Resistors, thermistors, LDRs and I–V characteristics (RP4) – GCSE Physics exam questions with mark scheme

1Grade 4–53 marks

The graphs show how the current $I$ varies with potential difference $V$ for three different components.

Identify the component that gives each graph. Choose from: diode, filament lamp, resistor at constant temperature. [3]

Graph X: filament lamp(1)
Graph Y: diode(1)
Graph Z: resistor (at constant temperature)(1)
Answer: X filament lamp; Y diode; Z resistor at constant temperature
Examiner tip: A diode only lets current flow one way — its graph is flat along the negative $V$ axis.

2Grade 6–7Higher3 marks

The resistance of a filament lamp increases as the current through it increases.

Explain why. [3]

(as current increases) the temperature of the filament increases(1)
the ions / atoms (in the metal) vibrate more(1)
so electrons collide with the ions more often / it is harder for electrons to flow (so resistance increases)(1)
Answer: Filament gets hotter, ions vibrate more, more collisions with electrons, so resistance rises.
Examiner tip: 'It gets hotter' is only 1 mark — the higher marks need ions vibrating and electrons colliding with them.

3Grade 6–7Higher5 marks

The diagram shows a temperature-sensing circuit. A thermistor is connected in series with a 300 Ω resistor and a 9.0 V battery. A voltmeter is connected across the 300 Ω resistor.

(a) At one temperature the current in the circuit is 0.012 A.

Calculate the resistance of the thermistor. [3]

(b) The temperature of the thermistor increases.

Explain what happens to the reading on the voltmeter. [2]

(a) total resistance $= \dfrac{9.0}{0.012}$ or pd across 300 Ω $= 0.012 \times 300 = 3.6$ V(1)
(a) $= 750$ (Ω) or pd across thermistor $= 9.0 - 3.6 = 5.4$ V(1)
(a) 450 Ω an answer of 450 (Ω) scores 3 marks(1)
(b) resistance of thermistor decreases (so total resistance decreases)(1)
(b) current increases so pd across the 300 Ω resistor / voltmeter reading increases allow thermistor takes a smaller share of the 9.0 V so the pd across the 300 Ω resistor increases(1)
Answer: (a) 450 Ω (b) reading increases: thermistor resistance falls, current rises
Examiner tip: Total resistance in series is the sum, so subtract the 300 Ω — 750 Ω is the whole circuit, not the thermistor.

4Grade 1–3Foundation4 marks

The diagram shows the circuit symbols for two components, P and Q.

(a) Name component P and component Q. [2]

(b) The light intensity on component Q increases.

What happens to the resistance of Q? [1]

(a) P: thermistor do not accept thermometer / variable resistor(1)
(a) Q: light dependent resistor accept LDR(1)
(b) (resistance) decreases allow goes down / gets lower(1)
(c) thermostat (in a heating system / oven / fridge / kettle) allow any device where it senses temperature, e.g. fire alarm, car engine temperature sensor, digital thermometer; ignore 'to measure temperature' alone if no device or context given(1)
Answer: (a) P thermistor; Q light dependent resistor (LDR) (b) it decreases (c) e.g. thermostat in a central-heating system
Examiner tip: Do not mix up the two symbols: the circle with arrows pointing in is the LDR; the bent line through the resistor box is the thermistor.

5Grade 4–54 marks

A student measured the current in a filament lamp at different values of potential difference.

Potential difference in V	1.0	2.0	4.0	6.0
Current in A	0.20	0.32	0.48	0.60

(a) Calculate the resistance of the lamp when the potential difference is 6.0 V. [2]

(b) An ohmic conductor has a constant resistance.

Use data from the table to show that the filament lamp is not an ohmic conductor. [2]

(a) $R = \dfrac{6.0}{0.60}$ allow $6.0 = 0.60 \times R$(1)
(a) $R = 10$ Ω an answer of 10 (Ω) scores 2 marks(1)
(b) a correct resistance calculated at a second pd, e.g. at 1.0 V $R = 5.0$ Ω or at 2.0 V $R = 6.25$ Ω or at 4.0 V $R = 8.3$ Ω allow when the pd doubles (from 2.0 V to 4.0 V) the current does not double (0.32 A to 0.48 A)(1)
(b) so the resistance is not constant / increases (as pd increases) or current is not directly proportional to pd dependent on a correct comparison using data(1)
Answer: (a) 10 Ω (b) e.g. R = 5.0 Ω at 1.0 V but 10 Ω at 6.0 V, so resistance is not constant
Examiner tip: 'Show that' needs numbers: work out two resistances (or show the current does not double when the pd doubles) — just saying 'the graph would curve' scores nothing.

6Grade 6–76 marks

A student wants to obtain the I–V characteristic of a filament lamp for both positive and negative values of potential difference.

Describe a method the student could use.

You should include a description of the circuit, the measurements she should take and how she should present her results. [6]

connect the lamp in series with a power supply / battery, an ammeter and a variable resistor allow use a variable pd supply instead of a variable resistor(1)
connect a voltmeter in parallel / across the lamp(1)
close the switch and record the readings on the ammeter and the voltmeter(1)
use the variable resistor / supply to change the pd across the lamp(1)
take readings for a range of values, e.g. in steps of 1 V up to the lamp's rated pd(1)
reverse the connections to the power supply (or the lamp) and repeat to get negative values of pd and current(1)
repeat each set of readings and calculate a mean(1)
switch off between readings / do not exceed the lamp's rated pd so the lamp is not damaged(1)
plot a graph of current (y-axis) against potential difference (x-axis), including negative values(1)
draw a smooth curve of best fit (through the origin)(1)

Level	Marks	What it looks like
Level 3	5–6	A coherent, logically ordered method that would produce a full I–V characteristic. Correct positions of both meters, a way of varying the pd, a range of readings and how negative values are obtained are all included, with results presented as a graph of current against pd.
Level 2	3–4	A method with the main steps but some detail missing or unclear, e.g. meters correctly placed and readings taken but no clear way of obtaining negative values, or no description of how results are presented.
Level 1	1–2	Some relevant, simple statements, e.g. 'use an ammeter and a voltmeter' or 'plot a graph', that do not form a workable method.
0	0	No relevant content.

Answer: Circuit: supply, switch, ammeter and variable resistor in series with the lamp; voltmeter across the lamp. Vary the pd, record I and V for a range of values; reverse the supply connections for negative values; repeat and average; plot I against V and draw a smooth curve.
Examiner tip: The 'negative values' instruction is in the question for a reason — top-band answers say exactly how to get them (reverse the supply connections).

7Grade 8–9Higher7 marks

An outside light switches on automatically when it gets dark.

The sensing circuit contains a light dependent resistor (LDR) connected in series with a 2.0 kΩ resistor and a 6.0 V battery. An electronic switch is connected across the LDR. The switch turns the light on when the potential difference across the LDR is greater than 3.0 V.

The table shows the resistance of the LDR at different light intensities.

Light intensity in lux	10	50	200	1000
Resistance of LDR in kΩ	20	4.0	1.0	0.25

Assume no current passes through the electronic switch.

(a) Calculate the potential difference across the LDR when the light intensity is 50 lux. [4]

(b) Determine whether the light is on or off when the light intensity is 200 lux. [3]

(a) total resistance $= 4.0 + 2.0 = 6.0$ kΩ allow 6000 Ω(1)
(a) $I = \dfrac{6.0}{6000}$ allow ecf from their total resistance(1)
(a) $I = 0.0010$ A allow 1.0 mA(1)
(a) $V = 0.0010 \times 4000 = 4.0$ V an answer of 4.0 (V) scores 4 marks; allow ecf from their current; allow a correct ratio method: $\dfrac{4.0}{6.0} \times 6.0$ V(1)
(b) total resistance $= 1.0 + 2.0 = 3.0$ kΩ and $I = \dfrac{6.0}{3000} = 0.0020$ A allow LDR has $\dfrac{1}{3}$ of the total resistance(1)
(b) pd across LDR $= 0.0020 \times 1000 = 2.0$ V(1)
(b) 2.0 V is less than 3.0 V so the light is off dependent on a calculated pd; allow ecf from their pd if the conclusion is consistent with it(1)
Answer: (a) 4.0 V (b) pd across LDR = 2.0 V, which is less than 3.0 V, so the light is off
Examiner tip: Use the LDR resistance, not the total resistance, in the final $V = IR$ step — and keep kΩ and A consistent (convert to Ω).

8Grade 4–5Foundation6 marks

The graph shows how the resistance of a thermistor changes with temperature.

(a) Describe how the resistance of the thermistor changes as the temperature increases. [1]

(b) Use the graph to find the resistance of the thermistor at 40 °C. [1]

Calculate the current in the thermistor. [3]

(d) The temperature increases to 60 °C. The potential difference across the thermistor stays at 3.0 V.

What happens to the current in the thermistor? [1]

(a) resistance decreases (as temperature increases) allow 'goes down'; ignore any reference to the rate of decrease(1)
(b) 1.5 kΩ accept 1500 Ω(1)
(c) $3.0 = I \times 1500$ allow ecf from (b)(1)
(c) $I = \dfrac{3.0}{1500}$(1)
(c) $I = 0.0020$ A allow 0.002 A; allow 2 mA; an answer of 0.0020 (A) scores 3 marks; an answer of 2 (A) from using 1.5 kΩ as 1.5 Ω scores 2 marks(1)
(d) (current) increases allow it doubles (to 0.0040 A)(1)
Answer: (a) it decreases (b) 1.5 kΩ (c) 0.0020 A (d) the current increases
Examiner tip: Resistance on this graph is in kΩ — multiply by 1000 before using $V = IR$, otherwise the current comes out 1000 times too big.

9Grade 4–55 marks

A student uses the circuit shown to find the I–V characteristic of a diode.

(a) Name component Y. [1]

(b) Explain why component Y is needed in this investigation. [1]

(d) The student also wants readings for negative values of potential difference.

State what she should change in the circuit. [1]

(e) When the potential difference across the diode is negative, the ammeter reads zero.

What does this show about the resistance of the diode in this direction? [1]

(a) variable resistor allow rheostat; do not accept 'resistor' alone(1)
(b) to change / vary the current in the circuit or the pd across the diode (so that a range of readings can be taken) ignore 'to control the current' alone(1)
(c) to limit the current (so the diode / ammeter is not damaged) allow to protect the diode; allow so the diode does not overheat(1)
(d) reverse the connections to the battery / power supply allow turn the battery round; allow turn the diode round (and the voltmeter connections)(1)
(e) (very) high resistance allow resistance is (almost) infinite; do not accept 'no resistance' / 'zero resistance'(1)
Answer: (a) variable resistor (b) to vary the current / pd across the diode (c) to limit the current and protect the diode (d) reverse the battery (supply) connections (e) the resistance is very high
Examiner tip: 'There is no current, so there is no resistance' is the commonest wrong answer — zero current with a pd across it means the resistance is very high.

10Grade 6–7Higher5 marks

The graph shows how the current in a diode varies with the potential difference across it.

(a) Determine the resistance of the diode when the potential difference across it is 0.70 V. [3]

(b) Use the graph to describe how the resistance of the diode changes as the potential difference increases from 0.70 V to 0.75 V.

Support your answer with a calculation. [2]

(a) current read from graph $= 20$ mA allow 0.020 A(1)
(a) $R = \dfrac{0.70}{0.020}$ allow ecf from their graph reading; allow $\dfrac{0.70}{20}$ for this mark (current not converted to A)(1)
(a) $R = 35$ Ω an answer of 35 (Ω) scores 3 marks; an answer of 0.035 (Ω) scores 2 marks (current not converted to A)(1)
(b) current at 0.75 V read as 45 mA and $R = \dfrac{0.75}{0.045} = 17$ Ω allow 16.7 Ω; allow 16–18 Ω from a correct graph reading(1)
(b) (so) the resistance decreases (as the pd increases) dependent on a calculation in (b) or comparison with (a)(1)
Answer: (a) 35 Ω (b) at 0.75 V, R = 0.75 ÷ 0.045 ≈ 17 Ω, so the resistance decreases as the pd increases
Examiner tip: Convert milliamps to amps before using $R = V \div I$; using 20 instead of 0.020 gives an answer 1000 times too small.

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