ExaminerPrep › Question bank › IAL Physics › 2.6 Current, potential difference and resistance
Edexcel International A Level Physics · Unit 2 · Waves and Electricity
40 free, high-quality past-paper-style questions on Current, potential difference and resistance 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.
The current in a bicycle rear light is 0.25 A. The light is left on for 2.0 minutes.
How much charge passes through the light?
| Answer | Mark | Notes |
|---|---|---|
| ΔQ = IΔt = 0.25 × (2.0 × 60) = 0.25 × 120 = 30 C. | ✓ 1 | Answer B |
Answer: B
Exam tip: Change the time to seconds before using ΔQ = IΔt; one coulomb is one ampere second.
A battery transfers 2.7 kJ of energy to the components of a circuit while 450 C of charge flows through it.
What is the potential difference across the components?
| Answer | Mark | Notes |
|---|---|---|
| V = W/Q = 2700 ÷ 450 = 6.0 V. | ✓ 1 | Answer C |
Answer: C
Exam tip: A volt is a joule per coulomb, so divide the energy in joules by the charge in coulombs.
Which of the following is a statement of Ohm's law?
| Answer | Mark | Notes |
|---|---|---|
| A. Ohm's law is the special case in which I ∝ V, which holds for a metal only if its temperature (and other physical conditions) stay constant. Statement B is the definition of resistance, which applies to every component. | ✓ 1 | Answer A |
Answer: A
Exam tip: A full statement of Ohm's law needs three things: a metallic conductor, constant temperature, and current directly proportional to p.d.
A portable speaker is powered from a 5.0 V USB supply. The current in the speaker is 0.48 A, and the speaker plays music for 40 minutes.
Define potential difference.
Calculate the charge that flows through the speaker in 40 minutes.
Calculate the energy transferred to the speaker in this time.
Calculate the power of the speaker.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| energy transferred per unit charge (between two points) | ✓ 1 | |
| Part (b) | ||
| ΔQ = IΔt = 0.48 × 40 × 60 | ✓ 1 | |
| = 1.2 × 103 C (1152 C) | ✓ 1 | |
| Part (c) | ||
| W = VQ = 5.0 × 1152 = 5.8 × 103 J (5760 J) (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| P = VI = 5.0 × 0.48 = 2.4 W | ✓ 1 | |
Exam tip: A definition of p.d. must say “per unit charge” (or per coulomb). “The energy used by a component” on its own is not enough.
In a factory that makes electronic components, a worker touches a grounded metal bench and a static discharge occurs. A charge of 3.0 µC flows in 120 ns. e = 1.60 × 10−19 C.
Calculate the mean current during the discharge.
Calculate the number of electrons that flow.
Explain why the charge on any object must be a whole-number multiple of e.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = ΔQ/Δt = 3.0 × 10−6 ÷ 120 × 10−9 | ✓ 1 | |
| = 25 A | ✓ 1 | |
| Part (b) | ||
| 3.0 × 10−6 ÷ 1.60 × 10−19 = 1.9 × 1013 (1.875 × 1013) | ✓ 1 | |
| Part (c) | ||
| charge is carried by particles such as electrons, which each carry charge e, so charge is gained or lost only in whole numbers of electrons (charge is quantised) | ✓ 1 | |
Exam tip: Write prefixes as powers of ten before dividing: µ = 10−6, n = 10−9.
The graph shows how the total charge Q that has passed through a motor changes with time t. Note that the time axis is in minutes.
What is the current in the motor?
| Answer | Mark | Notes |
|---|---|---|
| Current = gradient of the Q–t graph = ΔQ/Δt. Using the end point: 60 C ÷ (4.0 × 60 s) = 60 ÷ 240 = 0.25 A. | ✓ 1 | Answer B |
Answer: B
Exam tip: The gradient of a charge–time graph is the current, but only in amperes if the time is in seconds. Convert minutes first.
The graph shows the current in a small fan motor during its first 10 s after being switched on and then off again.
What is the total charge that flows through the motor?
| Answer | Mark | Notes |
|---|---|---|
| Charge = area under the I–t graph = (0.60 × 3.0) + (0.20 × 5.0) = 1.8 + 1.0 = 2.8 C. | ✓ 1 | Answer A |
Answer: A
Exam tip: Split the area under a current–time graph into rectangles (and triangles) and add them. Each strip is current × its own time interval.
Two components X and Y are connected in series to a supply. The graph shows how the energy W transferred in each component depends on the charge Q that has flowed through it.
What is the potential difference across Y?
| Answer | Mark | Notes |
|---|---|---|
| V = W/Q, the gradient of the line. For Y: 90 J ÷ 10 C = 9.0 V. | ✓ 1 | Answer C |
Answer: C
Exam tip: On an energy–charge graph the gradient is energy per unit charge, which is the potential difference in volts.
The graph shows how the current I in each of three conductors P, Q and R depends on the potential difference V across it. The current is in milliamperes.
What is the resistance of Q?
| Answer | Mark | Notes |
|---|---|---|
| Q passes through (5.0 V, 200 mA). R = V/I = 5.0 ÷ 0.200 = 25 Ω. | ✓ 1 | Answer C |
Answer: C
Common error: Dividing current by p.d. on an I–V graph gives 1/R, not R. Also convert mA to A before dividing.
A student measures the current I in a carbon resistor for different values of the potential difference V across it, keeping the current small. The graph shows her results and her line of best fit.
Explain how the graph shows that the resistor obeys Ohm's law.
Use the graph to determine the resistance of the resistor.
The resistor is replaced by one of resistance 240 Ω. State the current in it when the p.d. across it is 6.0 V.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the best-fit line is straight | ✓ 1 | |
| and passes through the origin, so I is directly proportional to V | ✓ 1 | |
| Part (b) | ||
| uses a large triangle on the line, e.g. 50 mA at 6.0 V | ✓ 1 | |
| gradient = 0.050 ÷ 6.0 = 8.3 × 10−3 A V−1 (or R = V/I = 6.0 ÷ 0.050) | ✓ 1 | |
| R = 1/gradient = 120 Ω (allow 115–125 Ω) | ✓ 1 | |
| Part (c) | ||
| 25 mA (0.025 A) | ✓ 1 | |
Common error: Taking the gradient of an I–V graph as the resistance. The gradient is I/V = 1/R, and the current must be in amperes.
A small heating resistor R is used to keep a fish-tank sensor warm. The circuit diagram shows the meter readings when the switch is closed.
Calculate the resistance of R.
The switch is closed for 4.0 minutes. Calculate the charge that flows through R.
Calculate the energy transferred in R in this time.
Calculate the power of R.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| R = V/I = 8.4 ÷ 0.75 = 11 Ω (11.2 Ω) | ✓ 1 | |
| Part (b) | ||
| ΔQ = IΔt = 0.75 × 240 | ✓ 1 | |
| = 1.8 × 102 C (180 C) | ✓ 1 | |
| Part (c) | ||
| W = VIt = 8.4 × 0.75 × 240 (or W = VQ) | ✓ 1 | |
| = 1.5 × 103 J (1512 J) | ✓ 1 | |
| Part (d) | ||
| P = VI = 8.4 × 0.75 = 6.3 W | ✓ 1 | |
Exam tip: Read each meter for the component it measures: the ammeter in series gives the current through R, and the voltmeter in parallel gives the p.d. across R.
A copper cable of cross-sectional area 1.5 mm2 carries a current of 4.0 A. The number of free electrons per unit volume in copper is 8.5 × 1028 m−3. e = 1.60 × 10−19 C.
What is the drift velocity of the electrons?
| Answer | Mark | Notes |
|---|---|---|
| v = I/nqA with A = 1.5 × 10−6 m2: 4.0 ÷ (8.5 × 1028 × 1.60 × 10−19 × 1.5 × 10−6) = 2.0 × 10−4 m s−1 (1.96 × 10−4 m s−1). | ✓ 1 | Answer C |
Answer: C
Common error: Converting mm2 to m2 with 10−3. A square millimetre is (10−3 m)2 = 10−6 m2.
The current in a sensor circuit is 3.2 mA. e = 1.60 × 10−19 C.
How many electrons pass a point in the circuit each second?
| Answer | Mark | Notes |
|---|---|---|
| Charge per second = 3.2 × 10−3 C. Number of electrons = 3.2 × 10−3 ÷ 1.60 × 10−19 = 2.0 × 1016. | ✓ 1 | Answer D |
Answer: D
Exam tip: A current in amperes is the number of coulombs per second. Divide by e to count the electrons.
In an X-ray tube, an electron is accelerated from rest through a potential difference of 5.0 kV. e = 1.60 × 10−19 C.
How much energy does the electron gain?
| Answer | Mark | Notes |
|---|---|---|
| W = QV = 1.60 × 10−19 × 5.0 × 103 = 8.0 × 10−16 J. | ✓ 1 | Answer A |
Answer: A
Exam tip: Rearrange V = W/Q to W = QV. For a single electron the charge is e.
The current–potential difference graph for a component is a curve. At a p.d. of 6.0 V the current is 0.40 A, and the gradient of the tangent to the curve at this point is 0.025 A V−1.
What is the resistance of the component at 6.0 V?
| Answer | Mark | Notes |
|---|---|---|
| Resistance is defined by R = V/I at the operating point: 6.0 ÷ 0.40 = 15 Ω. The tangent gradient does not give the resistance for a curved graph. | ✓ 1 | Answer C |
Answer: C
Common error: Using 1/gradient of the tangent. For a curved I–V graph, read V and I at the point and divide.
An electric oven element has a power of 2.0 kW when connected to a 230 V supply.
What is the resistance of the element when it is operating?
| Answer | Mark | Notes |
|---|---|---|
| P = V2/R, so R = V2/P = 2302 ÷ 2000 = 26 Ω (26.45 Ω). | ✓ 1 | Answer D |
Answer: D
Exam tip: When you know the power and the supply p.d., R = V2/P gets the resistance in one step.
A student investigates whether a length of nichrome wire obeys Ohm's law. She uses a variable d.c. supply, an ammeter in series with the wire and a voltmeter across it. Her results are:
p.d. / V: 0.40, 0.80, 1.20, 1.60, 2.00
current / A: 0.054, 0.107, 0.162, 0.214, 0.270
State Ohm's law.
Use the results to show that the wire obeys Ohm's law.
Determine the resistance of the wire.
Explain why the student keeps the currents small and switches off between readings.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| for a metallic conductor, the current is directly proportional to the potential difference | ✓ 1 | |
| provided the temperature (physical conditions) remains constant | ✓ 1 | |
| Part (b) | ||
| calculates V/I for at least three rows: 7.41, 7.48, 7.41, 7.48, 7.41 Ω | ✓ 1 | |
| the ratio is constant (to within 1%), so I ∝ V / a graph of I against V is a straight line through the origin | ✓ 1 | |
| Part (c) | ||
| mean of V/I = 7.4 Ω | ✓ 1 | |
| Part (d) | ||
| a large current (or a long time with current on) heats the wire | ✓ 1 | |
| its temperature and so its resistance would rise, and the condition for Ohm's law would not be met | ✓ 1 | |
Model sketch
Exam tip: To show proportionality from a table, calculate the ratio for every row and comment on whether it is constant — or plot the graph and check it passes through the origin.
A filament lamp for a model railway is marked “12 V, 24 W”. When the p.d. across the lamp is only 2.0 V, the current in it is 0.75 A.
Calculate the current in the lamp when it operates at its rated values.
Calculate the resistance of the lamp at 12 V and at 2.0 V.
Explain, using your answers, whether the lamp obeys Ohm's law.
Calculate the power dissipated in the lamp at 2.0 V.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = P/V = 24 ÷ 12 = 2.0 A | ✓ 1 | |
| Part (b) | ||
| at 12 V: 12 ÷ 2.0 = 6.0 Ω | ✓ 1 | |
| at 2.0 V: 2.0 ÷ 0.75 = 2.7 Ω (2.67 Ω) | ✓ 1 | |
| Part (c) | ||
| the resistance is not constant / V/I changes, so I is not proportional to V: it does not obey Ohm's law | ✓ 1 | |
| at the higher current the filament is much hotter, so the temperature is not constant | ✓ 1 | |
| Part (d) | ||
| P = VI = 2.0 × 0.75 = 1.5 W | ✓ 1 | |
Common error: Assuming the resistance at the rated values also applies at a lower p.d. A filament's resistance depends strongly on its temperature, so calculate V/I separately at each operating point.
An electric shower has a power of 9.5 kW when connected to a 230 V supply. A family uses the shower for a total of 36 minutes each day.
Calculate the current in the shower.
Calculate the resistance of the heating element.
Calculate the energy transferred by the shower in one day, in kWh.
Convert this energy to joules.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = P/V = 9500 ÷ 230 = 41 A (41.3 A) | ✓ 1 | |
| Part (b) | ||
| R = V2/P (or V/I) = 2302 ÷ 9500 | ✓ 1 | |
| = 5.6 Ω (5.57 Ω) | ✓ 1 | |
| Part (c) | ||
| time = 36 ÷ 60 = 0.60 h | ✓ 1 | |
| E = 9.5 × 0.60 = 5.7 kWh | ✓ 1 | |
| Part (d) | ||
| 5.7 × 3.6 × 106 = 2.1 × 107 J (2.05 × 107 J) | ✓ 1 | |
Exam tip: kW × hours gives kWh with no conversion. Only convert to joules (× 3.6 × 106) when the question asks.
The copper cable from a car battery to its starter motor has a cross-sectional area of 25 mm2 and a length of 0.80 m. While the engine is being started the current in it is 150 A. Copper has 8.5 × 1028 free electrons per m3. e = 1.60 × 10−19 C.
Show that the drift velocity of the electrons in the cable is about 4 × 10−4 m s−1.
Calculate the time an electron would take to drift the full length of the cable.
The starter motor begins to turn as soon as the key is turned. Explain why.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| v = I/nqA with A = 25 × 10−6 m2 | ✓ 1 | |
| = 150 ÷ (8.5 × 1028 × 1.60 × 10−19 × 25 × 10−6) = 4.4 × 10−4 m s−1 | ✓ 1 | |
| Part (b) | ||
| t = 0.80 ÷ 4.41 × 10−4 | ✓ 1 | |
| = 1.8 × 103 s (1810 s) | ✓ 1 | |
| Part (c) | ||
| the electric field (p.d.) is established along the whole circuit almost instantly | ✓ 1 | |
| free electrons already present throughout the cable and the motor all start drifting at once, so no electron needs to travel from the battery to the motor | ✓ 1 | |
Common error: Picturing electrons racing from the battery to the motor. The current starts everywhere at once because the conductors are already full of free electrons.
In a practical lesson a student measures the energy supplied to a 12 V immersion heater placed in a block of aluminium. The voltmeter reads 11.6 V and the ammeter, which has a resolution of 0.1 A, reads 3.9 A. The heater is switched on for 300 s, timed with a stopwatch.
Calculate the energy supplied to the heater.
Calculate the resistance of the heater during the experiment.
Calculate the percentage uncertainty in the ammeter reading.
Calculate the charge that passed through the heater.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| W = VIt = 11.6 × 3.9 × 300 | ✓ 1 | |
| = 1.4 × 104 J (13 572 J) | ✓ 1 | |
| Part (b) | ||
| R = V/I = 11.6 ÷ 3.9 = 3.0 Ω (2.97 Ω) | ✓ 1 | |
| Part (c) | ||
| uncertainty = half the resolution = ± 0.05 A | ✓ 1 | |
| 0.05 ÷ 3.9 × 100 = 1.3% | ✓ 1 | |
| (allow uncertainty = resolution = ± 0.1 A, giving 2.6%) | — | |
| Part (d) | ||
| Q = It = 3.9 × 300 = 1.2 × 103 C (1170 C) | ✓ 1 | |
Exam tip: For a single digital reading, take the uncertainty as half the resolution unless the reading fluctuates; then use half the range of the fluctuation.
Each sketch graph shows data for a different conductor.
Which graph shows a conductor that obeys Ohm's law?
| Answer | Mark | Notes |
|---|---|---|
| A. A constant resistance at every current means V/I is constant, so I ∝ V. In B the line does not pass through the origin, so I is not proportional to V; in C the ratio V/I changes; in D P ∝ I, which would need a fixed p.d. — for a constant resistance P = I2R, so P ∝ I2. | ✓ 1 | Answer A |
Answer: A
Exam tip: Ohm's law means I ∝ V: on an I–V graph that is a straight line through the origin, and on a resistance graph it is a horizontal line.
The graph shows how the power P dissipated in a resistor varies with the current I in it. The temperature of the resistor stays constant.
What is the resistance of the resistor?
| Answer | Mark | Notes |
|---|---|---|
| At X, P = 18 W and I = 1.5 A. P = I2R, so R = 18 ÷ 1.52 = 18 ÷ 2.25 = 8.0 Ω. | ✓ 1 | Answer B |
Answer: B
Common error: Taking the gradient of the curve at a point as the resistance. The gradient of a P–I graph is 2IR, not R; use R = P/I2 at one point.
A householder switches off everything in a flat except an electric heater and reads the energy meter twice, as shown.
What is the mean power of the heater?
| Answer | Mark | Notes |
|---|---|---|
| Energy = 4518.3 − 4512.6 = 5.7 kWh. Time = 18:40 to 21:10 = 2 h 30 min = 2.5 h. Mean power = 5.7 ÷ 2.5 = 2.3 kW (2.28 kW). | ✓ 1 | Answer A |
Answer: A
Common error: Writing 2 h 30 min as 2.3 h. Thirty minutes is 0.5 h, so the time is 2.5 h.
The chart shows the number density n of free charge carriers in three materials. Rods of X and of Y with the same cross-sectional area are connected in series, so they carry the same current. The charge on each carrier has magnitude e in both.
What is the value of drift velocity in Y ÷ drift velocity in X?
| Answer | Mark | Notes |
|---|---|---|
| From I = nqvA with I, q and A the same, v ∝ 1/n. vY/vX = nX/nY = 6.0 × 1028 ÷ 1.5 × 1022 = 4.0 × 106. | ✓ 1 | Answer D |
Answer: D
Exam tip: Fewer charge carriers must drift faster to carry the same current. Check that your ratio makes the semiconductor's carriers the faster ones.
A mobile phone battery is charged from flat. The graph shows how the charging current I changes with time t. e = 1.60 × 10−19 C.
State what is represented by the area under a current–time graph.
Use the graph to calculate the total charge that flows into the battery.
Calculate the number of electrons that this charge represents.
Assume the p.d. across the battery is 4.0 V throughout the charging. Calculate the energy transferred to the battery.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the charge that has flowed | ✓ 1 | |
| Part (b) | ||
| times converted to seconds: 45 min = 2700 s and 15 min = 900 s | ✓ 1 | |
| area = 2.0 × 2700 + ½ × 2.0 × 900 = 5400 + 900 | ✓ 1 | |
| = 6.3 × 103 C (6300 C) | ✓ 1 | |
| Part (c) | ||
| 6300 ÷ 1.60 × 10−19 = 3.9 × 1022 (3.94 × 1022) (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| W = QV = 6300 × 4.0 = 2.5 × 104 J (25 200 J) (allow error carried forward) | ✓ 1 | |
Exam tip: Charge from a current–time graph is the area, so the time must be in seconds to give coulombs. Split the shape into a rectangle and a triangle.
The graph shows the current–potential difference characteristics of a small torch lamp and of a fixed resistor.
Calculate the resistance of the lamp at point P.
State the p.d. at which the lamp and the resistor have the same resistance, and give the value of this resistance.
The p.d. across each component is 1.0 V. Calculate the ratio power dissipated in the lamp ÷ power dissipated in the resistor.
Explain why the lamp does not obey Ohm's law.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| R = V/I = 1.0 ÷ 0.15 = 6.7 Ω (6.67 Ω) | ✓ 1 | |
| Part (b) | ||
| 3.0 V, where the graphs cross at Q | ✓ 1 | |
| R = 3.0 ÷ 0.30 = 10 Ω | ✓ 1 | |
| Part (c) | ||
| current in resistor at 1.0 V = 0.10 A (from graph or 1.0 ÷ 10) | ✓ 1 | |
| ratio = (1.0 × 0.15) ÷ (1.0 × 0.10) = 1.5 | ✓ 1 | |
| Part (d) | ||
| the graph is a curve / V/I changes (from 6.7 Ω to 10 Ω), so I is not proportional to V | ✓ 1 | |
| the filament gets hotter as the current increases, so its temperature is not constant | ✓ 1 | |
Exam tip: Where two I–V graphs cross, the two components have the same V and I, so they have the same resistance at that point.
An electric car is charged overnight from a 230 V supply. The graph shows how the power drawn by the charger varies with time t after the car is plugged in.
Calculate the current in the supply cable during the first 5.0 hours.
Use the graph to determine the total energy supplied, in kWh.
Convert this energy to joules.
Assume that all this energy is transferred to the car's battery at a p.d. of 400 V. Calculate the charge that flows into the battery.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = P/V = 6000 ÷ 230 = 26 A (26.1 A) | ✓ 1 | |
| Part (b) | ||
| energy = area under the P–t graph | ✓ 1 | |
| 6.0 × 5.0 + ½ × 6.0 × 1.0 = 30 + 3.0 | ✓ 1 | |
| = 33 kWh | ✓ 1 | |
| Part (c) | ||
| 33 × 3.6 × 106 = 1.2 × 108 J (1.19 × 108 J) | ✓ 1 | |
| Part (d) | ||
| Q = W/V = 1.19 × 108 ÷ 400 | ✓ 1 | |
| = 3.0 × 105 C (2.97 × 105 C) | ✓ 1 | |
Exam tip: With power in kW and time in hours, the area under a power–time graph is already in kWh. 1 kWh = 3.6 × 106 J.
The diagram shows a copper track on a printed circuit board. The track has a uniform thickness and carries a current of 0.50 A. For copper, n = 8.5 × 1028 m−3. e = 1.60 × 10−19 C.
Calculate the cross-sectional area of the wide part of the track.
Calculate the drift velocity of the free electrons in the wide part of the track.
State and explain how the drift velocity in the narrow part compares with that in the wide part.
Calculate the time taken for a free electron to drift along the narrow part of the track.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 1.20 × 10−3 × 35 × 10−6 = 4.2 × 10−8 m2 | ✓ 1 | |
| Part (b) | ||
| v = I/nqA = 0.50 ÷ (8.5 × 1028 × 1.60 × 10−19 × 4.2 × 10−8) | ✓ 1 | |
| = 8.8 × 10−4 m s−1 (8.75 × 10−4 m s−1) | ✓ 1 | |
| Part (c) | ||
| the current, n and q are the same, so v ∝ 1/A; the area is one quarter, since the width is one quarter and the thickness is the same | ✓ 1 | |
| so the drift velocity is 4 times larger: 3.5 × 10−3 m s−1 | ✓ 1 | |
| Part (d) | ||
| t = 6.0 × 10−3 ÷ 3.50 × 10−3 = 1.7 s (1.71 s) (allow error carried forward) | ✓ 1 | |
Common error: Thinking the electrons slow down in the narrow part because it is harder to get through. The same current passes through a smaller area, so the carriers must drift faster.
The diagram shows the principle of an ion engine used to move a space probe. Singly charged xenon ions start almost at rest at the first grid and are accelerated between the grids. Mass of a xenon ion = 2.18 × 10−25 kg. e = 1.60 × 10−19 C.
Calculate the energy gained by one ion as it moves between the grids.
Calculate the number of ions leaving the engine each second.
Calculate the power transferred to the ion beam.
Calculate the speed of the ions as they leave the second grid.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| W = QV = 1.60 × 10−19 × 1200 = 1.9 × 10−16 J (1.92 × 10−16 J) | ✓ 1 | |
| Part (b) | ||
| 1.5 ÷ 1.60 × 10−19 = 9.4 × 1018 s−1 (9.38 × 1018 s−1) | ✓ 1 | |
| Part (c) | ||
| P = VI = 1200 × 1.5 (or energy per ion × ions per second) | ✓ 1 | |
| = 1.8 × 103 W | ✓ 1 | |
| Part (d) | ||
| ½mv2 = 1.92 × 10−16 J | ✓ 1 | |
| v = √(2 × 1.92 × 10−16 ÷ 2.18 × 10−25) = 4.2 × 104 m s−1 | ✓ 1 | |
Exam tip: A p.d. is energy per unit charge, so each ion of charge e gains eV of energy. The p.d. here is the difference between the two grid potentials.
The current in a resistor of constant resistance is increased by 50%.
By what percentage does the power dissipated in the resistor increase?
| Answer | Mark | Notes |
|---|---|---|
| P = I2R with R constant, so the power is multiplied by 1.52 = 2.25. The increase is 2.25 − 1 = 1.25 = 125%. | ✓ 1 | Answer B |
Answer: B
Exam tip: Find the multiplying factor first, then subtract 1 to turn it into a percentage increase.
Two wires X and Y made of the same metal are connected in series in a circuit. The diameter of Y is half the diameter of X.
What is the ratio drift velocity of electrons in Y ÷ drift velocity of electrons in X?
| Answer | Mark | Notes |
|---|---|---|
| The current is the same in both (series), and n and q are the same (same metal). From I = nqvA, v ∝ 1/A. Halving the diameter makes the area one quarter, so v is 4.0 times larger in Y. | ✓ 1 | Answer D |
Answer: D
Common error: Using the diameter ratio instead of the area ratio. Area depends on diameter squared.
A garden pond pump of power 750 W is used on a 230 V supply at the end of a long extension cable. The two copper conductors in the cable have a total resistance of 0.80 Ω. Assume the p.d. across any appliance connected at the end of the cable stays at 230 V.
Calculate the current in the cable.
Calculate the rate at which energy is dissipated in the cable.
A student calculates the power dissipated in the cable as 2302 ÷ 0.80. Explain why this is wrong.
The pump is replaced by a 3.0 kW pressure washer on the same cable. Calculate the factor by which the power dissipated in the cable increases.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = P/V = 750 ÷ 230 = 3.3 A (3.26 A) | ✓ 1 | |
| Part (b) | ||
| P = I2R = 3.262 × 0.80 | ✓ 1 | |
| = 8.5 W (8.51 W) | ✓ 1 | |
| Part (c) | ||
| in P = V2/R, V must be the p.d. across the cable itself | ✓ 1 | |
| the p.d. across the cable is only IR = 2.6 V, not the 230 V supply p.d. | ✓ 1 | |
| Part (d) | ||
| the current increases by a factor 3000 ÷ 750 = 4 | ✓ 1 | |
| the power in the cable is ∝ I2, so it increases by a factor of 16 | ✓ 1 | |
Exam tip: In P = I2R and P = V2/R, every quantity must belong to the same component. For a cable, use the current through it and its own resistance.
Copper, silicon and polythene have resistivities at room temperature of about 10−8 Ω m, 103 Ω m and 1016 Ω m respectively. They are classified as a conductor, a semiconductor and an insulator.
State what is meant by the number density of charge carriers in a material.
Use the equation I = nqvA to explain why the resistivities of these three materials are so different. Your answer should refer to the drift velocity needed to carry a given current. (Extended response, indicative content.)
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the number of free (mobile) charge carriers per unit volume | ✓ 1 | |
| Part (b) | ||
| in copper (a metal) n is very large, about one or more free electrons per atom | ✓ 1 | |
| in silicon (a semiconductor) n is many orders of magnitude smaller at room temperature | ✓ 1 | |
| in polythene (an insulator) almost all electrons are bound to atoms, so n is negligible | ✓ 1 | |
| for the same current in a sample of the same size, a smaller n requires a much larger drift velocity v (from I = nqvA) | ✓ 1 | |
| which needs a much larger p.d., so the resistance and resistivity are much larger when n is smaller | ✓ 1 | |
| the resistivities span about 24 orders of magnitude because n differs by a similarly enormous factor between the materials | ✓ 1 | |
Exam tip: In a six-mark explanation, link each step: how n differs between the materials, what I = nqvA then requires, and what that means for resistance. Separate, connected points are what earn the marks.
A student is given data on a thin rectangular slice of semiconductor used in a magnetic field sensor. The slice is 5.0 mm wide and 0.20 mm thick, and the current along it is 8.0 mA. The number density of charge carriers is 1.2 × 1022 m−3 and each carrier has a charge of magnitude 1.60 × 10−19 C. For copper, n = 8.5 × 1028 m−3.
Calculate the number of charge carriers passing through a cross-section of the slice each second.
Calculate the drift velocity of the charge carriers in the slice.
A copper slice of the same dimensions carries the same current. Calculate the ratio drift velocity in the semiconductor ÷ drift velocity in copper.
The temperature of the slice rises and its number density of charge carriers increases. The current is kept at 8.0 mA. State and explain the effect on the drift velocity.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 8.0 × 10−3 ÷ 1.60 × 10−19 = 5.0 × 1016 per second | ✓ 1 | |
| Part (b) | ||
| A = 5.0 × 10−3 × 0.20 × 10−3 = 1.0 × 10−6 m2 | ✓ 1 | |
| v = I/nqA = 8.0 × 10−3 ÷ (1.2 × 1022 × 1.60 × 10−19 × 1.0 × 10−6) | ✓ 1 | |
| = 4.2 m s−1 (4.17 m s−1) | ✓ 1 | |
| Part (c) | ||
| ratio = ncopper ÷ nsemiconductor = 8.5 × 1028 ÷ 1.2 × 1022 = 7.1 × 106 | ✓ 1 | |
| Part (d) | ||
| the drift velocity decreases | ✓ 1 | |
| I, q and A are unchanged, so v ∝ 1/n | ✓ 1 | |
Exam tip: When two conductors carry the same current through the same cross-section, the ratio of drift velocities is just the inverse ratio of the number densities — no need to recalculate each one.
The diagram shows part of a beam of identical ions, each with a charge of +2e, travelling at a speed of 2.0 × 105 m s−1. The ions are evenly spaced as shown. e = 1.60 × 10−19 C.
What is the current in the beam?
| Answer | Mark | Notes |
|---|---|---|
| Number of ions passing P per second = speed ÷ spacing = 2.0 × 105 ÷ 4.0 × 10−3 = 5.0 × 107 s−1. I = ΔQ/Δt = 5.0 × 107 × 2 × 1.60 × 10−19 = 1.6 × 10−11 A. | ✓ 1 | Answer D |
Answer: D
Exam tip: For a stream of charged particles, current = (number passing a point per second) × (charge on each particle). Count the charge as 2e for a 2+ ion.
The graph shows how the drift velocity v of the free electrons in a copper wire of circular cross-section depends on the current I. For copper, n = 8.5 × 1028 m−3. e = 1.60 × 10−19 C.
What is the diameter of the wire?
| Answer | Mark | Notes |
|---|---|---|
| At 6.0 A, v = 0.25 mm s−1 = 2.5 × 10−4 m s−1. A = I/nqv = 6.0 ÷ (8.5 × 1028 × 1.60 × 10−19 × 2.5 × 10−4) = 1.76 × 10−6 m2. d = √(4A/π) = 1.50 × 10−3 m = 1.5 mm. | ✓ 1 | Answer C |
Answer: C
Exam tip: After finding the area from I = nqvA, use A = πd2/4 to get the diameter. Do not stop at the area.
The graph shows how the number density n of free charge carriers in a sample of a semiconductor changes with its temperature θ. Each carrier has a charge of magnitude 1.60 × 10−19 C.
Use the graph to find the factor by which n increases between 20 °C and 50 °C.
At 20 °C the sample, of cross-sectional area 2.0 mm2, carries a current of 0.50 mA. Calculate the drift velocity of the charge carriers.
The sample is warmed to 50 °C and the current in it is kept at 0.50 mA. Calculate the drift velocity of the charge carriers at 50 °C.
A copper rod of the same cross-sectional area also carries a current of 0.50 mA. For copper, n = 8.5 × 1028 m−3. Calculate the drift velocity of the free electrons in the copper, and use I = nqvA to explain why copper has a far smaller resistivity than the semiconductor.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 8.0 × 1019 ÷ 1.0 × 1019 = 8.0 | ✓ 1 | |
| Part (b) | ||
| v = I/nqA = 0.50 × 10−3 ÷ (1.0 × 1019 × 1.60 × 10−19 × 2.0 × 10−6) | ✓ 1 | |
| = 1.6 × 102 m s−1 (156 m s−1) | ✓ 1 | |
| Part (c) | ||
| I, q and A are unchanged, so v ∝ 1/n: v = 156 ÷ 8.0 (or I/nqA with n = 8.0 × 1019 m−3) | ✓ 1 | |
| = 2.0 × 101 m s−1 (19.5 m s−1) (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| v = 0.50 × 10−3 ÷ (8.5 × 1028 × 1.60 × 10−19 × 2.0 × 10−6) = 1.8 × 10−8 m s−1 | ✓ 1 | |
| copper has about 1010 times more free charge carriers per unit volume | ✓ 1 | |
| so a much larger current flows for the same p.d. (the same current needs only a tiny drift velocity), so its resistance, and resistivity, is far smaller | ✓ 1 | |
Exam tip: With I, q and A fixed, v ∝ 1/n. A material with very few free carriers per unit volume needs a large drift velocity to carry even a small current, which is why its resistivity is high.
A student measures the power P dissipated in a resistor for several values of the current I in it, and plots P against I2. The graph shows the results.
Explain why a graph of P against I2 for a resistor of constant resistance is a straight line through the origin.
Use the gradient of the line to determine the resistance.
Calculate the p.d. across the resistor when I2 = 0.40 A2.
Point Z lies above the line. Calculate the resistance at Z and suggest why it differs from your answer to (b).
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| P = I2R with R constant, so P ∝ I2 (gradient = R) | ✓ 1 | |
| Part (b) | ||
| gradient from a large triangle, e.g. 6.0 W ÷ 0.40 A2 | ✓ 1 | |
| R = 15 Ω (allow 14.5–15.5 Ω) | ✓ 1 | |
| Part (c) | ||
| I = √0.40 = 0.632 A | ✓ 1 | |
| V = IR = 0.632 × 15 (or P/I = 6.0 ÷ 0.632) = 9.5 V (9.49 V) | ✓ 1 | |
| Part (d) | ||
| R = P/I2 = 7.8 ÷ 0.48 = 16 Ω (16.25 Ω) | ✓ 1 | |
| the resistance has increased | ✓ 1 | |
| at the largest current the resistor heats up, so its temperature rises and is no longer constant | ✓ 1 | |
Exam tip: Choosing what to plot so the graph is a straight line through the origin lets the gradient give the quantity you want: here P = R × I2, so the gradient is R.
A filament lamp is connected to a 12 V supply of negligible internal resistance and switched on. The graph shows how the current I in the lamp changes during the first 200 ms. The current then stays at 0.50 A.
Calculate the resistance of the filament at the instant of switching on and when the current is steady.
Explain why the current falls and then becomes steady.
Estimate the charge that flows through the lamp in the first 200 ms.
Calculate the energy transferred to the lamp in the first 200 ms.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| at switch-on: 12 ÷ 4.0 = 3.0 Ω | ✓ 1 | |
| steady: 12 ÷ 0.50 = 24 Ω | ✓ 1 | |
| Part (b) | ||
| the current heats the filament, so its temperature rises | ✓ 1 | |
| the resistance of the metal filament increases with temperature, so I = V/R falls | ✓ 1 | |
| the current becomes steady when the filament loses energy as fast as it is supplied, so its temperature and resistance stop changing | ✓ 1 | |
| Part (c) | ||
| charge = area under the graph, found by counting squares (one small square = 0.25 A × 10 ms = 2.5 × 10−3 C) | ✓ 1 | |
| 0.24 C (allow 0.22–0.26 C) | ✓ 1 | |
| Part (d) | ||
| W = VQ = 12 × 0.24 = 2.9 J (allow 2.6–3.1 J, error carried forward) | ✓ 1 | |
Common error: Multiplying the starting current by the time to find the charge. The current is changing, so the charge must come from the area under the curve.
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