ExaminerPrep › Question bank › IAL Physics › 2.8 Resistivity and the potential divider
Edexcel International A Level Physics · Unit 2 · Waves and Electricity
40 free, high-quality past-paper-style questions on Resistivity and the potential divider 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 the unit of resistivity?
| Answer | Mark | Notes |
|---|---|---|
| Rearranging R = ρl/A gives ρ = RA/l, with unit Ω × m2 ÷ m = Ω m. | ✓ 1 | Answer C |
Answer: C
Exam tip: Get the unit of any constant by making it the subject of its defining equation and substituting units. Ω m−1 is the unit of the gradient of an R–l graph, not of resistivity.
In an experiment to determine the resistivity ρ of the metal of a uniform wire of cross-sectional area A, a student plots the resistance R of different lengths l of the wire on the y-axis against l on the x-axis.
What is the gradient of the graph?
| Answer | Mark | Notes |
|---|---|---|
| R = (ρ/A) × l has the form y = mx, so the gradient is ρ/A and ρ = gradient × A. | ✓ 1 | Answer B |
Answer: B
Exam tip: Write the equation in the form y = mx + c with the plotted quantities as y and x; whatever multiplies x is the gradient.
A 3.0 kΩ resistor and a 1.0 kΩ resistor are connected in series across a 12 V supply of negligible internal resistance.
What is the p.d. across the 1.0 kΩ resistor?
| Answer | Mark | Notes |
|---|---|---|
| The same current flows through both, so the p.d. is shared in the ratio of the resistances: V = 1.0/(3.0 + 1.0) × 12 = 3.0 V. | ✓ 1 | Answer A |
Answer: A
Exam tip: Check that the p.d.s add up to the supply: 3.0 V + 9.0 V = 12 V.
A copper track on a printed circuit board is 12 cm long and 0.50 mm wide. The copper is 35 µm thick. The resistivity of copper is 1.7 × 10−8 Ω m.
State the unit of resistivity.
Calculate the cross-sectional area of the track.
Calculate the resistance of the track.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Ω m / ohm metre | ✓ 1 | |
| Part (b) | ||
| 0.50 × 10−3 × 35 × 10−6 = 1.75 × 10−8 m2 | ✓ 1 | |
| Part (c) | ||
| R = ρl/A = 1.7 × 10−8 × 0.12 ÷ 1.75 × 10−8 | ✓ 1 | |
| = 0.12 Ω (0.117 Ω) | ✓ 1 | |
Common error: Leaving the length in cm or the thickness in µm. Convert every dimension to metres before substituting into R = ρl/A.
A student is planning an experiment to determine the resistivity of the metal of a thin wire.
Explain why she should measure the diameter of the wire at several points along its length and in perpendicular directions.
A single micrometer reading of the diameter is 0.28 mm. The resolution of the micrometer is 0.01 mm. Calculate the percentage uncertainty in this reading.
Explain why the current in the wire should be kept small, with the circuit switched off between readings.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the wire may not be uniform / perfectly circular, so a mean of several readings gives a better (more accurate) value of the diameter | ✓ 1 | |
| Part (b) | ||
| uncertainty = half the resolution = 0.005 mm; 0.005 ÷ 0.28 × 100 = 1.8% (allow 3.6% if the full resolution, 0.01 mm, is used as the uncertainty) | ✓ 1 | |
| Part (c) | ||
| a large current heats the wire / raises its temperature | ✓ 1 | |
| the resistivity (and so the resistance) of a metal increases with temperature, so the measured values of R would be too high and would change during the experiment | ✓ 1 | |
Exam tip: For a single reading, the uncertainty is half the smallest scale division of the instrument; for repeated readings, half the range.
A potential divider is made from a 6.8 kΩ resistor R1 and a 3.3 kΩ resistor R2 connected in series across a 12 V supply of negligible internal resistance. The output p.d. Vout is taken across R2.
Show that Vout = R2Vin/(R1 + R2), where Vin is the supply p.d.
Calculate Vout.
Calculate the p.d. across R1.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the same current flows through both resistors: I = Vin/(R1 + R2) | ✓ 1 | |
| Vout = IR2 = R2Vin/(R1 + R2) | ✓ 1 | |
| Part (b) | ||
| 3.3 ÷ 10.1 × 12 = 3.9 V (3.92 V) | ✓ 1 | |
| Part (c) | ||
| 12 − 3.92 = 8.1 V (allow 6.8 ÷ 10.1 × 12) | ✓ 1 | |
Exam tip: In a “show that” derivation, state the physics (same current in series) before writing the equations — the algebra alone is not an explanation.
The figure shows four wires, W, X, Y and Z, all made of the same metal at the same temperature. Their lengths and diameters are given on the figure.
Which wire has the greatest resistance?
| Answer | Mark | Notes |
|---|---|---|
| R = ρl/A and A ∝ d2, so for one metal R ∝ l/d2. In units of m mm−2: W 1.00/0.25 = 4.0; X 2.00/0.1225 = 16; Y 1.40/0.040 = 35; Z 0.60/0.0144 = 42. Wire Z has the greatest resistance even though it is the shortest. | ✓ 1 | Answer D |
Answer: D
Exam tip: Compare l/d2 for each wire. Halving the diameter makes the resistance four times larger, which can outweigh a large difference in length.
A cell is connected across a uniform resistance wire PQ as shown. End Q is at 0 V.
Which graph shows how the potential V varies with the distance x from P along the wire?
| Answer | Mark | Notes |
|---|---|---|
| P is joined to the positive terminal, so it is at the highest potential. Equal lengths of a uniform wire have equal resistance and carry the same current, so equal lengths have equal p.d.s across them. The potential therefore falls linearly from its value at P to 0 V at Q: graph A. | ✓ 1 | Answer A |
Answer: A
Exam tip: Along a uniform wire carrying a steady current, potential changes by the same amount for every centimetre: the graph of potential against distance is a straight line.
The figure shows a potential divider connected to a supply of negligible internal resistance. The voltmeter has a very high resistance.
What is the resistance of R2?
| Answer | Mark | Notes |
|---|---|---|
| The p.d. across R1 is 9.0 − 6.0 = 3.0 V. The same current passes through both resistors, so R2/R1 = 6.0/3.0 and R2 = 2 × 1.5 = 3.0 kΩ. | ✓ 1 | Answer C |
Answer: C
Exam tip: In a potential divider the p.d.s are in the same ratio as the resistances. Find the p.d. across the known resistor first.
A technician finds a reel of unlabelled metal wire. A 2.50 m length of the wire has a resistance of 0.71 Ω at 20 °C. Its diameter is 0.50 mm. The bar chart gives the resistivities of five metals at 20 °C.
Calculate the cross-sectional area of the wire.
Calculate the resistivity of the metal.
Use the bar chart to identify the metal.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A = π(0.25 × 10−3)2 = 1.96 × 10−7 m2 | ✓ 1 | |
| Part (b) | ||
| ρ = RA/l = 0.71 × 1.96 × 10−7 ÷ 2.50 | ✓ 1 | |
| = 5.6 × 10−8 Ω m (5.58 × 10−8 Ω m) (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| tungsten (5.6 × 10−8 Ω m on the chart) (allow error carried forward from (b) if a metal is chosen whose bar is closest to the calculated value) | ✓ 1 | |
Common error: Using the 0.50 mm diameter as the radius. That makes the area four times too big and points to the wrong metal on the chart.
A uniform wire AB is stretched along a metre rule, with A at the 0 cm mark and B at the 100.0 cm mark. A steady current passes through the wire. The figure shows the circuit. Both voltmeters have very high resistance.
Read the position of the sliding contact from the metre rule.
Calculate the reading on voltmeter V2.
Calculate the distance from A at which the contact should be placed for V2 to read 1.26 V.
The wire is replaced by a wire of the same material and length but half the diameter. The resistor is adjusted until V1 again reads 1.80 V. State and explain the reading on V2 with the contact at the same position as in (a).
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 35.0 cm from A (allow 34.8–35.2 cm) | ✓ 1 | |
| Part (b) | ||
| 1.80 × 35.0/100.0 = 0.63 V (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| 1.26 ÷ 1.80 × 100.0 = 70.0 cm | ✓ 1 | |
| Part (d) | ||
| 0.63 V (unchanged): the new wire is still uniform, so the 1.80 V is still shared in proportion to length, whatever the wire's total resistance | ✓ 1 | |
Exam tip: For a uniform wire, p.d. ∝ length. Work out the fraction of the wire between the two contacts and multiply it by the p.d. across the whole wire.
A chain of four resistors is connected across a 12 V supply of negligible internal resistance to provide several output p.d.s, as shown. Terminal P is at 0 V.
Calculate the current in the resistors.
Calculate the potential at terminal S.
State the two terminals between which there is a p.d. of 6.0 V.
State which resistor has the largest p.d. across it and explain why.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = 12 ÷ (1.0 + 2.0 + 3.0 + 4.0) kΩ = 1.2 mA | ✓ 1 | |
| Part (b) | ||
| 1.2 mA × (1.0 + 2.0 + 3.0) kΩ = 7.2 V (allow 12 × 6.0/10) | ✓ 1 | |
| Part (c) | ||
| Q and S (potentials 1.2 V and 7.2 V) | ✓ 1 | |
| Part (d) | ||
| the 4.0 kΩ resistor: the current is the same in all of them, so the p.d. (= IR) is largest across the largest resistance | ✓ 1 | |
Exam tip: In a chain of series resistors, the potential at each junction is the supply p.d. multiplied by the fraction of the total resistance below that junction.
A heating coil is made from 0.75 m of nichrome wire of diameter 0.50 mm. The resistivity of nichrome is 1.1 × 10−6 Ω m.
What is the resistance of the wire?
| Answer | Mark | Notes |
|---|---|---|
| Radius = 0.25 × 10−3 m, so A = π(0.25 × 10−3)2 = 1.96 × 10−7 m2. R = ρl/A = 1.1 × 10−6 × 0.75 ÷ 1.96 × 10−7 = 4.2 Ω. | ✓ 1 | Answer C |
Answer: C
Common error: Putting the diameter into πr2. Halve the diameter first, then convert to metres, then square.
Wire Y is made of the same metal as wire X. Wire Y is twice as long as wire X and has twice the diameter.
What is the ratio RY/RX?
| Answer | Mark | Notes |
|---|---|---|
| R ∝ l/d2. Length × 2 and d2 × 4, so RY/RX = 2/4 = 0.50. | ✓ 1 | Answer B |
Answer: B
Exam tip: Apply one factor at a time: length factor in the numerator, (diameter factor)2 in the denominator. Same metal means the resistivity cancels.
A student measures the diameter of a wire at five places with a micrometer screw gauge of resolution 0.01 mm. Her readings are 0.31 mm, 0.33 mm, 0.32 mm, 0.32 mm and 0.30 mm. She estimates the uncertainty in the mean from the spread of her readings.
What is the percentage uncertainty in the mean diameter?
| Answer | Mark | Notes |
|---|---|---|
| Mean = 1.58 ÷ 5 = 0.316 mm. Uncertainty = half the range = (0.33 − 0.30) ÷ 2 = 0.015 mm. Percentage uncertainty = 0.015 ÷ 0.316 × 100 = 4.7%. | ✓ 1 | Answer C |
Answer: C
Exam tip: For repeated readings, the uncertainty is half the range; for a single reading, it is half the resolution of the instrument. Use the one the question describes.
A uniform resistance wire XY is 80.0 cm long. A steady current in the wire gives a p.d. of 2.4 V between X and Y, with X at 0 V. Point P on the wire is 30.0 cm from Y.
What is the potential at P?
| Answer | Mark | Notes |
|---|---|---|
| The potential rises uniformly from X. P is 80.0 − 30.0 = 50.0 cm from X, so V = 2.4 × 50.0/80.0 = 1.5 V. | ✓ 1 | Answer D |
Answer: D
Common error: Measuring the distance from the wrong end. Potential is measured from the 0 V end, so first find how far the point is from that end.
A fixed resistor and a light-dependent resistor (LDR) are connected in series across a battery. The output p.d. is taken across the LDR.
The light intensity on the LDR increases. Which statement is correct?
| Answer | Mark | Notes |
|---|---|---|
| Brighter light lowers the LDR's resistance. The current is the same in both components, so each p.d. is proportional to its resistance: the LDR's share of the battery p.d. falls and the output decreases. | ✓ 1 | Answer D |
Answer: D
Exam tip: Reason with shares of the supply p.d., not with current alone: in a series circuit both components carry the same current, so the larger resistance has the larger p.d.
The element of a small heater must have a resistance of 24 Ω. It is to be made from nichrome wire of diameter 0.32 mm. The resistivity of nichrome is 1.1 × 10−6 Ω m.
Calculate the cross-sectional area of the wire.
Calculate the length of wire needed.
Nichrome wire of diameter 0.64 mm is used instead. Explain what length of this wire is needed for the same resistance.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A = π(0.16 × 10−3)2 = 8.04 × 10−8 m2 | ✓ 1 | |
| Part (b) | ||
| l = RA/ρ = 24 × 8.04 × 10−8 ÷ 1.1 × 10−6 | ✓ 1 | |
| = 1.8 m (1.75 m) (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| doubling the diameter makes the cross-sectional area four times larger (A ∝ d2) | ✓ 1 | |
| so the length must be four times larger for the same R: 7.0 m (allow 4 × answer to (b)) | ✓ 1 | |
Exam tip: In a “what length is needed” question, rearrange R = ρl/A to l = RA/ρ before substituting, so each value goes in the right place.
A student is given a reel of thin metal wire, a micrometer screw gauge, a metre rule, an ammeter, a voltmeter, a battery, a switch, sticky tape and connecting leads with crocodile clips.
Describe how she could use this apparatus to determine the resistivity of the metal. Your answer should include how she would process her results. (6 marks: indicative content, one mark per point, up to 6)
| Answer | Mark | Notes |
|---|---|---|
| measure the diameter with the micrometer (after checking for zero error) at several places along the wire and in perpendicular directions, and calculate the mean | ✓ 1 | |
| calculate the cross-sectional area from A = πd2/4 | ✓ 1 | |
| tape the wire to the metre rule; connect it in series with the battery, switch and ammeter, with the voltmeter across the length between the clips | ✓ 1 | |
| move one clip to give a range of lengths l (e.g. 0.10 m to 1.00 m), measured with the metre rule, recording V and I and calculating R = V/I for each | ✓ 1 | |
| keep the current small / switch off between readings so that the temperature of the wire does not rise | ✓ 1 | |
| plot a graph of R against l; gradient = ρ/A, so ρ = gradient × A | ✓ 1 |
Exam tip: In a long method answer, cover each measured quantity with its instrument, say how you get several readings, and finish with exactly how the final value is calculated from the graph.
A 4.5 V supply of negligible internal resistance is connected in series with a 15 Ω resistor and a uniform wire of length 1.20 m and resistance 7.5 Ω. One end of the wire, end Q, is connected to the negative terminal of the supply, which is at 0 V.
Calculate the current in the circuit.
Calculate the p.d. across the wire.
Describe how the potential varies along the wire, starting from end Q.
Two sliding contacts are placed on the wire. Calculate the distance between them that gives a p.d. of 0.50 V between the contacts.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = 4.5 ÷ (15 + 7.5) = 0.20 A | ✓ 1 | |
| Part (b) | ||
| V = 0.20 × 7.5 = 1.5 V | ✓ 1 | |
| Part (c) | ||
| potential increases uniformly / linearly with distance from Q, from 0 V to 1.5 V at the other end | ✓ 1 | |
| Part (d) | ||
| potential gradient = 1.5 ÷ 1.20 = 1.25 V m−1 | ✓ 1 | |
| distance = 0.50 ÷ 1.25 = 0.40 m (allow error carried forward) | ✓ 1 | |
Exam tip: On a uniform wire the p.d. between two points depends only on how far apart they are: p.d. = potential gradient × separation.
A cooling fan in a computer is controlled by a thermistor and a variable resistor connected in series across a 6.0 V supply of negligible internal resistance. The thermistor is connected to the positive terminal; the output p.d. is taken across the variable resistor. A switching circuit turns the fan on when the output reaches 4.0 V. At 30 °C the thermistor's resistance is 3.6 kΩ.
Explain why the output p.d. increases as the temperature of the thermistor increases.
Calculate the resistance the variable resistor must be set to so that the fan switches on at 30 °C.
Explain how the variable resistor should be adjusted so that the fan switches on at a higher temperature.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the resistance of the thermistor decreases as its temperature increases | ✓ 1 | |
| so the thermistor takes a smaller share of the 6.0 V (same current, p.d. ∝ resistance) and the p.d. across the variable resistor increases | ✓ 1 | |
| Part (b) | ||
| thermistor p.d. = 6.0 − 4.0 = 2.0 V; R/3.6 = 4.0/2.0 | ✓ 1 | |
| R = 7.2 kΩ | ✓ 1 | |
| Part (c) | ||
| at a higher temperature the thermistor's resistance is lower | ✓ 1 | |
| so the variable resistor must be reduced, so that the output only reaches 4.0 V when the thermistor's resistance has fallen to this lower value | ✓ 1 | |
Exam tip: For an adjustment question, fix the switching p.d. and ask what the sensor's resistance will be at the new temperature; the other resistor must change in proportion.
The figure shows a rectangular block of a conducting material of resistivity 2.4 × 10−5 Ω m. Current enters and leaves the block through metal contacts that cover the two shaded end faces.
What is the resistance of the block between the contacts?
| Answer | Mark | Notes |
|---|---|---|
| The current flows along the 6.0 cm length, so l = 0.060 m and the cross-section is the end face: A = 0.020 × 0.0040 = 8.0 × 10−5 m2. R = ρl/A = 2.4 × 10−5 × 0.060 ÷ 8.0 × 10−5 = 0.018 Ω. | ✓ 1 | Answer C |
Answer: C
Exam tip: Decide which way the current flows first. The length in R = ρl/A is measured along the current; the area is the face the current passes through.
In an experiment to determine resistivity, a student measures the resistance R of different lengths ℓ of a uniform wire. The mean diameter of the wire is 0.36 mm. The graph shows her results and best-fit line.
What is the resistivity of the metal?
| Answer | Mark | Notes |
|---|---|---|
| Gradient = 4.0 Ω ÷ 0.80 m = 5.0 Ω m−1 = ρ/A. A = π(0.18 × 10−3)2 = 1.02 × 10−7 m2, so ρ = 5.0 × 1.02 × 10−7 = 5.1 × 10−7 Ω m. | ✓ 1 | Answer B |
Answer: B
Exam tip: Read the gradient from two points far apart on the best-fit line, not from a single data point. Then ρ = gradient × A.
A student measures the diameter of a wire with a micrometer screw gauge. The spindle moves 0.50 mm for each complete turn of the thimble, and the thimble scale has 50 divisions. The micrometer has no zero error. The figure shows the scales.
What is the diameter of the wire?
| Answer | Mark | Notes |
|---|---|---|
| One thimble division = 0.50 mm ÷ 50 = 0.01 mm. The sleeve shows 0 mm plus the 0.5 mm line (the 1 mm line is still hidden), giving 0.50 mm. The thimble line at the datum is 34, adding 0.34 mm. Diameter = 0.50 + 0.34 = 0.84 mm. | ✓ 1 | Answer C |
Answer: C
Common error: Missing the half-millimetre lines below the datum line. Check whether a 0.5 mm line is uncovered before adding the thimble reading.
Two potential dividers are connected in parallel across a 12 V supply of negligible internal resistance, as shown. A voltmeter of very high resistance is connected between X and Y.
What is the reading on the voltmeter?
| Answer | Mark | Notes |
|---|---|---|
| Taking the negative terminal as 0 V: VX = 12 × 8.0/12.0 = 8.0 V and VY = 12 × 3.0/12.0 = 3.0 V. The voltmeter reads the difference: 8.0 − 3.0 = 5.0 V. | ✓ 1 | Answer B |
Answer: B
Exam tip: Find the potential of each junction separately, measured from the same reference point, then subtract.
A thermistor at a constant temperature is connected in series with a variable resistor R across a 6.0 V supply of negligible internal resistance. The graph shows how the output p.d. across R varies as R is changed.
Use point K to determine the resistance of the thermistor.
| Answer | Mark | Notes |
|---|---|---|
| At K, R = 3.0 kΩ and Vout = 2.0 V, so the thermistor has 6.0 − 2.0 = 4.0 V across it. Same current: RT/3.0 = 4.0/2.0, so RT = 6.0 kΩ. (Check: the curve passes 3.0 V, half the supply, where R = 6.0 kΩ.) | ✓ 1 | Answer D |
Answer: D
Exam tip: When the output is exactly half the supply p.d., the two resistances in the divider are equal. This gives a quick check of any value you calculate.
A 3.0 m length of copper pipe is used as an earth connection. The figure shows the pipe and its cross-section. The resistivity of copper is 1.7 × 10−8 Ω m.
Show that the cross-sectional area of the copper is about 3.1 × 10−5 m2.
Calculate the resistance of the pipe.
A solid copper rod of the same length has the same resistance as the pipe. Calculate the diameter of the rod.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| inner diameter = 15 − 2 × 0.70 = 13.6 mm | ✓ 1 | |
| A = π(7.52 − 6.82) mm2 = 31.4 mm2 = 3.14 × 10−5 m2 | ✓ 1 | |
| Part (b) | ||
| R = ρl/A = 1.7 × 10−8 × 3.0 ÷ 3.14 × 10−5 | ✓ 1 | |
| = 1.6 × 10−3 Ω (1.62 × 10−3 Ω; 1.65 × 10−3 Ω using 3.1 × 10−5 m2) | ✓ 1 | |
| Part (c) | ||
| same ρ, l and R, so the same cross-sectional area: πd2/4 = 31.4 mm2 | ✓ 1 | |
| d = 6.3 mm (6.33 mm) (allow error carried forward) | ✓ 1 | |
Common error: Using the outer diameter alone, as if the pipe were solid. Only the ring of metal carries the current: subtract the area of the hollow centre.
A student cuts three lengths, 0.30 m, 0.60 m and 0.90 m, from the same reel of wire. For each length she measures the p.d. across it for several currents. The graph shows her results. The diameter of the wire is 0.30 mm.
Use the graph to determine the resistance of the 0.60 m length.
Explain how the graph shows that the resistance of the wire is proportional to its length.
Calculate the resistivity of the metal.
Suggest why the student kept the current no greater than 0.50 A.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| gradient, e.g. 2.0 V ÷ 0.50 A = 4.0 Ω (allow 3.9–4.1 Ω) | ✓ 1 | |
| Part (b) | ||
| the resistances (gradients) are 2.0 Ω, 4.0 Ω and 6.0 Ω for 0.30 m, 0.60 m and 0.90 m | ✓ 1 | |
| so R/l is the same (6.7 Ω m−1) for each length / doubling or tripling the length doubles or triples the resistance | ✓ 1 | |
| Part (c) | ||
| A = π(0.15 × 10−3)2 = 7.07 × 10−8 m2 | ✓ 1 | |
| ρ = RA/l = 4.0 × 7.07 × 10−8 ÷ 0.60 = 4.7 × 10−7 Ω m (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| larger currents would heat the wire, increasing its resistance (resistivity), so the readings would not be at a constant temperature | ✓ 1 | |
Exam tip: On a V–I graph the resistance is the gradient (or V/I for any point on a line through the origin). Read values where a line crosses gridlines exactly.
A steady current of 0.50 A passes along a long, uniform wire of diameter 0.40 mm. A voltmeter is connected between two contacts on the wire, a distance x apart. The graph shows how the voltmeter reading V varies with x.
Determine the potential gradient along the wire.
Calculate the resistance of 1.0 m of the wire.
Calculate the resistivity of the metal of the wire.
Explain why the graph is a straight line through the origin.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| gradient of the line, e.g. 2.0 V ÷ 0.80 m = 2.5 V m−1 (allow 2.4–2.6) | ✓ 1 | |
| Part (b) | ||
| R = V/I = 2.5 ÷ 0.50 = 5.0 Ω (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| A = π(0.20 × 10−3)2 = 1.26 × 10−7 m2 | ✓ 1 | |
| ρ = RA/l = 5.0 × 1.26 × 10−7 ÷ 1.0 = 6.3 × 10−7 Ω m (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| the wire is uniform and carries the same current everywhere, so every metre has the same resistance and the same p.d.: V ∝ x, with V = 0 when x = 0 | ✓ 1 | |
Exam tip: The gradient of a graph of p.d. against length is the potential gradient. Dividing by the current gives the resistance per unit length, which equals ρ/A.
A student compares two ways of controlling the brightness of a small lamp, using a 6.0 V supply of negligible internal resistance. In circuit X a variable resistor (0–24 Ω) is in series with the lamp. In circuit Y a 24 Ω resistance track is used as a potential divider. Assume the lamp has a constant resistance of 8.0 Ω.
Calculate the smallest p.d. that can be obtained across the lamp in circuit X.
State the range of p.d. across the lamp that can be obtained with circuit Y.
In circuit Y the slider is at the midpoint of the track, as shown. Calculate the p.d. across the lamp.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| variable resistor at 24 Ω: I = 6.0 ÷ (24 + 8.0) = 0.1875 A | ✓ 1 | |
| V = 0.1875 × 8.0 = 1.5 V | ✓ 1 | |
| Part (b) | ||
| from 0 (slider at the bottom) to 6.0 V (slider at the top) | ✓ 1 | |
| Part (c) | ||
| lower half (12 Ω) in parallel with the lamp: (1/12 + 1/8.0)−1 = 4.8 Ω | ✓ 1 | |
| V = 6.0 × 4.8/(12 + 4.8) = 1.7 V (1.71 V), not 3.0 V, because the lamp lowers the resistance of the lower half | ✓ 1 | |
Common error: Assuming the output of a potential divider is unchanged when a load is connected. Combine the load in parallel with its part of the track first.
Two 47 kΩ resistors are connected in series across a 9.0 V supply of negligible internal resistance. A voltmeter of resistance 100 kΩ is connected across one of the resistors.
What is the reading on the voltmeter?
| Answer | Mark | Notes |
|---|---|---|
| The voltmeter and resistor in parallel: 1/R = 1/47 + 1/100, so R = 32.0 kΩ. Reading = 32.0/(47 + 32.0) × 9.0 = 3.6 V. | ✓ 1 | Answer A |
Answer: A
Exam tip: A voltmeter is in parallel with the component it measures. Combine the two resistances first, then use the potential divider.
A fire alarm uses a thermistor in series with a fixed resistor R across a 5.0 V supply of negligible internal resistance. The output is taken across R. The alarm sounds when the output reaches 3.0 V. The thermistor's resistance at the alarm temperature is 1.2 kΩ.
What value of R is needed?
| Answer | Mark | Notes |
|---|---|---|
| At the alarm point R has 3.0 V and the thermistor has 5.0 − 3.0 = 2.0 V. Same current, so R/1.2 = 3.0/2.0 and R = 1.8 kΩ. | ✓ 1 | Answer B |
Answer: B
Exam tip: Find the p.d. across each component at the switching point, then use the ratio of p.d.s = ratio of resistances.
In a resistivity experiment, a student measures the diameter of a wire at five places: 0.25 mm, 0.26 mm, 0.25 mm, 0.24 mm and 0.25 mm. She then measures the resistance of different lengths of the wire.
Results: l / m — 0.150, 0.300, 0.450, 0.600, 0.750; R / Ω — 3.50, 6.74, (see part a), 13.15, 16.31.
For the 0.450 m length the voltmeter reads 3.96 V and the ammeter reads 0.400 A.
Calculate the resistance of the 0.450 m length.
Calculate the mean diameter and the cross-sectional area of the wire.
Her best-fit line of R against l passes through (0.100 m, 2.45 Ω) and (0.800 m, 17.40 Ω). Determine the resistivity of the metal.
The best-fit line does not pass through the origin. Explain why this does not affect the value of the resistivity.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| R = 3.96 ÷ 0.400 = 9.90 Ω | ✓ 1 | |
| Part (b) | ||
| mean d = 1.25 ÷ 5 = 0.250 mm | ✓ 1 | |
| A = π(0.125 × 10−3)2 = 4.91 × 10−8 m2 | ✓ 1 | |
| Part (c) | ||
| gradient = (17.40 − 2.45) ÷ (0.800 − 0.100) = 21.4 Ω m−1 | ✓ 1 | |
| ρ = gradient × A | ✓ 1 | |
| = 21.4 × 4.91 × 10−8 = 1.0 × 10−6 Ω m (1.05 × 10−6 Ω m) (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| the intercept is due to a constant extra resistance (e.g. at the clips / contacts) or a zero error, which shifts every point by the same amount and does not change the gradient | ✓ 1 | |
Exam tip: Take the gradient from two points read off the best-fit line that are far apart — not from two data points — and show the subtraction in full.
A student builds a reference circuit for a comparator from two resistors, 220 kΩ and 330 kΩ, in series with a 9.0 V battery whose internal resistance is negligible. To test her circuit, she measures the p.d. across the 330 kΩ resistor using a voltmeter whose resistance is 1.0 MΩ.
Calculate the p.d. across the 330 kΩ resistor before the voltmeter is connected.
Calculate the combined resistance of the voltmeter and the 330 kΩ resistor.
Calculate the reading on the voltmeter.
Explain why the reading is less than your answer to (a), and suggest how the student could obtain a reading closer to it.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 330 ÷ 550 × 9.0 = 5.4 V | ✓ 1 | |
| Part (b) | ||
| 1/R = 1/330 + 1/1000 | ✓ 1 | |
| R = 248 kΩ | ✓ 1 | |
| Part (c) | ||
| V = 248 ÷ (220 + 248) × 9.0 | ✓ 1 | |
| = 4.8 V (4.77 V) (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| the voltmeter in parallel lowers the resistance of that part of the circuit, so it takes a smaller share of the supply p.d. | ✓ 1 | |
| use a voltmeter whose resistance is much larger than 330 kΩ, e.g. a digital voltmeter of about 10 MΩ | ✓ 1 | |
Common error: Treating the voltmeter as ideal when its resistance is given. A meter resistance only about three times the resistor it is across changes the reading noticeably.
A student investigates an LDR. She connects the LDR in series with a fixed 4.7 kΩ resistor across a 6.0 V supply of negligible internal resistance and measures the output p.d. Vout across the fixed resistor with a high-resistance voltmeter. She places a lamp at different distances from the LDR.
Results: distance 60 cm — Vout = 0.85 V; 30 cm — 2.10 V; 15 cm — 3.60 V.
State what the results show about how the resistance of the LDR changes as the light intensity increases.
Show that the resistance of the LDR is given by RLDR = R(Vin − Vout)/Vout, where R = 4.7 kΩ and Vin = 6.0 V.
Calculate the resistance of the LDR when the lamp is 30 cm away.
The student wants an output of 3.0 V with the lamp 30 cm away. State the value of the fixed resistor she should use.
Suggest one variable she should control, other than the supply p.d.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| as the lamp gets closer (greater intensity) the p.d. across the fixed resistor rises, so the LDR's resistance decreases | ✓ 1 | |
| Part (b) | ||
| p.d. across the LDR = Vin − Vout | ✓ 1 | |
| same current in both, so RLDR/R = (Vin − Vout)/Vout | ✓ 1 | |
| Part (c) | ||
| 4.7 × (6.0 − 2.10) ÷ 2.10 = 8.7 kΩ (8.73 kΩ) | ✓ 1 | |
| Part (d) | ||
| half the supply p.d. needs equal resistances: 8.7 kΩ (allow error carried forward from (c)) | ✓ 1 | |
| Part (e) | ||
| the background light level (e.g. work in a darkened room) / the power of the lamp / the LDR facing the lamp at the same angle | ✓ 1 | |
Exam tip: In a potential divider the two p.d.s add up to the supply p.d. Use that first to find the p.d. across the component you are not measuring across.
Two wires, PQ and QR, made of the same metal, are joined at Q and connected in series to a supply. The graph shows how the potential varies with distance from P. The diameter of wire PQ is 0.45 mm.
What is the diameter of wire QR?
| Answer | Mark | Notes |
|---|---|---|
| Potential gradient = IR/l = Iρ/A, and I and ρ are the same for both wires, so the gradient ∝ 1/d2. PQ: 1.2 V ÷ 0.60 m = 2.0 V m−1. QR: (3.0 − 1.2) V ÷ 0.40 m = 4.5 V m−1. dQR = 0.45 × √(2.0/4.5) = 0.30 mm. | ✓ 1 | Answer A |
Answer: A
Exam tip: The same current passes through both wires, so a steeper potential gradient means more resistance per metre, which means a thinner wire.
In the circuit shown, the supply has negligible internal resistance. At the present light level the resistance of the light-dependent resistor (LDR) is 6.0 kΩ.
What is the p.d. across the relay coil?
| Answer | Mark | Notes |
|---|---|---|
| The coil and the LDR in parallel have resistance (1/6.0 + 1/6.0)−1 = 3.0 kΩ. This is in series with the 3.0 kΩ resistor, so the 12 V is shared equally: p.d. across the coil = 12 × 3.0/(3.0 + 3.0) = 6.0 V. | ✓ 1 | Answer B |
Answer: B
Common error: Using the divider formula with the LDR alone. A load connected across part of a divider lowers the resistance of that part, so the output falls.
A student determines the resistivity of the metal of a wire. With the micrometer jaws closed and nothing between them, the micrometer reads +0.02 mm. She then measures the diameter of the wire at five places: 0.29 mm, 0.30 mm, 0.30 mm, 0.31 mm and 0.30 mm (readings not corrected). The graph shows her measurements of resistance R against length ℓ and her best-fit line.
Calculate the mean diameter of the wire, corrected for the zero error.
Calculate the cross-sectional area of the wire.
Determine the gradient of the best-fit line.
Calculate the resistivity of the metal.
Calculate the percentage by which her value of resistivity would have been too large if she had not corrected for the zero error.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| mean reading = 1.50 ÷ 5 = 0.300 mm | ✓ 1 | |
| corrected diameter = 0.300 − 0.02 = 0.28 mm | ✓ 1 | |
| Part (b) | ||
| A = π(0.14 × 10−3)2 = 6.16 × 10−8 m2 (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| large triangle using points on the line, e.g. (0.10 m, 2.5 Ω) and (0.85 m, 17.5 Ω) | ✓ 1 | |
| gradient = 15.0 ÷ 0.75 = 20.0 Ω m−1 (allow 19.5–20.5) | ✓ 1 | |
| Part (d) | ||
| ρ = gradient × A = 20 × 6.16 × 10−8 = 1.2 × 10−6 Ω m (1.23 × 10−6 Ω m; allow 1.2–1.3 × 10−6 Ω m from a gradient in the allowed range) (allow error carried forward) | ✓ 1 | |
| Part (e) | ||
| ρ ∝ d2: (0.300/0.28)2 = 1.15, so about 15% too large | ✓ 1 | |
Common error: Adding a positive zero error instead of subtracting it. If the closed jaws already read +0.02 mm, every reading is 0.02 mm too large.
A frost alarm uses a thermistor and a variable resistor connected in series across a 6.0 V supply of negligible internal resistance. The alarm unit is connected across the thermistor and sounds when the p.d. across its input reaches 4.0 V. The graph shows how the resistance of the thermistor varies with temperature. In parts (a)–(c), assume the alarm unit draws no current.
Use the graph to state the resistance of the thermistor at 0 °C.
Explain why the p.d. across the alarm unit increases as the temperature falls.
Calculate the resistance of the variable resistor needed for the alarm to sound at 0 °C.
In fact the alarm unit has an input resistance of 120 kΩ. With the variable resistor set as in (c), calculate the p.d. across the alarm unit at 0 °C and state the effect on the alarm.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 30 kΩ (3.0 × 104 Ω) | ✓ 1 | |
| Part (b) | ||
| as the temperature falls the resistance of the thermistor increases (graph) | ✓ 1 | |
| the current is the same in both components, so the thermistor takes a larger share of the 6.0 V (p.d. ∝ resistance) | ✓ 1 | |
| Part (c) | ||
| p.d. across the variable resistor = 6.0 − 4.0 = 2.0 V; R/30 = 2.0/4.0 | ✓ 1 | |
| R = 15 kΩ (allow error carried forward from (a)) | ✓ 1 | |
| Part (d) | ||
| thermistor and alarm in parallel: (1/30 + 1/120)−1 = 24 kΩ; V = 6.0 × 24/(15 + 24) = 3.7 V (3.69 V) | ✓ 1 | |
| this is below 4.0 V, so the alarm does not sound at 0 °C; it sounds only at a lower temperature | ✓ 1 | |
Exam tip: A sensor's output rises across whichever component's resistance rises. For a cold alarm the output is taken across the thermistor, whose resistance increases as it cools.
A street lamp is controlled by a light-sensing circuit. A variable resistor and a light-dependent resistor (LDR) are connected in series across a 5.0 V supply of negligible internal resistance. The switching circuit, which draws negligible current, turns the lamp on when the p.d. across the LDR rises to 2.5 V. The graph shows how the resistance of the LDR varies with illuminance.
Explain how the circuit switches the lamp on as it gets dark in the evening.
Calculate the resistance of the variable resistor needed for the lamp to switch on at an illuminance of 10 lux.
The variable resistor is set to 10 kΩ instead. Determine the illuminance at which the lamp now switches on, and state whether this is earlier or later in the evening.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| as the illuminance falls, the resistance of the LDR increases (graph) | ✓ 1 | |
| the current is the same in both components, so the LDR takes a larger share of the 5.0 V and the p.d. across it rises | ✓ 1 | |
| when this p.d. reaches 2.5 V the switching circuit turns the lamp on | ✓ 1 | |
| Part (b) | ||
| from the graph, RLDR = 20 kΩ at 10 lux | ✓ 1 | |
| 2.5 V is half of 5.0 V, so the resistances are equal: 20 kΩ (2.0 × 104 Ω) | ✓ 1 | |
| Part (c) | ||
| switches on when RLDR = 10 kΩ, which the graph gives at 20 lux | ✓ 1 | |
| 20 lux is reached while it is still brighter, so the lamp switches on earlier | ✓ 1 | |
Exam tip: For a switching threshold of half the supply p.d., the sensor's resistance at the switching point equals the setting of the variable resistor, so the graph can be read directly.
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