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GCSE · Physics

P2a Circuits, Resistance and Components

AQA 8464 · Calculator allowed · about 95 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Sensing circuits in everyday life

Original text written for Revision Library.

Many circuits found in homes, schools and greenhouses depend on components whose resistance changes with the environment around them. A streetlight that switches on automatically at dusk uses a light-dependent resistor (LDR) to sense falling light levels, while a frost alarm or an automatic heater often relies on a thermistor to sense a change in temperature. In both cases, a change in resistance changes the current flowing in a circuit, and this change can be used to trigger a switch. Before any of this is possible, though, an engineer has to understand the basics: how current, potential difference and resistance are related in series and parallel circuits, and how the resistance of ordinary components such as resistors, filament lamps and diodes can be investigated and compared.

The diagram shows three standard circuit symbols, labelled X, Y and Z.

(a)State the name of the component represented by symbol X.(1)

(b)State the name of the component represented by symbol Y.(1)

(c)State the name of the component represented by symbol Z.(1)

(Total for Question 1 is 3 marks)

A phone charger delivers a current of 2.5 A to a phone battery for 3.0 minutes. Use the equation charge = current x time (Q = I x t).

(a)Convert the charging time into seconds.(1)

(b)Calculate the charge that flows into the battery during the 3.0 minutes.(2)

(Total for Question 2 is 3 marks)

The diagram shows two circuits, A and B, each containing a cell, a lamp and a resistor.

(a)State whether circuit A is a series circuit or a parallel circuit.(1)

(b)State whether circuit B is a series circuit or a parallel circuit.(1)

(c)In circuit B, state how the potential difference across the lamp compares with the potential difference across the resistor, and explain why.(1)

(Total for Question 3 is 3 marks)

Use the equation potential difference = current x resistance (V = I x R) to answer the following.

(a)A resistor has resistance 15 ohm. A current of 0.40 A flows through it. Calculate the potential difference across the resistor.(2)

(b)A different resistor has a potential difference of 12 V across it, and a current of 0.80 A flows through it. Calculate the resistance of this resistor.(2)

(Total for Question 4 is 4 marks)

Figure 1 shows two resistors, R1 (8 ohm) and R2 (12 ohm), connected in series with a 10 V cell.

(a)In a series circuit, total resistance = R1 + R2. Calculate the total resistance of the circuit.(1)

(b)Use V = I x R and your answer to part (a) to calculate the current supplied by the cell.(2)

(c)State the current through R2, giving a reason for your answer.(2)

(Total for Question 5 is 5 marks)

Figure 2 shows a 20 ohm resistor and a 5 ohm resistor connected in parallel across a 10 V battery.

(a)State the potential difference across the 5 ohm resistor, giving a reason.(2)

(b)Use I = V / R to calculate the current in the 20 ohm resistor.(2)

(c)The current in the 5 ohm resistor is 2.0 A. Calculate the total current supplied by the battery.(2)

(Total for Question 6 is 6 marks)

A resistor is connected in a circuit and its current-potential difference (I-V) graph is plotted at constant temperature.

(a)State the shape of the I-V graph for a resistor at constant temperature.(1)

(b)Explain what this graph shape shows about the resistance of the resistor as the potential difference increases.(2)

(Total for Question 7 is 3 marks)

A filament lamp is connected in the same type of circuit and its I-V graph is plotted.

(a)Describe the shape of the I-V graph for a filament lamp.(1)

(b)Explain why the graph has this shape, referring to the temperature and resistance of the filament.(3)

(Total for Question 8 is 4 marks)

A diode is connected into a test circuit and its I-V characteristics are investigated in both directions.

(a)State what happens to the current in the diode when it is forward biased and the potential difference is increased from zero.(1)

(b)State what happens to the current in the diode when it is reverse biased.(1)

(c)Give one practical use of a diode, and explain why the property described in parts (a) and (b) makes it useful for this.(2)

(Total for Question 9 is 4 marks)

A student carries out the required practical to investigate how the length of a wire affects its resistance. The apparatus consists of a low-voltage power supply, a length of resistance wire fixed to a metre ruler, a fixed crocodile clip at one end, a movable crocodile clip used to select the test length, and meters to measure current and potential difference.

(a)State the name of the meter connected in parallel across the test length of wire, and the name of the meter connected in series in the circuit.(2)

(b)Identify the independent variable and the dependent variable in this investigation.(2)

(c)State one variable that should be controlled during this investigation, and explain why.(2)

(d)A student obtains these results for three test lengths: 0.20 m gives V = 1.0 V and I = 0.50 A; 0.40 m gives V = 2.0 V and I = 0.50 A; 0.60 m gives V = 3.0 V and I = 0.50 A. Calculate the resistance for each length (using R = V / I), and describe the relationship between length and resistance.(3)

(Total for Question 10 is 9 marks)

A student carries out the required practical to investigate the I-V characteristics of a resistor, a filament lamp and a diode, using a circuit containing a power supply, a variable resistor, an ammeter and a voltmeter.

(a)State the equipment needed in addition to the component being tested, and describe the role of the variable resistor in this investigation.(2)

(b)For the filament lamp, explain why the circuit is adjusted to also take readings with the current flowing in the opposite direction.(1)

(c)Explain one precaution the student should take when investigating the filament lamp, other than reversing the current direction, to obtain a valid I-V graph.(2)

(d)State one difference you would expect to see between the I-V graph for the resistor and the I-V graph for the diode.(1)

(Total for Question 11 is 6 marks)

A light-dependent resistor (LDR) is used in a circuit that switches on a light automatically when it gets dark.

(a)State what happens to the resistance of an LDR as light intensity increases.(1)

(b)Explain how the LDR's changing resistance can be used to switch the light on automatically as it gets dark, referring to how the current in the circuit changes.(2)

(Total for Question 12 is 3 marks)

A thermistor is a component whose resistance depends on its temperature.

(a)State what happens to the resistance of a thermistor as its temperature increases.(1)

(b)Give one everyday application of a thermistor, and state the property that makes a thermistor suitable for this use.(2)

(Total for Question 13 is 3 marks)

A greenhouse heater is controlled by a circuit containing a thermistor connected in series with a fixed resistor and a relay switch, powered by a constant potential difference supply. When the temperature in the greenhouse falls, the heater should switch on. Explain, using ideas about resistance, current and potential difference, how the thermistor allows this circuit to detect a fall in temperature and trigger the heater switching on.

(Total for Question 14 is 6 marks)

Higher tier only. Figure 3 shows a 6 ohm resistor (R(A)) connected in series with a section containing a 4 ohm resistor (R(B)) and a 12 ohm resistor (R(C)) connected in parallel with each other. The combination is connected across a 12 V supply.

(a)Use 1 / R(parallel) = 1 / R(B) + 1 / R(C) to calculate the combined resistance of R(B) and R(C).(3)

(b)Calculate the total resistance of the whole circuit.(1)

(c)Calculate the current supplied by the 12 V supply.(2)

(Total for Question 15 is 6 marks)

A resistor is connected across a battery. A second, identical resistor is then added in parallel with the first, connected directly across the same two points.

(a)State what happens to the total resistance of the circuit when the second resistor is added in parallel.(1)

(b)Explain, in terms of the paths available for the flow of charge, why adding a resistor in parallel decreases the total resistance of the circuit.(2)

(c)State what happens to the total current drawn from the battery when the second resistor is added, assuming the battery's potential difference stays the same.(1)

(Total for Question 16 is 4 marks)

Higher tier only. Figure 4 shows a thermistor connected in series with a fixed 15 ohm resistor, across a 6.0 V supply. At a particular temperature, the thermistor has a resistance of 45 ohm.

(a)Calculate the total resistance of the circuit at this temperature.(1)

(b)Calculate the current in the circuit at this temperature.(2)

(c)The temperature rises and the resistance of the thermistor falls to 15 ohm. Calculate the new current in the circuit, and use it to describe how the potential difference across the fixed resistor changes as the temperature rises.(3)

(Total for Question 17 is 6 marks)

A student sets up a circuit to measure the current through, and the potential difference across, a resistor. By mistake, they connect the ammeter in parallel with the resistor, and the voltmeter in series in the main circuit loop.

(a)State the correct way to connect an ammeter when measuring the current through a component.(1)

(b)State the correct way to connect a voltmeter when measuring the potential difference across a component.(1)

(c)Explain one problem this incorrect wiring would cause with the student's results.(1)

(Total for Question 18 is 3 marks)

In an I-V characteristics investigation, a student obtains the following readings for a resistor at constant temperature: at V = 1.0 V, I = 0.20 A; at V = 2.0 V, I = 0.40 A; at V = 3.0 V, I = 0.60 A; at V = 4.0 V, I = 0.80 A.

(a)Use two pairs of readings to show that the resistance of the resistor is constant.(3)

(b)The student repeats the investigation with a filament lamp and obtains one further pair of readings: V = 4.0 V, I = 0.50 A. Calculate the resistance of the filament lamp at this potential difference.(2)

(c)Explain why the resistance of the filament lamp calculated in part (b) is greater than the resistance of the resistor at the same potential difference (4.0 V, in part (a)).(1)

(Total for Question 19 is 6 marks)

Higher tier only. Figure 5 shows a 10 ohm resistor (R1) connected in series with a parallel combination of a 20 ohm resistor (R2) and a 30 ohm resistor (R3). The circuit is connected to a 12 V supply.

(a)Show that the combined resistance of R2 and R3 in parallel is 12 ohm.(2)

(b)Calculate the total resistance of the circuit.(1)

(c)Calculate the current supplied by the 12 V supply, giving your answer to 2 significant figures.(2)

(d)Calculate the potential difference across R1.(1)

(Total for Question 20 is 6 marks)

Mark scheme · P2a Circuits, Resistance and Components

Question 1

(a) B1 variable resistor (accept rheostat)
(a) Answer: Variable resistor
(b) B1 light-dependent resistor (accept LDR)
(b) Answer: Light-dependent resistor (LDR)
(c) B1 cell (single cell, not a battery)
(c) Answer: Cell

Question 2

(a) B1 180 s (3.0 x 60)
(a) Answer: 180 s
(b) M1 substitutes: Q = 2.5 x 180
(b) A1 450 C cao
(b) Answer: 450 C

Question 3

(a) B1 series
(a) Answer: Series
(b) B1 parallel
(b) Answer: Parallel
(c) B1 they are the same (equal), because components connected in parallel are connected across the same two points, so share the same potential difference, oe
(c) Answer: They are the same, because the lamp and resistor are connected across the same two points.

Question 4

(a) M1 substitutes: V = 0.40 x 15
(a) A1 6.0 V cao
(a) Answer: 6.0 V
(b) M1 rearranges and substitutes: R = 12 / 0.80
(b) A1 15 ohm cao
(b) Answer: 15 ohm

Question 5

(a) B1 20 ohm (8 + 12)
(a) Answer: 20 ohm
(b) M1 ft rearranges and substitutes: I = 10 / 20
(b) A1 0.50 A cao (ft from (a))
(b) Answer: 0.50 A
(c) B1 0.50 A ft from (b)
(c) B1 reason: current is the same at every point in a series circuit, because there is only one path for charge to flow, oe
(c) Answer: 0.50 A, because there is only one loop, so the current is the same everywhere in a series circuit.

Question 6

(a) B1 10 V
(a) B1 reason: components connected in parallel share the same potential difference as the supply, oe
(a) Answer: 10 V, the same as the battery, because parallel branches share the supply potential difference.
(b) M1 substitutes: I = 10 / 20
(b) A1 0.50 A cao
(b) Answer: 0.50 A
(c) M1 ft adds branch currents: 0.50 + 2.0
(c) A1 2.5 A cao (ft from (b))
(c) Answer: 2.5 A

Question 7

(a) B1 a straight line through the origin
(a) Answer: A straight line through the origin.
(b) B1 the resistance stays constant, oe
(b) B1 because current is directly proportional to potential difference (a straight line through the origin), oe
(b) Answer: The resistance stays constant, because current is directly proportional to potential difference.

Question 8

(a) B1 a curve that becomes less steep (bends towards the potential difference axis) as potential difference increases, oe
(a) Answer: A curve that gets less steep as the potential difference increases.
(b) B1 as current increases, the filament gets hotter
(b) B1 the resistance of the filament increases as its temperature increases
(b) B1 so for each additional volt, a smaller increase in current occurs, curving the graph, oe
(b) Answer: The filament heats up as current increases; its resistance rises with temperature, so the current increases less for each extra volt.

Question 9

(a) B1 almost no current flows until a threshold (turn-on) voltage is reached, then current increases rapidly, oe
(a) Answer: Almost no current flows until a threshold voltage is reached, then current rises rapidly.
(b) B1 almost no (negligible) current flows, because the diode has a very high resistance in this direction, oe
(b) Answer: Almost no current flows; the diode has a very high resistance in reverse bias.
(c) B1 use, e.g. in a rectifier circuit converting a.c. to d.c.
(c) B1 linked reason: the diode only allows current to flow in one direction, so it can be used to block current flowing the wrong way, oe
(c) Answer: Used in rectifier circuits, because a diode only allows current to flow in one direction.

Question 10

(a) B1 voltmeter, connected in parallel across the length of wire being tested
(a) B1 ammeter, connected in series in the circuit
(a) Answer: Voltmeter (parallel, across the test length); ammeter (series, in the main circuit).
(b) B1 independent variable: length of the wire
(b) B1 dependent variable: resistance of the wire (calculated from V and I)
(b) Answer: Independent: length of wire. Dependent: resistance of the wire.
(c) B1 e.g. the temperature of the wire, or the current kept low (so the wire does not heat up)
(c) B1 reason: resistance changes with temperature, so an uncontrolled temperature change would introduce an error unrelated to length, oe
(c) Answer: The temperature of the wire, because resistance changes with temperature and this would introduce an error unrelated to length.
(d) M1 correct method R = V / I applied to at least one length
(d) A1 all three correct: 2.0 ohm, 4.0 ohm, 6.0 ohm
(d) B1 resistance is directly proportional to length (as length doubles, resistance doubles), oe
(d) Answer: 2.0 ohm, 4.0 ohm, 6.0 ohm; resistance is directly proportional to length.

Question 11

(a) B1 any two of: power supply, ammeter, voltmeter, variable resistor, connecting leads
(a) B1 the variable resistor changes the current (and potential difference) in the circuit, so several pairs of readings can be taken across a range of values, oe
(a) Answer: Power supply, ammeter, voltmeter, connecting leads; the variable resistor changes the current so a range of readings can be taken.
(b) B1 to check whether the lamp behaves in the same way when the potential difference is reversed (to see if the graph is symmetrical about the origin), oe
(b) Answer: To check whether the lamp behaves the same way with the current reversed, so the graph can be checked for symmetry.
(c) B1 take each reading quickly (do not leave the lamp switched on for long before recording current and potential difference)
(c) B1 reason: the lamp heats up over time, changing its resistance, which would affect the measured relationship, oe
(c) Answer: Take readings quickly, because the lamp heats up over time and its resistance changes, affecting the results.
(d) B1 the resistor's graph is a straight line through the origin, symmetrical in both directions; the diode only conducts appreciably in one direction (forward bias), oe
(d) Answer: The resistor gives a straight line through the origin in both directions; the diode only conducts in one direction.

Question 12

(a) B1 resistance decreases
(a) Answer: Resistance decreases.
(b) B1 as light intensity falls (it gets dark), the resistance of the LDR increases
(b) B1 this decreases the current in the circuit, and the change in current (or potential difference) is used to trigger a switch that turns the light on, oe
(b) Answer: As it gets dark, the LDR's resistance increases, decreasing the current; this change triggers a switch to turn the light on.

Question 13

(a) B1 resistance decreases
(a) Answer: Resistance decreases.
(b) B1 e.g. a car engine temperature sensor, a digital thermostat, or a fire/frost alarm
(b) B1 because its resistance changes in a predictable way with temperature, allowing temperature to be monitored or a circuit to respond to it, oe
(b) Answer: E.g. a digital thermostat, because a thermistor's resistance changes predictably with temperature.

Question 14

Level 1 (1-2): One or two isolated, correct statements are made, for example that thermistor resistance changes with temperature, with little or no reference to current, potential difference or the switching action.
Level 2 (3-4): Describes that the thermistor's resistance increases as temperature falls, and links this to a change in current or potential difference somewhere in the circuit, but the explanation of how this triggers the relay/heater is only partial or lacks clear logical sequencing.
Level 3 (5-6): A clear, logically sequenced explanation that links the fall in temperature to an increase in the thermistor's resistance, to an increase in the total resistance of the series circuit, to a decrease in the current in the circuit, to a resulting change in potential difference across the fixed resistor, and finally to this change being what triggers the relay to switch the heater on. Correct scientific terminology is used throughout.
Indicative content:
- as the greenhouse temperature falls, the resistance of the thermistor increases
- the thermistor and fixed resistor are in series, so the total resistance of the circuit increases
- since the supply potential difference is constant, and current = potential difference / total resistance, the current in the circuit decreases as total resistance increases
- components in series share the supply potential difference between them, so as the current changes, the potential difference across the fixed resistor also changes
- this change in current or potential difference is what is detected by the relay/switching part of the circuit
- when the change reaches a threshold, the relay switches the heater circuit on
- the direction of the resistance change (increases as temperature falls) must be linked correctly for full credit

Question 15

(a) M1 substitutes: 1/R = 1/4 + 1/12
(a) M1 combines correctly: 1/R = 3/12 + 1/12 = 4/12 = 1/3
(a) A1 3.0 ohm cao
(a) Answer: 3.0 ohm
(b) B1 9.0 ohm ft (6 + their (a))
(b) Answer: 9.0 ohm
(c) M1 ft substitutes: I = 12 / 9.0
(c) A1 1.3 A (awrt, 2 s.f.) ft
(c) Answer: 1.3 A

Question 16

(a) B1 total resistance decreases
(a) Answer: Total resistance decreases.
(b) B1 adding a second resistor in parallel provides an additional path for charge to flow, oe
(b) B1 this means a larger total current can flow from the supply for the same potential difference, which is equivalent to a lower total resistance, oe
(b) Answer: There are now two paths for charge to flow, so a larger total current flows for the same potential difference, meaning the total resistance is lower.
(c) B1 total current increases
(c) Answer: Total current increases.

Question 17

(a) B1 60 ohm (45 + 15)
(a) Answer: 60 ohm
(b) M1 ft substitutes: I = 6.0 / 60
(b) A1 0.10 A cao (ft)
(b) Answer: 0.10 A
(c) M1 new total resistance = 15 + 15 = 30 ohm
(c) dM1 new current I = 6.0 / 30 = 0.20 A
(c) A1 correct comparison: potential difference across the fixed resistor increases, from 1.5 V (0.10 x 15) to 3.0 V (0.20 x 15), as temperature rises, oe
(c) Answer: New current = 0.20 A; the potential difference across the fixed resistor increases from 1.5 V to 3.0 V as temperature rises.

Question 18

(a) B1 in series, in the main circuit loop, so all the current to be measured flows through it, oe
(a) Answer: In series in the main circuit loop.
(b) B1 in parallel, directly across the component, oe
(b) Answer: In parallel, directly across the component.
(c) B1 the ammeter (very low resistance) connected in parallel would short-circuit/bypass the resistor, giving a much larger current reading than the true value in the resistor; and/or the voltmeter (very high resistance) connected in series would greatly reduce the current in the circuit, giving an inaccurate (near-zero) reading, oe
(c) Answer: The low-resistance ammeter in parallel would bypass the resistor, and the high-resistance voltmeter in series would drastically reduce the current, so neither reading would be accurate.

Question 19

(a) M1 calculates R for one pair, e.g. R = 1.0 / 0.20 = 5.0 ohm
(a) M1 calculates R for a second pair, e.g. R = 4.0 / 0.80 = 5.0 ohm
(a) A1 both values are equal (5.0 ohm), so resistance is constant, since current is directly proportional to potential difference, cso
(a) Answer: 5.0 ohm for both pairs, showing constant resistance.
(b) M1 substitutes: R = 4.0 / 0.50
(b) A1 8.0 ohm cao
(b) Answer: 8.0 ohm
(c) B1 the filament lamp gets hot at higher current/potential difference, and the resistance of a filament lamp increases as its temperature increases, unlike the resistor, whose temperature (and resistance) stays constant, oe
(c) Answer: The filament lamp heats up at higher pd, and its resistance rises with temperature, unlike the resistor's constant resistance.

Question 20

(a) M1 substitutes: 1/R = 1/20 + 1/30 = 3/60 + 2/60 = 5/60
(a) A1 R = 12 ohm cso (answer given)
(a) Answer: 12 ohm (shown)
(b) B1 22 ohm (10 + 12)
(b) Answer: 22 ohm
(c) M1 ft substitutes: I = 12 / 22
(c) A1 0.55 A (awrt, 2 s.f.) ft
(c) Answer: 0.55 A
(d) B1 5.5 V (awrt, ft: their (c) x 10)
(d) Answer: 5.5 V

Circuits, Resistance and Components - Worksheets, Questions and Revision