Edexcel International A Level Physics · Unit 2 · Waves and Electricity
Resistance, temperature and light
36 free, high-quality past-paper-style questions on Resistance, temperature and light 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.36 new this month · 124 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IALP-U2-10-A-01Unit 2 · ASEasy1 mark
The temperature of a metal wire increases while the potential difference across it is kept constant. Which statement explains why the resistance of the wire increases?
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Answer
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Notes
In a metal the number of conduction electrons per unit volume does not change significantly with temperature. The hotter lattice ions vibrate with a larger amplitude, so electron–ion collisions are more frequent, the drift velocity for a given p.d. falls, the current falls and the resistance rises.
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Answer A
Answer: A
Exam tip: For a metal, the explanation is about the drift velocity v in I = nqvA; the number density n stays the same.
The intensity of the light falling on a light-dependent resistor (LDR) increases. Which row describes the changes in the number of charge carriers per unit volume in the LDR and in its resistance?
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Answer
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Notes
Energy absorbed from the light releases more electrons to act as charge carriers, so n increases. With more charge carriers the current at a given p.d. is larger, so the resistance decreases.
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Answer A
Answer: A
Common error: Linking “more charge carriers” with “more resistance”. More carriers means more current for the same p.d., so R = V/I is smaller.
An NTC thermistor is connected in series with an ammeter and a cell of constant e.m.f. and negligible internal resistance. The thermistor is warmed gently with a hair dryer. Which statement is correct?
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Answer
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Notes
Warming the semiconductor releases many more charge carriers, so n increases greatly. This outweighs the small fall in drift velocity caused by the larger lattice vibrations, so the resistance falls and the current increases.
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Answer C
Answer: C
Exam tip: The lattice always vibrates more when it is hotter. For a thermistor, the explanation must say that the rise in n outweighs this effect.
A resistance thermometer uses a coil of fine platinum wire as its sensing element. The coil is placed in a liquid whose temperature is to be measured.
(a)
State how the resistance of the platinum coil changes as the temperature of the liquid increases.
(1)
(b)
Explain, in terms of the lattice ions and the conduction electrons, why the resistance changes in this way.
(3)
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Answer
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Notes
Part (a)
the resistance increases
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Part (b)
the lattice ions vibrate with a larger amplitude
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conduction electrons collide with the ions more frequently, so their drift velocity (for a given p.d.) decreases
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the number of conduction electrons per unit volume is unchanged, so the current decreases / R = V/I increases
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Exam tip: A three-mark explanation needs three linked steps: vibrations → collisions and drift velocity → current and resistance. Mention that n is unchanged for a metal.
An LDR and a 10 kΩ fixed resistor are connected in series across a 6.0 V battery of negligible internal resistance. The output p.d. is measured across the fixed resistor.
Figure: LDR and fixed resistor, with the output across the fixed resistor
(a)
State how the resistance of an LDR changes as the intensity of the light falling on it increases.
(1)
(b)
In bright light the resistance of the LDR is 2.0 kΩ. Calculate the output p.d.
(2)
(c)
In dim light the resistance of the LDR is 40 kΩ. Calculate the output p.d.
(1)
(d)
Suggest one use for this circuit, given the way its output changes with light level.
(1)
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Answer
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Notes
Part (a)
the resistance decreases
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Part (b)
current = 6.0 ÷ (10 + 2.0) kΩ = 0.50 mA, or share = 10 ÷ 12
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Vout = 5.0 V
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Part (c)
6.0 × 10 ÷ 50 = 1.2 V
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Part (d)
the output is high in bright light, so it could switch on something needed in bright conditions, e.g. close window blinds / switch on a sun-shade or a fan
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Common error: Calculating the p.d. across the LDR when the output is across the fixed resistor. Check which component the output terminals are connected to.
Figure 1 shows four sketch graphs, A to D, of resistance R against Celsius temperature θ. Which graph shows how the resistance of a copper wire varies between 0 °C and 100 °C?
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Answer
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Notes
As a metal is heated, its lattice ions vibrate more and the resistance rises steadily, almost linearly over this range. At 0 °C the wire still has a resistance, because 0 °C is not absolute zero, so the line starts from a positive value on the R axis: graph D.
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Answer D
Answer: D
Common error: Drawing the metal's line through the origin. On a Celsius scale the line has a positive intercept; the wire does not lose its resistance at 0 °C.
In the circuit shown in Figure 1, the battery has negligible internal resistance and the voltmeter has a very high resistance. In dim light the resistance of the LDR is 30 kΩ. What is the reading on the voltmeter?
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Answer
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Notes
The two components carry the same current, so the 9.0 V is shared in proportion to their resistances. The voltmeter is across the LDR: V = 9.0 × 30 ÷ (30 + 15) = 6.0 V.
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Answer C
Answer: C
Exam tip: Trace the voltmeter's two connections on the diagram before calculating; here it measures the LDR's share of the supply p.d., not the fixed resistor's.
Figure 1 shows a simple model of the same electrical component at 15 °C and at 75 °C. Which row identifies the component and describes how its resistance changes between 15 °C and 75 °C?
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Answer
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Notes
At 75 °C there are many more conduction electrons (about ten times as many), so n has increased greatly: this is a semiconductor, an NTC thermistor. The larger vibrations reduce the drift velocity a little, but the rise in n outweighs this, so the current at a given p.d. rises and the resistance decreases.
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Answer D
Answer: D
Exam tip: Look at the number of charge carriers first. A large rise in n with temperature identifies a semiconductor; in a metal n stays the same.
Figure 1 shows a circuit that switches on a street lamp automatically. The battery has negligible internal resistance and the switch unit draws negligible current. The switch unit turns the lamp on when the output p.d. Vout reaches 4.0 V.
(a)
State how the resistance of the LDR changes as it gets darker.
(1)
(b)
Explain, in terms of charge carriers, why the resistance of the LDR changes in this way.
(2)
(c)
Explain why Vout increases as it gets darker.
(2)
(d)
Calculate the resistance of the LDR at the moment the lamp is switched on.
(2)
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Answer
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Notes
Part (a)
the resistance increases
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Part (b)
less light energy is absorbed, so fewer electrons are released as charge carriers / the number of charge carriers per unit volume decreases
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so for a given p.d. the current is smaller and R = V/I is larger
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Part (c)
the current is the same in both components, so the 6.0 V is shared in the ratio of their resistances
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the LDR's resistance rises while the fixed resistor's stays at 10 kΩ, so the LDR's share of the 6.0 V, Vout, increases
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Part (d)
p.d. across the 10 kΩ resistor = 6.0 − 4.0 = 2.0 V, so current = 2.0 ÷ 10 kΩ = 0.20 mA
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R = 4.0 V ÷ 0.20 mA = 20 kΩ
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Exam tip: In a sensor circuit, first find which component the output is taken across. Its p.d. rises when its share of the total resistance rises.
Figure 1 shows a simple model of the material of an LDR at the same temperature, first in the dark and then in bright light.
(a)
Use Figure 1 to state which quantity in the equation I = nqvA is different in bright light, and the factor by which it changes.
(2)
(b)
Explain why the resistance of the LDR is smaller in bright light. Refer to Figure 1 in your answer.
(3)
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Answer
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Notes
Part (a)
n, the number of charge carriers (conduction electrons) per unit volume
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it increases by a factor of 6 (12 ÷ 2)
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Part (b)
energy absorbed from the light releases more electrons to act as charge carriers, so n increases
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the ions' range of vibration is the same in both diagrams because the temperature has not changed, so the drift velocity v for a given p.d. is unchanged
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so I = nqvA is larger for the same p.d., and R = V/I is smaller
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Common error: Saying that light makes the electrons move faster. In an LDR the change is in the number of charge carriers; at constant temperature the drift velocity for a given p.d. stays the same.
A small filament lamp takes a current of 0.50 A when the potential difference across it is 6.0 V, and a current of 0.30 A when the potential difference is 2.0 V. What is the ratio (resistance at 6.0 V) ÷ (resistance at 2.0 V)?
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Answer
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Notes
R = V/I: at 6.0 V, 6.0 ÷ 0.50 = 12 Ω; at 2.0 V, 2.0 ÷ 0.30 = 6.67 Ω. Ratio = 12 ÷ 6.67 = 1.8. The hotter filament has the greater resistance.
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Answer C
Answer: C
Exam tip: Resistance at a point on an I–V graph is V/I for that point, not the gradient of the curve.
A length of metal wire is connected to a supply of constant potential difference. As the wire warms up, the current in it falls from 2.40 A to 1.80 A. The number of conduction electrons per unit volume and the dimensions of the wire do not change. What is the ratio (final drift velocity of the electrons) ÷ (initial drift velocity)?
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Answer
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Notes
I = nqvA with n, q and A constant, so v ∝ I. Ratio = 1.80 ÷ 2.40 = 0.75.
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Answer B
Answer: B
Exam tip: When some factors in I = nqvA are fixed, write the proportionality first (here v ∝ I), then use the ratio of the values.
A light-dependent resistor and a fixed resistor of resistance 2.2 kΩ are connected in series across a 6.0 V supply of negligible internal resistance. In daylight the resistance of the LDR is 3.3 kΩ. What is the potential difference across the LDR in daylight?
Figure: LDR and fixed resistor in series across the supply
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Answer
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Notes
The current is the same in both, so the p.d. divides in the ratio of the resistances: VLDR = 6.0 × 3.3 ÷ (3.3 + 2.2) = 3.6 V.
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Answer B
Answer: B
Common error: Using the ratio of the two resistances instead of each resistance's share of the total. The denominator must be the total resistance, 5.5 kΩ.
An NTC thermistor and a fixed resistor are connected in series across a battery of constant e.m.f. and negligible internal resistance. A voltmeter is connected across the fixed resistor. The temperature of the thermistor increases. Which row describes the changes in the current in the circuit and in the voltmeter reading?
Figure: thermistor and fixed resistor, with the voltmeter across the fixed resistor
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Answer
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Notes
The thermistor's resistance falls, so the total resistance falls and the current increases. The p.d. across the fixed resistor is IR with R constant, so the voltmeter reading increases; equivalently, the fixed resistor's share of the total resistance has grown.
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Answer D
Answer: D
Exam tip: For the fixed resistor, use V = IR with R constant: if the current rises, its p.d. must rise.
The coil of an electromagnet is wound from copper wire of cross-sectional area 1.5 mm2. It is connected to a supply of constant potential difference. When the coil is first switched on, at room temperature, the current is 12.0 A. Copper has 8.5 × 1028 conduction electrons per m3. e = 1.60 × 10−19 C.
(a)
Calculate the drift velocity of the conduction electrons when the coil is first switched on.
(2)
(b)
After some time the coil has warmed up and its resistance has increased by 15%. Calculate the new current.
(2)
(c)
Calculate the new drift velocity.
(1)
(d)
Explain why the change in current is due entirely to the change in drift velocity.
(1)
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Answer
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Notes
Part (a)
v = I ÷ nqA = 12.0 ÷ (8.5 × 1028 × 1.60 × 10−19 × 1.5 × 10−6)
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= 5.9 × 10−4 m s−1 (5.88 × 10−4 m s−1)
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Part (b)
new resistance = 1.15 × original, and I = V/R with V constant
A student increases the potential difference across an NTC thermistor in steps from zero and records the current. For small potential differences the current is proportional to the p.d. At larger potential differences the graph of current against p.d. curves upwards: its gradient increases. The thermistor is in air at room temperature throughout.
(a)
State what is meant by an NTC thermistor.
(1)
(b)
Explain the shape of the graph at larger potential differences. Refer to I = nqvA in your answer.
(6)
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Answer
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Notes
Part (a)
a (semiconductor) resistor whose resistance decreases as its temperature increases
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Part (b)
Indicative content, one mark for each point:
—
at larger currents, the power dissipated in the thermistor (VI) heats it above room temperature
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as its temperature rises, more charge carriers are released, so n increases
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the lattice ions vibrate more, so carriers collide more often and v decreases slightly
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the increase in n outweighs the decrease in v, so nv increases
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so the resistance decreases as the p.d. increases
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each further increase in p.d. gives a larger increase in current, so the gradient ΔI/ΔV increases
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Exam tip: Start the chain with what causes the temperature to change: here it is the thermistor's own current. Then link temperature → charge carriers → resistance → gradient.
A student investigates how the resistance of an NTC thermistor depends on temperature, using a water bath, a thermometer with a resolution of 1 °C and a digital ohmmeter. Her results are: 20 °C: 5.74 kΩ; 40 °C: 2.68 kΩ; 60 °C: 1.37 kΩ; 80 °C: 0.755 kΩ.
(a)
Describe how she should obtain these readings so that the thermistor is at the temperature shown by the thermometer.
(3)
(b)
Calculate the decrease in resistance between 20 °C and 40 °C and between 60 °C and 80 °C. Use your answers to describe the shape of the graph of resistance against temperature.
(2)
(c)
Taking the uncertainty in a single reading as half the resolution of the thermometer, calculate the percentage uncertainty in the reading of 60 °C.
(1)
(d)
The ohmmeter passes only a very small current through the thermistor. Explain why this is an advantage.
(1)
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Answer
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Notes
Part (a)
thermistor and thermometer placed close together in the water, with the thermistor's leads/connections kept dry (or the thermistor in a sealed thin tube)
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heat the water (from cold) and stir it before each reading
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wait until the thermometer and ohmmeter readings are steady before recording both
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Part (b)
decreases of 3.06 kΩ and 0.615 kΩ
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resistance decreases non-linearly: a steep fall at low temperatures, levelling off at higher temperatures
a larger current would heat the thermistor above the water temperature / the power dissipated is negligible, so the measured resistance corresponds to the thermometer reading
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Model sketch
Figure: resistance against temperature: a curve whose gradient becomes less steep as the temperature rises
Exam tip: Unless told otherwise, take the uncertainty in a single scale reading as half the resolution of the instrument; then divide by the reading for the percentage.
A student measures the current in a 6.0 V filament lamp for a range of potential differences. Her results are: p.d. / V: 0.50, 1.0, 2.0, 4.0, 6.0 — current / A: 0.212, 0.324, 0.418, 0.478, 0.500.
Figure: the student's results for the 6.0 V filament lamp
(a)
Calculate the resistance of the lamp at 1.0 V and at 6.0 V.
(2)
(b)
Explain why the resistance of the lamp increases as the p.d. increases.
(3)
(c)
The student reverses the connections to the lamp and repeats the measurements. State how the results compare with those in the table.
(1)
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Answer
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Notes
Part (a)
1.0 ÷ 0.324 = 3.1 Ω (3.09 Ω)
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6.0 ÷ 0.500 = 12 Ω
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Part (b)
a greater p.d. gives a greater current, so more power is dissipated and the filament temperature rises
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the lattice ions vibrate with a larger amplitude, so electrons collide with them more frequently
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the drift velocity (for a given p.d.) is lower while n is unchanged, so R = V/I is larger
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Part (c)
same size of current at each size of p.d., with both negative / the graph is symmetrical through the origin
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Common error: Using the gradient of the I–V curve to find the resistance. Divide the p.d. by the current at that point.
Figure 1 shows how the resistance R of a filament lamp varies with the potential difference V across it. What is the power dissipated in the lamp when the p.d. across it is 6.0 V?
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Answer
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Notes
From the graph, R = 12 Ω at 6.0 V. P = V2/R = 6.02 ÷ 12 = 3.0 W (equivalently I = 0.50 A and P = VI).
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Answer B
Answer: B
Common error: Using the cold resistance (the value at 0 V) for a lamp that is lit. Read the resistance at the p.d. actually across the lamp.
Figure 1 shows how the resistance of a coil of metal wire varies with temperature. The coil is connected to a supply that keeps the potential difference across it constant at 12.0 V. By how much does the current in the coil decrease when the coil is heated from 0 °C to 100 °C?
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Answer
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Notes
At 0 °C, I = 12.0 ÷ 20 = 0.600 A. At 100 °C, I = 12.0 ÷ 28 = 0.429 A. Decrease = 0.600 − 0.429 = 0.17 A (0.171 A).
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Answer A
Answer: A
Exam tip: Find each current from I = V/R and then subtract. Dividing the p.d. by the change in resistance does not give the change in current.
Figure 1 shows how the resistance R of an LDR varies with the illuminance E of the light falling on it. The LDR is moved from a place where the illuminance is 25 lux to a place where it is 400 lux. Its temperature does not change, so the drift velocity of the charge carriers for a given p.d. is unchanged. By what factor does the number of charge carriers per unit volume in the LDR increase?
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Answer
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Notes
For a given p.d., I = nqvA with q, v and A unchanged, so I ∝ n and R = V/I ∝ 1/n. From the graph the resistance falls from 24.0 kΩ to 3.0 kΩ, so n increases by 24.0 ÷ 3.0 = 8.0.
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Answer B
Answer: B
Exam tip: Write the proportionality before using numbers: with v, q and A fixed, R ∝ 1/n, so the factor for n is the inverse of the factor for R.
A filament lamp is connected to a 6.0 V supply of negligible internal resistance at time t = 0. Figure 1 shows how the current in the lamp varies with time. What is the increase in the resistance of the filament between t = 0 and the time when the current has become steady?
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Answer
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Notes
At switch-on the filament is cold: R = 6.0 ÷ 3.0 = 2.0 Ω. When the current is steady (0.50 A) the filament is hot: R = 6.0 ÷ 0.50 = 12 Ω. Increase = 12 − 2.0 = 10 Ω. The current falls because the heated lattice ions vibrate more and reduce the drift velocity.
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Answer C
Answer: C
Exam tip: Find the resistance at each moment with R = V/I, then subtract. Dividing the p.d. by the change in current does not give a resistance of the lamp.
A coil of metal wire is placed first in melting ice at 0 °C and then in boiling water at 100 °C. At each temperature the current in the coil is measured for several values of potential difference. The current is kept small, so the coil stays at the temperature of its surroundings. Figure 1 shows the two I–V graphs, labelled P and Q.
(a)
Calculate the resistance of the coil for line P and for line Q.
(2)
(b)
State which line was obtained at 100 °C. Explain your answer in terms of the lattice ions and the conduction electrons.
(4)
(c)
Calculate the percentage increase in the resistance of the coil between 0 °C and 100 °C.
(1)
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Answer
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Notes
Part (a)
P: 6.0 ÷ 1.5 = 4.0 Ω
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Q: 6.0 ÷ 1.2 = 5.0 Ω
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Part (b)
Q, the line with the smaller gradient / smaller current at each p.d.
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at 100 °C the lattice ions vibrate with a larger amplitude
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the conduction electrons collide with the ions more frequently, so their drift velocity at a given p.d. is smaller
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the number of conduction electrons per unit volume is unchanged in a metal, so I = nqvA is smaller at each p.d. and the resistance is larger
Figure 1 shows the I–V characteristic of a 12 V filament lamp for positive potential differences.
(a)
Use Figure 1 to calculate the resistance of the lamp at 2.0 V and at 12.0 V.
(2)
(b)
Explain the shape of the graph. Refer to the lattice ions and the conduction electrons in the filament.
(4)
(c)
Two of these lamps are connected in series across a 12 V supply of negligible internal resistance. Use Figure 1 to find the current in the lamps.
(1)
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Answer
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Notes
Part (a)
2.0 ÷ 0.50 = 4.0 Ω
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12.0 ÷ 1.00 = 12 Ω
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Part (b)
as the p.d. increases, the current and so the power dissipated increase, and the temperature of the filament rises
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the lattice ions vibrate with a larger amplitude, so the conduction electrons collide with them more frequently
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the number of conduction electrons per unit volume is unchanged, so the drift velocity, and so the current, increase less for each further increase in p.d.
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the resistance V/I increases, so the gradient of the graph decreases / the curve flattens
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Part (c)
each lamp has 6.0 V across it, so I = 0.80 A (allow 0.79–0.81 A)
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Common error: In the last part, using the lamp's resistance at 12 V for a lamp that has only 6.0 V across it. The filament is cooler at 6.0 V, so read the current from the graph.
A student investigates how the resistance R of an LDR depends on the illuminance E. She places a lamp at different distances from the LDR, measures E with a light meter placed next to the LDR and measures R with an ohmmeter. She works in a darkened room, and the temperature of the LDR stays constant. Figure 1 shows her values of 1/R plotted against E, with a line of best fit.
(a)
Explain, in terms of the charge carriers in the LDR, why 1/R increases as E increases.
(2)
(b)
Determine the gradient of the line of best fit.
(2)
(c)
Use Figure 1 to find the resistance of the LDR when the illuminance is 250 lux.
(1)
(d)
Suggest why the investigation is carried out in a darkened room.
(1)
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Answer
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Notes
Part (a)
more light energy is absorbed per second, so more electrons are released and the number of charge carriers per unit volume n increases
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v, q and A are unchanged at constant temperature, so the current for a given p.d. is proportional to n and 1/R = I/V increases
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Part (b)
uses a large triangle, e.g. from (0, 0.10) to (500, 2.10)
1/R = 1.10 kΩ−1, so R = 0.91 kΩ (allow 0.89–0.93 kΩ)
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Part (d)
so that background light, which may change during the investigation, does not add to the illuminance on the LDR; the illuminance is then controlled by the lamp's distance alone
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Exam tip: For a gradient, draw a triangle that covers more than half of the line and read its corners where the line crosses gridlines.
Figure 1 shows how the resistance of an NTC thermistor varies with temperature, and the circuit of a temperature warning device for a greenhouse that uses it. The battery has negligible internal resistance and the voltmeter has a very high resistance.
(a)
Explain why the voltmeter reading increases as the temperature falls.
(2)
(b)
Use Figure 1 to calculate the voltmeter reading when the temperature is 20 °C.
(2)
(c)
On a cold night the voltmeter reads 3.0 V. Use Figure 1 to determine the temperature of the thermistor.
(3)
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Answer
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Notes
Part (a)
as the temperature falls, the resistance of the thermistor increases
✓ 1
so the thermistor's fraction of the total resistance increases and it takes a larger share of the 5.0 V
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Part (b)
from the graph, R = 10 kΩ at 20 °C
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V = 5.0 × 10 ÷ (10 + 10) = 2.5 V
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Part (c)
p.d. across the fixed resistor = 5.0 − 3.0 = 2.0 V, so current = 2.0 ÷ 10 kΩ = 0.20 mA
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resistance of thermistor = 3.0 ÷ 0.20 mA = 15 kΩ
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from the graph, θ = 10 °C (allow 9–11 °C)
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Exam tip: When a meter reading is given, work back to the sensor's resistance first, then read the temperature from the graph.
A semiconductor component is connected to a supply of constant potential difference. When it is heated from 20 °C to 50 °C, the number of charge carriers per unit volume increases by a factor of 6.0 and, because of increased lattice vibrations, the drift velocity of the carriers falls by 20%. The dimensions of the component do not change. By what factor does the resistance of the component change?
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Answer
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Notes
I = nqvA, so the current changes by 6.0 × 0.80 = 4.8. At constant p.d., R = V/I changes by 1 ÷ 4.8 = 0.21 (0.208).
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Answer C
Answer: C
Exam tip: Combine the two factors in I = nqvA by multiplying them, then invert for the resistance, because R ∝ 1/I at constant p.d.
A filament lamp is connected in series with a 10 Ω resistor and a supply of negligible internal resistance. Measurements on the lamp alone show that the p.d. across it is 1.0 V when the current is 0.20 A, and 2.5 V when the current is 0.30 A. In the series circuit the current is 0.30 A. What is the e.m.f. of the supply?
Figure: the lamp in series with the 10 Ω resistor
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Answer
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Notes
p.d. across the resistor = 0.30 × 10 = 3.0 V. At 0.30 A the lamp's p.d. is 2.5 V (read at the actual current; the lamp's resistance is not constant). e.m.f. = 3.0 + 2.5 = 5.5 V.
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Answer D
Answer: D
Common error: Treating the lamp's resistance as constant. Its resistance at 0.20 A (5.0 Ω) is smaller than at 0.30 A (8.3 Ω), because the filament is cooler.
An LDR is connected in series with a 1.0 kΩ resistor across a 5.0 V supply of negligible internal resistance. Over the range used, the resistance of the LDR is inversely proportional to the light intensity. At a certain light intensity the LDR's resistance is 4.0 kΩ. The light intensity is then doubled. What is the new p.d. across the 1.0 kΩ resistor?
Figure: LDR and 1.0 kΩ resistor in series across the supply
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Answer
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Notes
Doubling the intensity halves the LDR's resistance to 2.0 kΩ. p.d. across the fixed resistor = 5.0 × 1.0 ÷ (1.0 + 2.0) = 1.7 V (1.67 V). It was 1.0 V before.
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Answer B
Answer: B
Exam tip: Find the new resistance first, then recalculate the share of the supply p.d. The output of a potential divider is not proportional to the sensor's resistance.
A constant potential difference of 5.0 V is applied across an LDR. In dim light the current is 0.45 mA. When a lamp is switched on nearby, the current rises to 2.5 mA. The temperature of the LDR does not change significantly.
(a)
Calculate the resistance of the LDR in dim light and when the lamp is on.
(2)
(b)
Explain, with reference to I = nqvA, why the current increases when the lamp is switched on.
(4)
(c)
Assuming the drift velocity of the charge carriers does not change, calculate the factor by which the number of charge carriers per unit volume increases.
(1)
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Answer
Mark
Notes
Part (a)
5.0 ÷ 0.45 × 10−3 = 11 kΩ (11.1 kΩ)
✓ 1
5.0 ÷ 2.5 × 10−3 = 2.0 kΩ
✓ 1
Part (b)
the LDR absorbs more light energy (per second)
✓ 1
this releases more electrons, which become charge carriers
✓ 1
so the number of charge carriers per unit volume n increases
✓ 1
the temperature, so the drift velocity v (and q, A), are unchanged, so I = nqvA increases
✓ 1
Part (c)
n ∝ I, so 2.5 ÷ 0.45 = 5.6 (5.56)
✓ 1
Exam tip: For an LDR the change is in n only. Say explicitly that the drift velocity is unchanged because the temperature is unchanged.
A heat alarm for a kitchen uses an NTC thermistor in series with a fixed resistor across a 5.0 V supply of negligible internal resistance. The alarm sounds when the p.d. across its input reaches 3.0 V. The resistance of the thermistor is 10.0 kΩ at 25 °C and 1.76 kΩ at 70 °C. The alarm input draws negligible current.
Figure: thermistor and fixed resistor in series across the supply
(a)
Explain whether the alarm input should be connected across the thermistor or across the fixed resistor.
(2)
(b)
Calculate the resistance of the fixed resistor needed for the alarm to sound at 70 °C.
(3)
(c)
Calculate the p.d. across the fixed resistor at 25 °C.
(2)
(d)
Suggest why a variable resistor is often used in place of the fixed resistor.
(1)
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Answer
Mark
Notes
Part (a)
as the temperature rises the thermistor's resistance falls, so its share of the 5.0 V falls
✓ 1
so the p.d. across the fixed resistor rises: connect the input across the fixed resistor
✓ 1
Part (b)
p.d. across the thermistor = 5.0 − 3.0 = 2.0 V
✓ 1
current = 2.0 ÷ 1760 = 1.14 × 10−3 A, or ratio R ÷ 1.76 = 3.0 ÷ 2.0
✓ 1
R = 2.6 kΩ (2.64 kΩ)
✓ 1
Part (c)
5.0 × 2.64 ÷ (2.64 + 10.0)
✓ 1
= 1.0 V (1.04 V) (allow error carried forward)
✓ 1
Part (d)
so that the temperature at which the alarm sounds can be adjusted / calibrated
✓ 1
Model sketch
Figure: alarm input connected across the fixed resistor
Exam tip: At the switching point, write the p.d. across each component first (here 3.0 V and 2.0 V). The resistances are then in the same ratio as these p.d.s.
Figure 1 shows the I–V graphs for an LDR in dim light and in bright light. In bright light, the LDR is connected in series with a 2.0 kΩ resistor and a 6.0 V supply of negligible internal resistance. What is the current in the circuit?
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Answer
Mark
Notes
From the bright-light line, R = V/I = 4.0 V ÷ 2.0 mA = 2.0 kΩ. Total resistance = 2.0 + 2.0 = 4.0 kΩ, so I = 6.0 ÷ 4.0 kΩ = 1.5 mA.
✓ 1
Answer A
Answer: A
Common error: Reading the current from the graph at the supply p.d. The LDR shares the 6.0 V with the resistor, so find its resistance from the graph first.
Figure 1 shows the I–V graphs for an NTC thermistor in air and for a resistor. The thermistor and the resistor are connected in parallel across a supply of negligible internal resistance. The current from the supply is 30 mA. What is the p.d. of the supply?
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Answer
Mark
Notes
In parallel both components have the supply p.d. across them and the currents add. At 4.0 V the resistor carries 10 mA and the thermistor 20 mA: total 30 mA, so the supply p.d. is 4.0 V. The thermistor's curve bends upwards because its current heats it, releasing more charge carriers.
✓ 1
Answer B
Answer: B
Exam tip: For components in parallel, look for the p.d. at which the currents read from the two graphs add up to the total; do not assume the current divides equally.
Figure 1 shows how the resistance of an NTC thermistor varies with temperature, and a temperature-sensing circuit that uses it. The battery has negligible internal resistance and the voltmeter has a very high resistance. The voltmeter reading must be 3.0 V when the thermistor is at 40 °C. What value of R is needed?
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Answer
Mark
Notes
From the graph the thermistor's resistance at 40 °C is 2.0 kΩ. The p.d. across it is 9.0 − 3.0 = 6.0 V. The current is the same in both, so R ÷ 2.0 kΩ = 3.0 ÷ 6.0, giving R = 1.0 kΩ.
✓ 1
Answer A
Answer: A
Exam tip: Write the p.d. across each component first (here 6.0 V and 3.0 V); the resistances are in the same ratio.
Figure 1 shows how the ratio R/R20 varies with temperature θ for a metal wire and for an NTC thermistor, where R20 is each component's resistance at 20 °C. Both components have a resistance of 500 Ω at 20 °C.
(a)
Use Figure 1 to calculate the resistance of each component at 60 °C.
(2)
(b)
Explain, with reference to I = nqvA, why the resistances of the two components change in opposite directions as the temperature rises, and why the change is much larger for the thermistor.
(6)
Show mark scheme
Answer
Mark
Notes
Part (a)
metal wire: 1.16 × 500 = 580 Ω
✓ 1
thermistor: 0.24 × 500 = 120 Ω
✓ 1
Part (b)
Indicative content, one mark for each point:
—
in both components the lattice ions vibrate with a larger amplitude as the temperature rises
✓ 1
so the charge carriers collide with them more frequently and the drift velocity v for a given p.d. decreases
✓ 1
in the metal the number of charge carriers per unit volume n is (effectively) constant, so I decreases and the resistance increases
✓ 1
in the thermistor (a semiconductor) more charge carriers are released as the temperature rises, so n increases
✓ 1
the increase in n is much greater than the decrease in v, so nv and the current increase and the resistance decreases
✓ 1
the metal's resistance changes only through v, a small effect, while the thermistor's n changes by a large factor, so its fractional change is much larger
✓ 1
Exam tip: For a comparison, cover both factors in I = nqvA for each component: say what happens to n and to v, then which effect wins.
Figure 1 shows how the number of charge carriers per unit volume n in the material of a thermistor varies with temperature, and the dimensions of a thermistor rod made from this material. A constant potential difference is applied between the ends of the rod. At 20 °C the current is 3.6 mA. Each charge carrier has a charge of 1.60 × 10−19 C.
(a)
Use Figure 1 to find the factor by which n increases between 20 °C and 60 °C.
(1)
(b)
Calculate the drift velocity of the charge carriers at 20 °C.
(2)
(c)
When the rod is heated to 60 °C, increased lattice vibrations reduce the drift velocity at this p.d. by 25%. Calculate the new current.
(2)
(d)
Calculate the ratio (resistance at 60 °C) ÷ (resistance at 20 °C), and explain why the resistance falls even though the drift velocity decreases.
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