AS & A Level Physics (9702)
Capacitance Banker
24 questions · about 3% of the marks we indexed · examined in 9 of 9 sessions.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Capacitors, combinations & energy stored · Core · Paper 4 style · 4 marksA 470 µF capacitor is connected to a 12.0 V d.c. supply until fully charged.
- (a) Define capacitance. [1]
- (b) Calculate the charge stored. [1]
- (c) Calculate the energy stored. [2]
Show answer
- (a) Charge stored per unit potential difference (C = Q/V).
B1: ratio of charge (on one plate) to potential difference between the plates.
- (b) 0.00564 C
Q = CV = 470 × 10⁻⁶ × 12 = 0.00564 C (A1).
- (c) 0.0338 J
W = ½CV² (M1) = ½ × 470 × 10⁻⁶ × 144 = 0.0338 J (A1).
Two capacitors have capacitances 2.0 µF and 6.0 µF.
- (a) Calculate the combined capacitance when the capacitors are connected in series. [2]
- (b) Calculate the combined capacitance when they are connected in parallel. [1]
- (c) When connected in series to a battery, state which capacitor has the larger potential difference across it, giving a reason. [2]
Show answer
- (a) 1.5 µF
1/C = 1/2.0 + 1/6.0 (M1) → C = 1.5 µF (A1).
- (b) 8.0 µF
C = 2.0 + 6.0 = 8.0 µF (A1).
- (c) The 2.0 µF capacitor
B1: capacitors in series carry the same charge; B1: V = Q/C so the smaller capacitance has the larger pd.
The flash unit in a camera uses a 150 µF capacitor charged to 330 V.
- (a) Calculate the energy stored in the capacitor. [2]
- (b) The capacitor discharges through the flash tube in 1.0 ms. Calculate the average power delivered. [1]
- (c) Explain why a capacitor, rather than the 1.5 V battery directly, is used to power the flash. [1]
Show answer
- (a) 8.17 J
W = ½CV² (M1) = ½ × 150 × 10⁻⁶ × 330² = 8.17 J (A1).
- (b) 8170 W
P = W/t = 8.17/1.0 × 10⁻³ = 8170 W (A1).
- (c) The capacitor can release its energy very quickly (very high power/current) at high pd; the battery cannot.
B1: capacitor discharges rapidly, giving high power that a battery cannot deliver.
- (a) Express the farad in terms of SI base units. [2]
- (b) A 2200 µF capacitor stores a charge of 0.0198 C. Calculate the potential difference across it. [1]
Show answer
- (a) kg⁻¹ m⁻² s⁴ A²
M1: F = C V⁻¹ = A s/(J C⁻¹) = A² s²/J (or V = kg m² s⁻³ A⁻¹); A1: A² s⁴ kg⁻¹ m⁻².
- (b) 9.0 V
V = Q/C = 0.0198/2200 × 10⁻⁶ = 9.0 V (A1).
- (a) Show that the combined capacitance C of capacitors C₁ and C₂ in series is given by 1/C = 1/C₁ + 1/C₂. [3]
- (b) State two practical uses of capacitors in electrical circuits. [2]
Show answer
- (a) Same charge Q on each; V = V₁ + V₂; Q/C = Q/C₁ + Q/C₂; divide by Q.
B1: charge on each capacitor is the same, Q; B1: total pd = sum of pds; B1: V = Q/C substitution and division by Q.
- (b) e.g. smoothing a rectified supply; storing energy for a flash or back-up supply; timing circuits.
B1 each for any two: smoothing; energy store (flash, back-up memory, defibrillator); timing (RC); blocking d.c.
A 3.0 µF capacitor is connected in series with a parallel combination of a 4.0 µF capacitor and a 2.0 µF capacitor. The network is connected to a 9.0 V battery and the capacitors are fully charged.
- (a) Calculate the total capacitance of the network. [2]
- (b) Calculate the charge on the 3.0 µF capacitor. [1]
- (c) Calculate the potential difference across the 4.0 µF capacitor and the charge on it. [3]
Show answer
- (a) 2.0 µF
Parallel pair = 6.0 µF (M1); series with 3.0 µF: 1/C = 1/3 + 1/6 → 2.0 µF (A1).
- (b) 18 µC
Q = CV = 2.0 µF × 9.0 V = 18 µC (A1).
- (c) 3.0 V; 12 µC
pd across 3.0 µF = 18/3.0 = 6.0 V (M1); pd across parallel pair = 9.0 − 6.0 = 3.0 V (A1); Q = 4.0 × 3.0 = 12 µC (A1).
- (a) Use a graph of potential difference V against charge Q for a capacitor to show that the energy stored is W = ½QV. [2]
- (b) A capacitor is charged to pd V by a battery of e.m.f. V through a resistor. Show that only half the energy supplied by the battery is stored in the capacitor, and state what happens to the rest. [3]
Show answer
- (a) V ∝ Q gives a straight line through the origin; work done = area under the V–Q graph = ½QV.
M1: work done adding small charge ΔQ at pd V is VΔQ, so energy = area under V–Q graph; A1: graph is a straight line through the origin, area of triangle = ½QV.
- (b) Battery supplies QV (charge Q moved through e.m.f. V); capacitor stores ½QV; the other ½QV is dissipated as thermal energy in the resistance.
B1: energy from battery = QV (constant e.m.f.); B1: stored = ½QV, so half; B1: rest dissipated as heat in the resistor/wires.
A solar garden light stores energy in a 50 F supercapacitor rated at 2.7 V.
- (a) Calculate the energy stored when the supercapacitor is fully charged. [1]
- (b) The lamp circuit works until the pd falls to 1.0 V. Calculate the energy available to the circuit. [2]
- (c) The circuit takes a constant power of 0.50 W. Calculate how long it can run. [1]
Show answer
- (a) 182 J
½ × 50 × 2.7² = 182 J (A1).
- (b) 157 J
Energy at 1.0 V = ½ × 50 × 1.0² = 25 J (M1); available = 182 − 25 = 157 J (A1).
- (c) 315 s
t = 157.25/0.50 = 314.5 ≈ 315 s (A1; accept 314 s).
As a capacitor is charged, the charge Q on it and the potential difference V across it are recorded:Q / mC: 0, 1.5, 3.0, 4.5V / V: 0, 2.0, 4.0, 6.0
- (a) Determine the capacitance. [1]
- (b) Determine the energy stored when the pd is 6.0 V. [1]
- (c) Calculate the energy transferred to the capacitor as its pd rises from 4.0 V to 6.0 V. [2]
Show answer
- (a) 750 µF
C = Q/V = 4.5 × 10⁻³/6.0 = 7.5 × 10⁻⁴ F (A1).
- (b) 13.5 mJ
Area = ½ × 4.5 × 10⁻³ × 6.0 = 13.5 mJ (A1).
- (c) 7.5 mJ
ΔW = ½C(6.0² − 4.0²) (M1) = ½ × 7.5 × 10⁻⁴ × 20 = 7.5 mJ (A1).
A capacitor stores 0.40 J of energy when the pd across it is 20 V.
- (a) Calculate the capacitance. [2]
- (b) Calculate the pd needed for the capacitor to store 1.6 J. [1]
- (c) Calculate the charge stored at this pd. [1]
Show answer
- (a) 0.0020 F (2.0 mF)
W = ½CV² → C = 2W/V² (M1) = 0.80/400 = 0.0020 F (A1).
- (b) 40 V
Energy ∝ V²: four times the energy needs twice the pd → 40 V (A1).
- (c) 0.080 C
Q = CV = 0.0020 × 40 = 0.080 C (A1).
A 4.7 µF capacitor and a 10 µF capacitor are connected in series to a 15 V supply.
- (a) Calculate the combined capacitance. [1]
- (b) Calculate the charge on each capacitor. [1]
- (c) Calculate the pd across each capacitor. [2]
- (d) Calculate the energy stored in the 4.7 µF capacitor. [1]
Show answer
- (a) 3.20 µF
1/C = 1/4.7 + 1/10 → C = 3.20 µF (A1).
- (b) 48.0 µC
Q = CV = 3.20 × 15 = 48.0 µC (same on both) (A1).
- (c) 10.2 V across 4.7 µF; 4.80 V across 10 µF
V = Q/C: 48.0/4.7 = 10.2 V (A1); 48.0/10 = 4.80 V (A1).
- (d) 2.45 × 10⁻⁴ J
½ × 4.7 × 10⁻⁶ × 10.2² = 2.45 × 10⁻⁴ J (A1).
A 20 µF capacitor is charged to 12 V and then disconnected from the supply. It is then connected across an uncharged 10 µF capacitor, positive plate to one plate and negative plate to the other.
- (a) Calculate the final potential difference across the capacitors. [2]
- (b) Calculate the total energy stored before and after the connection. [2]
- (c) Suggest what has happened to the missing energy. [1]
Show answer
- (a) 8.0 V
Charge conserved: Q = 20 × 10⁻⁶ × 12 = 240 µC (M1); V = 240 µC/30 µF = 8.0 V (A1).
- (b) 1.44 mJ before; 0.960 mJ after
½ × 20 × 10⁻⁶ × 12² = 1.44 mJ (A1); ½ × 30 × 10⁻⁶ × 8.0² = 0.960 mJ (A1).
- (c) Dissipated as thermal energy in the connecting wires (and some as electromagnetic radiation) as charge flowed.
B1: heating in resistance of wires (as current flows) / radiated.
A technician has a box of identical 100 µF capacitors, each with a maximum safe working voltage of 25 V.
- (a) Describe, without a diagram, an arrangement of the smallest number of these capacitors that has a total capacitance of 100 µF and can safely be used at 50 V. Justify your answer. [4]
- (b) Calculate the total energy stored by the arrangement at 50 V, and the energy stored in each capacitor. [2]
Show answer
- (a) Four capacitors: two in series (50 µF, 50 V rating), and two such series pairs connected in parallel (100 µF).
B1: two in series share 50 V, so 25 V each (safe); B1: series pair has capacitance 50 µF; B1: two pairs in parallel → 100 µF; B1: 4 is the minimum (one or two cannot give both 50 V rating and 100 µF; three in series give 33 µF). Also accept two parallel pairs (200 µF each) connected in series: 100 µF, 25 V across each.
- (b) 0.125 J total; 0.0313 J each
W = ½ × 100 × 10⁻⁶ × 50² = 0.125 J (A1); each has 25 V: ½ × 100 × 10⁻⁶ × 25² = 0.03125 ≈ 0.0313 J (A1; = 0.125/4).
A 1000 µF capacitor, initially uncharged, is charged by a constant current of 2.0 mA for 15 s.
- (a) Calculate the pd across the capacitor after 15 s. [2]
- (b) Calculate the energy stored after 15 s. [1]
- (c) Calculate the rate at which energy is being stored at t = 15 s, and explain why this rate is not constant. [2]
Show answer
- (a) 30 V
Q = It = 2.0 × 10⁻³ × 15 = 0.030 C (M1); V = Q/C = 30 V (A1).
- (b) 0.45 J
½CV² = ½ × 10⁻³ × 30² = 0.45 J (A1).
- (c) 0.060 W; the pd rises steadily while the current is constant, so P = VI increases.
P = VI = 30 × 2.0 × 10⁻³ = 0.060 W (A1); B1: V ∝ t so power ∝ t (energy ∝ t²).
A heart defibrillator uses a 32 µF capacitor charged to 5.0 kV.
- (a) Calculate the energy stored in the capacitor when fully charged. [1]
- (b) During a pulse 300 J is delivered. Calculate the pd across the capacitor at the end of the pulse. [3]
- (c) State the fraction of the original charge still on the capacitor. [1]
Show answer
- (a) 400 J
½ × 32 × 10⁻⁶ × (5.0 × 10³)² = 400 J (A1).
- (b) 2500 V
Remaining energy = 400 − 300 = 100 J (M1); V = √(2W/C) (M1) = √(200/32 × 10⁻⁶) = 2500 V (A1).
- (c) 0.50
Q ∝ V: 2500/5000 = 0.50 (A1).
A 470 µF capacitor charged to 9.0 V is discharged through a 22 kΩ resistor.
- (a) Calculate the time constant of the circuit. [1]
- (b) Calculate the pd across the capacitor 5.0 s after the discharge starts. [2]
Show answer
- (a) 10.3 s
τ = RC = 22 × 10³ × 470 × 10⁻⁶ = 10.3 s (A1).
- (b) 5.55 V
V = V₀e^(−t/RC) (M1) = 9.0 × e^(−5.0/10.3) = 5.55 V (A1).
- (a) Define the time constant of a capacitor–resistor discharge circuit. [1]
- (b) Show that the product RC has the unit of seconds. [2]
- (c) State the percentage of the initial charge remaining after a time equal to three time constants. [1]
Show answer
- (a) The time taken for the charge (or pd or current) to fall to 1/e (about 37%) of its initial value; equal to RC.
B1: time for Q to fall to 1/e of initial value (accept τ = RC with symbols explained).
- (b) Ω × F = (V A⁻¹)(C V⁻¹) = C A⁻¹ = (A s) A⁻¹ = s.
M1: Ω = V A⁻¹ and F = C V⁻¹; A1: C = A s so units reduce to s.
- (c) 5.0%
e⁻³ = 0.0498 → 5.0% (A1).
A charged capacitor is discharged through a resistor.
- (a) Explain why the current in the resistor decreases with time. [2]
- (b) Explain why the charge decreases exponentially rather than linearly. [2]
- (c) State the effect on the discharge time of increasing the resistance. [1]
Show answer
- (a) As charge leaves the capacitor, the pd across it (V = Q/C) falls; the current I = V/R therefore falls.
B1: charge decreases so pd decreases; B1: I = V/R decreases.
- (b) The rate of loss of charge (current) is proportional to the charge remaining, so equal fractions are lost in equal times.
B1: rate of discharge ∝ Q (since I = Q/RC); B1: constant fraction lost in equal time intervals → exponential.
- (c) It increases (time constant RC larger).
B1: slower discharge, τ = RC larger.
A 1000 µF capacitor, charged to 6.0 V, is discharged through a 3.3 kΩ resistor.
- (a) Calculate the initial current. [1]
- (b) Calculate the time constant. [1]
- (c) Calculate the current 6.6 s after the discharge begins. [2]
Show answer
- (a) 1.82 mA
I₀ = 6.0/3300 = 0.00182 A (A1).
- (b) 3.3 s
RC = 3300 × 1000 × 10⁻⁶ = 3.3 s (A1).
- (c) 0.246 mA
6.6 s = 2τ (M1); I = I₀e⁻² = 0.246 mA (A1).
A capacitor of capacitance 330 µF discharges through a 15 kΩ resistor.
- (a) Calculate the time constant. [1]
- (b) Calculate the percentage of the initial charge remaining after 10 s. [2]
Show answer
- (a) 4.95 s
RC = 15 × 10³ × 330 × 10⁻⁶ = 4.95 s (A1).
- (b) 13.3%
Q/Q₀ = e^(−10/4.95) (M1) = 0.133 → 13.3% (A1).
A 100 µF capacitor is charged to 12.0 V and then discharged through a 47 kΩ resistor.
- (a) Calculate the initial discharge current. [1]
- (b) Calculate the charge on the capacitor after 3.0 s. [2]
- (c) Calculate the time taken for the pd to halve. [2]
Show answer
- (a) 0.255 mA
I = V/R = 12/47000 = 2.55 × 10⁻⁴ A (A1).
- (b) 6.34 × 10⁻⁴ C
Q₀ = CV = 1.2 × 10⁻³ C, τ = 4.70 s (M1); Q = Q₀e^(−3.0/4.70) = 6.34 × 10⁻⁴ C (A1).
- (c) 3.26 s
½ = e^(−t/RC) → t = RC ln 2 (M1) = 3.26 s (A1).
A capacitor discharges through a 50 kΩ resistor. The pd V across the capacitor is recorded at time t:t = 0 s, V = 10.00 Vt = 10 s, V = 6.70 Vt = 20 s, V = 4.49 Vt = 30 s, V = 3.01 Vt = 40 s, V = 2.02 V
- (a) Show that a graph of ln(V / V) against t should be a straight line, and state how the time constant can be found from it. (ln(V / V) means the natural log of the number of volts.) [2]
- (b) Calculate ln(V / V) for each reading. [1]
- (c) Determine the gradient of the line and hence the capacitance. [3]
Show answer
- (a) ln V = ln V₀ − t/RC; gradient = −1/RC.
M1: take logs of V = V₀e^(−t/RC); A1: straight line with gradient −1/RC (τ = −1/gradient).
- (b) ln(V / V): 2.303, 1.902, 1.502, 1.102, 0.703
B1: all correct to 3 d.p.
- (c) gradient = −0.040 s⁻¹; C = 500 µF
Gradient = (0.703 − 2.303)/40 = −0.040 s⁻¹ (M1 A1); C = −1/(gradient × R) = 5.0 × 10⁻⁴ F (A1).
A 2200 µF capacitor is charged to 16 V and then discharged through a resistor.
- (a) Calculate the initial energy stored. [1]
- (b) Show that after one time constant the energy stored is e⁻² of its initial value. [2]
- (c) Calculate the energy stored after one time constant. [1]
Show answer
- (a) 0.282 J
½ × 2200 × 10⁻⁶ × 16² = 0.282 J (A1).
- (b) V falls to V₀/e; W ∝ V², so W falls to W₀/e².
M1: V = V₀e⁻¹ after t = RC; A1: W ∝ V² → factor e⁻².
- (c) 0.0381 J
0.282 × e⁻² = 0.0381 J (A1).
A capacitor charged to 6.0 V discharges through a 1.2 kΩ resistor. The time constant of the circuit is 1.8 s.
- (a) Calculate the fraction of the initial charge remaining after 2.5 s. [2]
- (b) Calculate the current at t = 2.5 s. [2]
- (c) Calculate the capacitance. [1]
Show answer
- (a) 0.249
Q/Q₀ = e^(−t/τ) (M1) = e^(−2.5/1.8) = 0.249 (A1).
- (b) 1.25 mA
I₀ = 6.0/1200 = 5.0 mA (M1); I = 5.0 × 0.249 = 1.25 mA (A1).
- (c) 1500 µF
C = τ/R = 1.8/1200 = 0.00150 F (A1).
A student charges a capacitor to 8.0 V and lets it discharge through different resistors R. For each she times how long the pd takes to fall to 4.0 V (t½):R = 10 kΩ, t½ = 7.0 sR = 22 kΩ, t½ = 15.0 sR = 33 kΩ, t½ = 23.1 sR = 47 kΩ, t½ = 32.5 sR = 68 kΩ, t½ = 47.3 s
- (a) Show that t½ = (C ln 2) R. [1]
- (b) A graph of t½ against R is a straight line through the origin with gradient 0.694 s kΩ⁻¹. Determine C. [2]
- (c) The capacitor is labelled 1000 µF ± 20%. State whether the result is consistent with the label. [1]
- (d) Suggest why timing with a stopwatch is less satisfactory for the smallest resistor, and an improvement. [2]
Show answer
- (a) ½V₀ = V₀e^(−t½/RC) → t½ = RC ln 2.
B1: derivation shown.
- (b) 1000 µF
Gradient = C ln 2 (M1); C = 6.94 × 10⁻⁴ s Ω⁻¹/ln 2 = 0.00100 F (A1).
- (c) Yes, it lies within 800–1200 µF.
B1: comparison with range.
- (d) The half-time is short (about 7 s), so reaction time is a large fraction of it; use a data logger with a voltage sensor, or use a larger capacitor/resistor.
B1: short time → large % uncertainty from reaction time; B1: data logger / repeat and average / larger RC.
A capacitor of capacitance C discharges through a resistor of resistance R.
- (a) By applying Kirchhoff's second law to the circuit, show that dQ/dt = −Q/(RC), where Q is the charge on the capacitor. [2]
- (b) State the solution of this equation for a capacitor with initial charge Q₀. [1]
- (c) For C = 680 µF and R = 4.7 kΩ, calculate the time for the charge to fall to 10% of Q₀. [2]
Show answer
- (a) Q/C = IR with I = −dQ/dt (charge decreasing), so dQ/dt = −Q/RC.
M1: pd across C = pd across R: Q/C = IR; A1: I = −dQ/dt → result.
- (b) Q = Q₀e^(−t/RC)
B1.
- (c) 7.36 s
0.10 = e^(−t/RC) → t = RC ln 10 (M1) = 3.20 × 2.303 = 7.36 s (A1).
In a stairwell light timer, a 220 µF capacitor is charged to 6.0 V and then discharges through a resistor. The light switches off when the capacitor pd falls to 2.0 V. The delay required is 15 s.
- (a) Show that the required resistance is given by R = t/(C ln(V₀/V)). [2]
- (b) Calculate the resistance needed. [2]
- (c) Resistors are available in values of 47 kΩ, 56 kΩ, 68 kΩ and 82 kΩ. State which one gives a delay closest to 15 s and calculate that delay. [2]
Show answer
- (a) V = V₀e^(−t/RC) → ln(V₀/V) = t/RC → R = t/(C ln(V₀/V)).
M1: take natural logs; A1: rearranged correctly.
- (b) 62100 Ω
R = 15/(220 × 10⁻⁶ × ln 3) (M1) = 62100 Ω (A1).
- (c) 68 kΩ; 16.4 s
Closest to 62.1 kΩ is 56 kΩ or 68 kΩ: delays 13.5 s and 16.4 s (M1); 68 kΩ gives the closer delay (A1).
A 0.47 F capacitor charged to 5.0 V keeps the clock memory of a digital radio working during a power cut. The memory circuit behaves like a 10 kΩ resistor and works until the pd falls to 2.0 V.
- (a) Calculate the time constant. [1]
- (b) Calculate how long the memory is kept, in minutes. [3]
- (c) Calculate the fraction of the stored energy that has been used when the memory fails. [1]
Show answer
- (a) 4700 s
τ = RC = 10 × 10³ × 0.47 = 4700 s (A1).
- (b) 71.8 min
2.0 = 5.0e^(−t/4700) (M1); t = 4700 ln 2.5 (M1) = 4310 s = 71.8 min (A1).
- (c) 0.84
1 − (2.0/5.0)² = 0.84 (A1).
To find an unknown capacitance, a student charges the capacitor and lets it discharge through a 100 kΩ resistor, timing how long the voltmeter reading across the capacitor takes to halve. The time is 6.9 s.
- (a) Calculate the capacitance, assuming the voltmeter takes no current. [2]
- (b) The voltmeter actually has a resistance of 10 MΩ and is connected across the capacitor. Calculate the corrected capacitance. [2]
- (c) Comment on whether the voltmeter resistance matters in this experiment. [1]
Show answer
- (a) 99.5 µF
t½ = RC ln 2 (M1); C = 6.9/(1.00 × 10⁵ × ln 2) = 9.95 × 10⁻⁵ F (A1).
- (b) 101 µF
Discharge resistance = 100 kΩ ∥ 10 MΩ = 99.0 kΩ (M1); C = 6.9/(99000 × ln 2) = 1.01 × 10⁻⁴ F (A1).
- (c) Only slightly: about a 1% difference, small compared with typical capacitor tolerances.
B1: difference ≈ 1%, so negligible for most purposes (but would matter if R were comparable to the meter resistance).
A 470 µF capacitor charged to 12.0 V begins to discharge at t = 0 through a 10 kΩ resistor.
- (a) Calculate the pd across the capacitor at t = 3.0 s. [2]
- (b) At t = 3.0 s a switch opens so that a 20 kΩ resistor is added in series with the 10 kΩ resistor. Calculate the pd at t = 8.0 s. [3]
- (c) Calculate the charge that has flowed through the 10 kΩ resistor between t = 0 and t = 8.0 s. [1]
Show answer
- (a) 6.34 V
τ₁ = 10 × 10³ × 470 × 10⁻⁶ = 4.70 s (M1); V = 12e^(−3.0/4.70) = 6.34 V (A1).
- (b) 4.45 V
New R = 30 kΩ, τ₂ = 14.1 s (M1); start from 6.34 V (M1); V = 6.34e^(−5.0/14.1) = 4.45 V (A1).
- (c) 0.00355 C
ΔQ = C(12 − 4.45) = 0.00355 C (A1).
