ln V = ln V0 − t/RC / 2.6 = 4.2e−t/220 | ✓ 1 | |
| t = 220 × ln(4.2 ÷ 2.6) | ✓ 1 | |
| = 220 × 0.480 = 106 s (≈ 110 s) | ✓ 1 | |
| Part (c) | ||
| use a capacitor of larger capacitance (so RC is larger) / use a capacitor with a higher voltage rating, charged to a higher p.d. | ✓ 1 | |
| Part (d) | ||
| W = ½ × 1.0 × (4.22 − 2.62) | ✓ 1 | |
| = 5.4 J (5.44 J) | ✓ 1 | |
Exam tip: To find a time, take natural logs: t = RC ln(V0/V). Check that the answer is less than one halving time when V is more than half of V0.
In an experiment, a capacitor is discharged through a 22.0 kΩ resistor and a data logger records the potential difference V across the capacitor.
Results: t = 0 s, V = 5.00 V; 10 s, 3.17 V; 20 s, 2.01 V; 30 s, 1.28 V; 40 s, 0.81 V.
Explain why a graph of ln(V/V) against t should be a straight line, and state what its gradient represents.
The line of best fit passes through the points for t = 0 and t = 40 s. Use these points to determine the time constant.
Determine the capacitance.
State one advantage of using a data logger rather than a voltmeter and a stopwatch.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| V = V0e−t/RC gives ln V = ln V0 − t/RC, which has the form y = c + mx | ✓ 1 | |
| gradient = −1/RC | ✓ 1 | |
| Part (b) | ||
| ln 5.00 = 1.609 and ln 0.81 = −0.211, so gradient = (−0.211 − 1.609) ÷ 40 = −0.0455 s−1 | ✓ 1 | |
| RC = 1 ÷ 0.0455 = 22 s (22.0 s) | ✓ 1 | |
| Part (c) | ||
| C = 22.0 ÷ 22.0 × 103 = 1.0 × 10−3 F (1000 µF) (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| readings taken at precise / equal / short time intervals, including at the start when V changes fastest / no reaction-time error / many more readings | ✓ 1 | |
Model sketch
Exam tip: Label the axis ln(V/V): you take the log of a pure number. Quote ln values to 3 decimal places so the gradient keeps its precision.
A 4700 µF capacitor, initially uncharged, is charged through a resistor by a 12 V battery until the current is zero.
Calculate the charge that flows from the battery.
Calculate the energy transferred by the battery.
Calculate the energy stored by the capacitor.
Explain what happens to the rest of the energy, and why the answer does not depend on the resistance of the resistor.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Q = CV = 4700 × 10−6 × 12 = 0.056 C (0.0564 C) | ✓ 1 | |
| Part (b) | ||
| W = QV = 0.0564 × 12 = 0.68 J (0.677 J) | ✓ 1 | |
| Part (c) | ||
| W = ½QV = 0.34 J (0.338 J) | ✓ 1 | |
| Part (d) | ||
| it is dissipated as thermal energy in the resistor (and connecting wires / battery) | ✓ 1 | |
| a smaller resistance gives a larger current for a shorter time, but the same charge moves and the same ½QV is dissipated / half the energy is always dissipated | ✓ 1 | |
Common error: Assuming all the battery's energy is stored. Each coulomb gets 12 J from the battery, but charge added early is stored at a low p.d.
A charged capacitor discharges through a 4.0 kΩ resistor. The current falls from 2.5 mA to 0.80 mA in 6.0 s.
Determine the time constant of the circuit.
Calculate the capacitance.
Calculate the p.d. across the capacitor at the start of the discharge.
Calculate the initial charge on the capacitor, and state how it could also be found from a graph of current against time for the whole discharge.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 0.80 = 2.5e−6.0/RC / ln(2.5 ÷ 0.80) = 6.0/RC | ✓ 1 | |
| RC = 6.0 ÷ ln 3.125 = 6.0 ÷ 1.139 = 5.3 s (5.27 s) | ✓ 1 | |
| Part (b) | ||
| C = 5.27 ÷ 4.0 × 103 = 1.3 × 10−3 F (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| V0 = I0R = 2.5 × 10−3 × 4.0 × 103 = 10 V | ✓ 1 | |
| Part (d) | ||
| Q0 = CV0 = 1.32 × 10−3 × 10 = 0.013 C | ✓ 1 | |
| it equals the (total) area under the current–time graph | ✓ 1 | |
Model sketch
Exam tip: The current obeys I = I0e−t/RC with the same time constant as the charge and p.d., so a ratio of currents gives RC directly.
The graph shows how the energy W stored in a capacitor varies with the potential difference V across it. Point P lies on the curve.
What is the capacitance of the capacitor?
| Answer | Mark | Notes |
|---|---|---|
| At P, V = 4.0 V and W = 20 mJ. W = ½CV2, so C = 2W/V2 = 2 × 0.020 ÷ 4.02 = 2.5 × 10−3 F = 2.5 mF. | ✓ 1 | Answer B |
Answer: B
Common error: Treating the curve as if W ∝ V. The energy depends on V2, which is why the graph is a curve and not a straight line.
An uncharged capacitor is charged by a constant current of 50 µA. The graph shows how the potential difference V across the capacitor varies with time t.
What is the energy stored in the capacitor at t = 40 s?
| Answer | Mark | Notes |
|---|---|---|
| Q = It = 50 × 10−6 × 40 = 2.0 × 10−3 C. From the graph V = 8.0 V at 40 s. W = ½QV = ½ × 2.0 × 10−3 × 8.0 = 8.0 × 10−3 J = 8.0 mJ. | ✓ 1 | Answer C |
Answer: C
Exam tip: With a constant current, the charge is simply It. Pair it with the p.d. read from the graph at the same instant and use ½QV.
A charged capacitor discharges through a 20 kΩ resistor. The graph shows how the potential difference V across the capacitor varies with time t. The tangent to the curve at t = 0 is drawn.
What is the capacitance?
| Answer | Mark | Notes |
|---|---|---|
| If the p.d. kept falling at its initial rate, the capacitor would be fully discharged after one time constant, so the tangent at t = 0 meets the time axis at t = RC = 4.0 s. C = 4.0 ÷ 20 × 103 = 2.0 × 10−4 F = 200 µF. | ✓ 1 | Answer B |
Answer: B
Exam tip: Two quick ways to find RC from a discharge curve: the time to fall to 37% of the starting value, or where the initial tangent cuts the time axis.
A capacitor discharges through a resistor. The graph shows how ln(V/V) varies with time t, where V is the p.d. across the capacitor.
What is the time constant of the circuit?
| Answer | Mark | Notes |
|---|---|---|
| ln V = ln V0 − t/RC, so the gradient is −1/RC. Gradient = (0 − 2.0) ÷ 50 = −0.040 s−1, so RC = 1 ÷ 0.040 = 25 s. | ✓ 1 | Answer A |
Answer: A
Common error: Reading the time where ln V = 0 as the end of the discharge. ln V = 0 means V = 1 V, not zero; ln V never reaches a final value.
A data logger records the potential difference V across an uncharged capacitor as it is charged through a 47 kΩ resistor from a 6.0 V supply. The graph shows the results.
What is the capacitance?
| Answer | Mark | Notes |
|---|---|---|
| V reaches 3.0 V (half of 6.0 V) at t = 8.0 s. At that moment the p.d. across the resistor, which decays as 6.0e−t/RC, has also halved, so 8.0 s = RC ln 2 and RC = 11.5 s. C = 11.5 ÷ 47 × 103 = 2.5 × 10−4 F = 250 µF. | ✓ 1 | Answer C |
Answer: C
Exam tip: On a charging curve, the time to reach half the final p.d. is also RC ln 2, because the p.d. across the resistor halves in that time.
In a practical investigation, a capacitor charged from a 6.0 V supply is discharged through a 47 kΩ resistor. A data logger records the p.d. V across the capacitor every 10 s. The graph shows ln(V/V) plotted against time t, with a line of best fit.
Identify the anomalous result and state how it was dealt with when the line of best fit was drawn.
Use the graph to determine the capacitance.
Use the intercept on the ln(V/V) axis to determine the p.d. across the capacitor at t = 0, and comment on your answer.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the point at t = 50 s | ✓ 1 | |
| it was ignored / not used when drawing the line (the reading could be repeated) | ✓ 1 | |
| Part (b) | ||
| gradient from a large triangle, e.g. (−0.60 − 1.80) ÷ 80 = −0.030 s−1 (allow −0.029 to −0.031) | ✓ 1 | |
| gradient = −1/RC, so RC = 1 ÷ 0.030 = 33 s | ✓ 1 | |
| C = 33.3 ÷ 47 × 103 = 7.1 × 10−4 F (allow 6.8–7.4 × 10−4 F) | ✓ 1 | |
| Part (c) | ||
| V0 = e1.80 = 6.0 V (6.05 V), equal to the supply p.d., so the capacitor was fully charged at the start | ✓ 1 | |
Exam tip: Use a triangle that covers most of the line to find a gradient, and read its corners where the line crosses grid lines. Never use two of the plotted points unless they lie on the line.
A capacitor is charged to 9.0 V and discharged through a resistor RX. It is then charged to 9.0 V again and discharged through a different resistor RY. The graph shows how the current I varies with time t for each discharge.
Use the graph to determine the resistances of RX and RY.
Use point P to determine the time constant for discharge X, and hence the capacitance.
Explain why the areas under the two curves are equal.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I0 for X = 3.0 mA, so RX = 9.0 ÷ 3.0 × 10−3 = 3.0 × 103 Ω | ✓ 1 | |
| I0 for Y = 1.0 mA, so RY = 9.0 × 103 Ω | ✓ 1 | |
| Part (b) | ||
| 1.10 = 3.0 e−3.0/RC, so RC = 3.0 ÷ ln(3.0 ÷ 1.10) = 3.0 ÷ 1.00 = 3.0 s | ✓ 1 | |
| C = 3.0 ÷ 3.0 × 103 = 1.0 × 10−3 F (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| the area under an I–t graph is the charge that flows | ✓ 1 | |
| the same capacitor charged to the same p.d. holds the same charge Q = CV, and all of it flows off in each discharge | ✓ 1 | |
Exam tip: A larger resistance gives a smaller starting current but a longer time constant; the area under the I–t curve stays equal to CV0.
A student displays the p.d. across a capacitor on an oscilloscope while the capacitor discharges through a 4.7 kΩ resistor. The figure shows the trace and the oscilloscope settings.
State the p.d. across the capacitor at the start of the discharge.
Use the trace to determine the time for the p.d. to fall to half its initial value.
Calculate the capacitance.
Use the trace to check that the p.d. halves again in the next 1.0 ms.
Suggest why a time-base setting of 5.0 ms div−1 would be unsuitable.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 6.0 div × 1.0 V div−1 = 6.0 V | ✓ 1 | |
| Part (b) | ||
| the trace reaches 3.0 div after 2.0 div, so t½ = 2.0 × 0.50 ms = 1.0 ms | ✓ 1 | |
| Part (c) | ||
| RC = t½/ln 2 = 1.0 × 10−3 ÷ 0.693 = 1.44 × 10−3 s | ✓ 1 | |
| C = 1.44 × 10−3 ÷ 4.7 × 103 = 3.1 × 10−7 F (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| at 4.0 div (2.0 ms) the trace is at 1.5 div = 1.5 V, half of 3.0 V, so the halving time is constant, as expected for an exponential decay | ✓ 1 | |
| Part (e) | ||
| the halving time would be only 0.2 div / the whole decay would fit in about one division, so times could not be read precisely | ✓ 1 | |
Exam tip: Convert every reading from divisions using the settings: horizontal × time-base gives time, vertical × Y-gain gives p.d.
An uncharged capacitor is charged through a resistor of resistance R from a 12 V supply of negligible internal resistance. The graph shows how the charge Q on the capacitor varies with time t. The tangent to the curve at t = 0 is drawn.
Use the tangent to determine the current at t = 0.
Calculate R.
Use the graph to determine the capacitance.
Show that the tangent reaches the final charge at a time equal to the time constant.
Calculate the energy stored when the capacitor is fully charged.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| current = gradient of the Q–t graph = 2.4 × 10−3 ÷ 5.0 | ✓ 1 | |
| = 4.8 × 10−4 A | ✓ 1 | |
| Part (b) | ||
| R = 12 ÷ 4.8 × 10−4 = 2.5 × 104 Ω (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| final charge = 2.4 mC, so C = 2.4 × 10−3 ÷ 12 = 2.0 × 10−4 F | ✓ 1 | |
| Part (d) | ||
| RC = 2.5 × 104 × 2.0 × 10−4 = 5.0 s, the same time at which the tangent reaches 2.4 mC | ✓ 1 | |
| Part (e) | ||
| W = ½QV = ½ × 2.4 × 10−3 × 12 = 0.014 J (0.0144 J) | ✓ 1 | |
Exam tip: The gradient of a Q–t graph is the current. The initial gradient gives the largest current, V/R, because the capacitor p.d. is zero at the start.
A student records the discharge of a capacitor through a 100 kΩ resistor with a data logger. A graph of ln(V/V) against time t is a straight line with gradient −0.025 s−1.
What is the capacitance?
| Answer | Mark | Notes |
|---|---|---|
| ln V = ln V0 − t/RC, so gradient = −1/RC. RC = 1 ÷ 0.025 = 40 s, and C = 40 ÷ 100 × 103 = 4.0 × 10−4 F = 400 µF. | ✓ 1 | Answer D |
Answer: D
Exam tip: The gradient of a ln–t graph is −1/RC, not −RC. Check the unit: s−1, so invert to get a time.
A charged capacitor discharges through a resistor.
What percentage of the initial stored energy remains after one time constant?
| Answer | Mark | Notes |
|---|---|---|
| V = V0e−1 after one time constant, and W ∝ V2, so W/W0 = e−2 = 0.135 ≈ 14%. | ✓ 1 | Answer A |
Answer: A
Exam tip: Charge, p.d. and current fall to 37% in one time constant; energy, which depends on the square of the p.d., falls to 37% squared.
An uncharged 1000 µF capacitor is connected in series with a 4.7 kΩ resistor, a switch and a 6.0 V supply of negligible internal resistance. The switch is closed at time t = 0.
Calculate the current immediately after the switch is closed.
Explain why the current decreases to zero and why the potential difference across the capacitor rises less and less quickly. Explain also how charging would differ if the resistor were replaced by one of larger resistance.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the capacitor p.d. is zero, so all 6.0 V is across the resistor: I = 6.0 ÷ 4.7 × 103 = 1.3 × 10−3 A | ✓ 1 | |
| Part (b) | ||
| Indicative content: | — | |
| at first the capacitor is uncharged, so the p.d. across it is zero and the whole supply p.d. is across the resistor, giving the maximum current | ✓ 1 | |
| charge flows onto one plate and off the other, so the plates gain equal and opposite charges | ✓ 1 | |
| as the charge increases, the p.d. across the capacitor (V = Q/C) increases | ✓ 1 | |
| so the p.d. across the resistor (6.0 V − VC) decreases and the current decreases | ✓ 1 | |
| a smaller current adds charge more slowly, so VC rises less quickly; the current is zero when VC = 6.0 V | ✓ 1 | |
| with a larger resistance the current is smaller at each stage, so charging takes longer (larger RC) but the final charge and p.d. are the same | ✓ 1 | |
Model sketch
Exam tip: In an extended explanation, link each step: charge builds up → VC rises → p.d. across R falls → current falls → charge builds up more slowly.
A student finds the capacitance of a small capacitor using an oscilloscope. A signal generator supplies a square wave to the capacitor in series with a 1.0 kΩ ± 2% resistor, and the oscilloscope displays the p.d. across the capacitor. From the trace, the time for the p.d. to fall to half its initial value during a discharge is 3.2 ms ± 0.1 ms.
State why the period of the square wave must be much longer than the time constant of the circuit.
Calculate the capacitance.
Calculate the absolute uncertainty in the capacitance.
The student repeats the experiment with a 1.0 MΩ resistor. The oscilloscope has an input resistance of 1.0 MΩ. Explain why the value of C obtained would now be too small.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| so that the capacitor charges / discharges fully in each half-cycle, giving a complete decay that starts from the full p.d. | ✓ 1 | |
| Part (b) | ||
| t½ = RC ln 2, so C = 3.2 × 10−3 ÷ (1.0 × 103 × ln 2) | ✓ 1 | |
| = 4.6 × 10−6 F (4.62 µF) | ✓ 1 | |
| Part (c) | ||
| percentage uncertainty in t½ = 0.1 ÷ 3.2 × 100 = 3.1%; total = 3.1% + 2% = 5.1% | ✓ 1 | |
| ΔC = 0.051 × 4.62 µF = 0.24 µF | ✓ 1 | |
| Part (d) | ||
| the oscilloscope provides a second discharge path in parallel with the resistor, so the effective resistance is about 0.5 MΩ | ✓ 1 | |
| so the discharge is faster / the measured t½ is smaller, and dividing by 1.0 MΩ gives C about half the true value | ✓ 1 | |
Exam tip: When a result is a product or quotient, add the percentage uncertainties of the measured quantities; ln 2 is exact and adds no uncertainty.
A greenhouse fan runs on for a set time after a door is closed. The delay is produced by an uncharged 220 µF capacitor charging through a resistor R from a 5.0 V supply. The fan switches off when the p.d. across the capacitor reaches 3.5 V, which must happen 45 s after charging starts. While charging, the p.d. across the capacitor is VC = V0(1 − e−t/RC).
State the p.d. across the resistor at the moment the fan switches off.
Calculate the resistance R needed.
The 5.0 V supply is replaced by a 9.0 V supply, with the same R, C and switch-off p.d. Calculate the new delay and explain why it is shorter.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 5.0 − 3.5 = 1.5 V | ✓ 1 | |
| Part (b) | ||
| e−45/RC = 1.5 ÷ 5.0 = 0.30 | ✓ 1 | |
| RC = 45 ÷ ln(1 ÷ 0.30) = 45 ÷ 1.204 = 37.4 s | ✓ 1 | |
| R = 37.4 ÷ 220 × 10−6 = 1.7 × 105 Ω | ✓ 1 | |
| Part (c) | ||
| e−t/RC = (9.0 − 3.5) ÷ 9.0 = 0.611 | ✓ 1 | |
| t = 37.4 × ln(9.0 ÷ 5.5) = 18 s (18.4 s) (allow error carried forward) | ✓ 1 | |
| 3.5 V is now a smaller fraction of the final p.d., reached earlier on the charging curve, where the p.d. rises fastest | ✓ 1 | |
Common error: Using the discharge equation for a charging capacitor. While charging, it is the p.d. across the resistor that decays: VR = V0e−t/RC.
A charged capacitor discharges through a resistor. The graph shows how the current I varies with time t.
What was the charge on the capacitor at t = 0?
| Answer | Mark | Notes |
|---|---|---|
| At 5.0 s the current is 0.74 ÷ 2.0 = 0.37 = e−1 of its initial value, so RC = 5.0 s. The charge is the total area under the curve: Q0 = ∫I0e−t/RC dt = I0RC = 2.0 × 10−3 × 5.0 = 1.0 × 10−2 C = 10 mC. | ✓ 1 | Answer D |
Answer: D
Exam tip: The whole area under a discharge I–t curve equals I0 × RC, the area of a rectangle of height I0 and width RC.
A capacitor charged to 20 V discharges through a resistor of resistance R. The graph shows how the energy W stored in the capacitor varies with time t.
Use the graph to determine the capacitance.
Show that the energy stored halves in a time ½RC ln 2.
Use the graph to determine RC, and hence calculate R.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| W0 = 0.80 J | ✓ 1 | |
| C = 2W/V2 = 2 × 0.80 ÷ 202 = 4.0 × 10−3 F | ✓ 1 | |
| Part (b) | ||
| W = ½CV2 and V = V0e−t/RC, so W = W0e−2t/RC | ✓ 1 | |
| W = ½W0 when 2t/RC = ln 2, so t = ½RC ln 2 | ✓ 1 | |
| Part (c) | ||
| W = 0.40 J at t = 3.0 s (allow 2.9–3.1 s), so RC = 2 × 3.0 ÷ ln 2 = 8.7 s | ✓ 1 | |
| R = 8.66 ÷ 4.0 × 10−3 = 2.2 × 103 Ω (allow 2.1–2.2 × 103 Ω; allow error carried forward) | ✓ 1 | |
Common error: Using the energy graph as if it were a p.d. graph. Energy depends on V2, so it decays twice as fast: its halving time is half that of the p.d.
A capacitor is charged to 12.0 V and then discharged through a 20 kΩ resistor R1. At t = 20 s a switch S is closed, which connects a second resistor R2 in parallel with R1. The graph shows how the p.d. V across the capacitor varies with time t.
Use points A and B to show that the time constant before S is closed is about 20 s.
Calculate the capacitance.
Use points B and C to determine the time constant after S is closed.
Determine the resistance of R2.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 4.41 = 12.00 e−20/RC, so ln(12.00 ÷ 4.41) = 20/RC | ✓ 1 | |
| RC = 20 ÷ 1.00 = 20 s (20.0 s) | ✓ 1 | |
| Part (b) | ||
| C = 20 ÷ 20 × 103 = 1.0 × 10−3 F | ✓ 1 | |
| Part (c) | ||
| ln(4.41 ÷ 0.60) = 10/τ, so 1.99 = 10/τ | ✓ 1 | |
| τ = 5.0 s | ✓ 1 | |
| Part (d) | ||
| total resistance = 5.0 ÷ 1.0 × 10−3 = 5.0 kΩ | ✓ 1 | |
| 1/R2 = 1/5.0 − 1/20, so R2 = 6.7 × 103 Ω (allow error carried forward) | ✓ 1 | |
Exam tip: Any two points on the same exponential section give the time constant: RC = Δt ÷ ln(V1/V2). They do not need to start at t = 0.
An uncharged capacitor is charged through a resistor from a 6.0 V supply of negligible internal resistance, starting at A. At C a switch disconnects the supply and connects the capacitor in a loop with the same resistor, so it discharges. The graphs show the p.d. V across the capacitor and the current I in the resistor.
Use the graphs to determine the resistance of the resistor.
Explain the shapes of both graphs from A to D. Your answer should refer to the labelled points and explain why the current changes direction at C.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| at A, all 6.0 V is across the resistor and I = 0.60 mA, so R = 6.0 ÷ 0.60 × 10−3 = 1.0 × 104 Ω | ✓ 1 | |
| Part (b) | ||
| Indicative content: | — | |
| at A the capacitor is uncharged, so VC = 0, the whole 6.0 V is across the resistor and the current is a maximum | ✓ 1 | |
| as charge builds up, VC = Q/C rises, so the p.d. across the resistor (6.0 − VC) and the current fall; the capacitor therefore charges ever more slowly | ✓ 1 | |
| the current and the rate of rise of VC decay exponentially; by B (several time constants) VC ≈ 6.0 V and I ≈ 0 | ✓ 1 | |
| from B to C the capacitor is fully charged: no current, VC constant at 6.0 V | ✓ 1 | |
| at C the capacitor acts as the source: charge flows off the positive plate round the loop through the resistor in the opposite direction, so I is negative, starting at −0.60 mA because all 6.0 V is across the resistor | ✓ 1 | |
| from C to D, as charge leaves, VC falls, so the current falls in size; both decay exponentially with the same time constant as charging, because R and C are unchanged | ✓ 1 | |
Exam tip: In an extended answer, link cause and effect at each step and use the letters on the graphs, so every statement is tied to a stage of the process.
A defibrillator delivers a pulse of energy to a patient's heart. Its capacitor is charged to 2.0 kV and then discharged through the patient's chest, which behaves as a resistance R. The graph shows how the current I varies with time t during the discharge.
Use the graph to determine R.
Use the graph to determine the time constant of the discharge.
Calculate the capacitance.
Calculate the energy stored in the capacitor before the discharge.
Calculate the energy delivered during the first 8.0 ms.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I0 = 40 A, so R = 2.0 × 103 ÷ 40 = 50 Ω (5.0 × 101 Ω) | ✓ 1 | |
| Part (b) | ||
| after one time constant I = 0.37 × 40 = 14.8 A | ✓ 1 | |
| reading the graph at 14.8 A gives 8.0 ms (allow 7.6–8.4 ms) | ✓ 1 | |
| Part (c) | ||
| C = 8.0 × 10−3 ÷ 50 = 1.6 × 10−4 F (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| W = ½CV2 = ½ × 1.6 × 10−4 × (2.0 × 103)2 = 320 J | ✓ 1 | |
| Part (e) | ||
| energy remaining = 320 × e−2 = 43 J (the p.d. has fallen to e−1, and W ∝ V2) | ✓ 1 | |
| energy delivered = 320 − 43 = 280 J (277 J) | ✓ 1 | |
Common error: Assuming 63% of the energy is delivered in one time constant. The p.d. falls by 63%, but the energy depends on V2, so 1 − e−2 ≈ 86% of it is delivered.
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