Quickfire: for each question, identify the one correct answer.
(a)Identify the correct SI unit for resistivity.(1)
A) ohm
B) ohm metre
C) ohm per metre
D) ohm metre squared
(b)Identify the equation that correctly defines electrical power P in terms of current I and potential difference V.(1)
A) P = I V
B) P = I / V
C) P = V / I
D) P = I + V
(c)Identify which statement correctly defines the volt.(1)
A) the potential difference across a component when 1 coulomb of charge transfers 1 joule of energy
B) the current in a component when 1 joule of energy is transferred per second
C) the charge that flows in 1 second when the current is 1 ampere
D) the resistance of a component that produces 1 watt of power per ampere
(d)Identify which component has a resistance that decreases as its temperature increases.(1)
A) a resistor at constant current
B) an NTC thermistor
C) a filament lamp
D) a length of copper wire carrying an increasing current
(e)Identify the correct expression for the combined resistance R of two resistors R1 and R2 connected in parallel.(1)
A) R = R1 + R2
B) R = (R1 + R2) / (R1 R2)
C) R = (R1 R2) / (R1 + R2)
D) R = R1 - R2
(Total for Question 1 is 5 marks)
2
A student uses a data logger to measure the charge that flows through a filament lamp over a short period of time after it is switched on.
(a)State what is meant by electric current.(1)
(b)A charge of 480 C flows through the lamp in 4.0 minutes. Calculate the current in the lamp. Use Q = I t.(3)
(c)The charge on one electron is 1.60 x 10-19 C. Calculate the number of electrons that pass a point in the filament each second.(3)
(Total for Question 2 is 7 marks)
3
A piece of copper wire used in a laboratory power supply has a cross-sectional area of 2.0 mm2. The number density of free (charge-carrying) electrons in copper is 8.5 x 1028 m-3.
(a)The current in the wire is 3.0 A. Calculate the mean drift velocity of the free electrons. Use I = n A v q, where q is the charge on an electron (1.60 x 10-19 C).(3)
(b)Explain why the drift velocity of the electrons would be greater in a thinner wire of the same material carrying the same current.(2)
(c)The wire is 5.0 m long and has a diameter of 0.60 mm. The resistivity of copper is 1.7 x 10-8 ohm metre. Calculate the resistance of the wire. Use R = resistivity x L / A.(3)
(Total for Question 3 is 8 marks)
4
A 3.0 kW electric kettle is connected to the UK 230 V mains supply.
(a)Calculate the current drawn by the kettle when it is operating normally. Use P = I V.(2)
(b)Calculate the energy transferred by the kettle in 4.0 minutes of continuous use. Use E = P t.(2)
(c)Electricity costs 28p per kilowatt-hour. Show that the cost of using the kettle for 4.0 minutes is about 5.6p.(3)
(d)State one advantage, in terms of energy cost, of only boiling the amount of water actually needed rather than filling the kettle completely.(1)
(Total for Question 4 is 8 marks)
5
Required practical: a student investigates the I-V characteristic of a filament lamp using a circuit containing a variable resistor connected as a potential divider across the battery, with the lamp, an ammeter and a voltmeter connected appropriately to obtain a series of current and potential difference readings across the lamp.
(a)Describe how the student could use the potential divider arrangement to obtain a full set of I-V readings for the lamp, including readings for both directions of current through the lamp.(4)
(b)Two of the student's readings are: at V = 2.0 V, I = 0.40 A; at V = 6.0 V, I = 0.60 A. Calculate the resistance of the lamp at each of these two pd values, and state how the resistance of the lamp changes between them.(3)
(c)Explain, in terms of the filament, why the resistance of the lamp increases as the current through it increases.(2)
(d)The student then investigates the I-V characteristic of a silicon diode using the same method. Explain why a protective resistor should be included in series with the diode, and suggest one other precaution the student should take.(3)
(Total for Question 5 is 12 marks)
6
Required practical: a student determines the resistivity of a sample of nichrome resistance wire of diameter 0.32 mm. The student measures the resistance R of different lengths L of the wire, using a crocodile clip to select each length, and plots a graph of R against L.
(a)Describe how the student should carry out this experiment to obtain a reliable set of R and L measurements, including how the diameter of the wire is measured.(5)
(b)Two of the student's results are: at L = 0.200 m, R = 2.80 ohm; at L = 1.000 m, R = 13.60 ohm. Show that the gradient of the graph of R against L is approximately 13.5 ohm per metre.(2)
(c)Calculate the resistivity of the nichrome wire, using the gradient found in part (b) and the diameter of the wire (0.32 mm). The gradient of R against L is equal to resistivity/A, where A is the cross-sectional area of the wire.(3)
(d)Suggest one reason why using the gradient of a graph of R against L gives a more accurate value of resistivity than calculating resistivity from a single pair of R and L readings.(2)
(Total for Question 6 is 12 marks)
7
A battery has EMF 6.0 V and internal resistance 0.50 ohm. It is connected to an external resistor of resistance 2.5 ohm. Use the equation EMF = I(R + r), where R is the external resistance and r is the internal resistance.
(a)Calculate the current in the circuit.(3)
(b)Calculate the terminal potential difference of the battery.(2)
(c)Calculate the power dissipated inside the battery, in its internal resistance.(2)
(d)Explain what happens to the terminal potential difference of the battery if the external resistor is replaced with one of much smaller resistance, such as in a short circuit.(3)
(Total for Question 7 is 10 marks)
8
A battery of EMF 12 V with negligible internal resistance is connected in series with a resistor R1 = 4.0 ohm and a parallel combination of two resistors, R2 = 6.0 ohm and R3 = 12 ohm.
(a)Calculate the combined resistance of the parallel combination of R2 and R3.(2)
(b)Calculate the total resistance of the circuit.(1)
(c)Calculate the total current supplied by the battery.(2)
(d)Calculate the potential difference across the parallel combination.(2)
(e)Calculate the current in the R3 branch.(2)
(f)Use Kirchhoff's first law to show that the current in the R2 branch is 1.0 A, and confirm that the currents in R2 and R3 sum to the total current found in part (c).(3)
(Total for Question 8 is 12 marks)
9
A light-dependent resistor (LDR) is connected in series with a fixed resistor R = 3.0 kilohm across a 9.0 V supply, forming a potential divider. The output voltage, Vout, is taken across the LDR. Use Vout = Vin x RLDR / (R + RLDR).
(a)In bright light the resistance of the LDR is 500 ohm. Calculate Vout.(3)
(b)In darkness the resistance of the LDR increases to 12 kilohm. Calculate the new value of Vout.(2)
(c)Explain how this potential divider circuit, together with a transistor switch connected to Vout, could be used to switch a garden lamp on automatically at night.(3)
(d)State and explain what would happen to Vout in darkness (RLDR = 12 kilohm) if the positions of R and the LDR in the circuit were swapped, so that Vout is now taken across R instead.(3)
(Total for Question 9 is 11 marks)
10
Compare and explain the I-V characteristics of three components: a metal wire resistor at constant temperature, a filament lamp, and a silicon diode. Your answer should refer to the effect of temperature (where relevant) and the behaviour of charge carriers in each case.
(Total for Question 10 is 6 marks)
11
A battery of EMF 12 V and internal resistance 1.5 ohm is connected to an external circuit made from two resistors, R1 = 8.0 ohm and R2 = 8.0 ohm, connected in parallel with each other. This parallel combination is connected in series with a third resistor, R3 = 2.0 ohm.
(a)Calculate the combined resistance of R1 and R2 in parallel.(2)
(b)Calculate the total resistance of the complete circuit, including the internal resistance of the battery.(2)
(c)Calculate the current supplied by the battery.(2)
(d)Calculate the terminal potential difference of the battery.(2)
(e)Show that the efficiency of energy transfer from the battery to the external circuit is 80%. Efficiency = useful power delivered to the external circuit / total power supplied by the battery, which is equal to the external resistance divided by the total resistance.(3)
(f)Calculate the current in resistor R1.(2)
(Total for Question 11 is 13 marks)
12
Some materials become superconductors when cooled below a critical temperature, at which their electrical resistance drops suddenly to zero.
(a)State what is meant by the critical temperature of a superconductor.(1)
(b)State two practical applications that make use of superconductors.(2)
(c)A superconducting cable and a copper cable each carry a current of 150 A. The copper cable has a resistance of 0.40 ohm per kilometre. Calculate the power dissipated per kilometre of the copper cable. Use P = I2 R.(2)
(d)Explain why using a superconducting cable rather than a copper cable to transmit electrical power over long distances is more efficient, and suggest one practical difficulty in using superconducting cables for this purpose.(3)
(Total for Question 12 is 8 marks)
Mark scheme · AP5 Electricity
Question 1
(a) B1 correct option identified: B, cao
(a) Answer: B (ohm metre)
(b) B1 correct option identified: A, cao
(b) Answer: A (P = I V)
(c) B1 correct option identified: A, cao
(c) Answer: A
(d) B1 correct option identified: B, cao
(d) Answer: B (NTC thermistor)
(e) B1 correct option identified: C, cao
(e) Answer: C
Question 2
(a) B1 the rate of flow of (positive) charge, oe
(a) Answer: Current is the rate of flow of charge.
(b) B1 convert time to seconds: 4.0 min = 240 s
(b) M1 correct rearrangement I = Q/t and substitution
(b) A1 I = 2.0 A, cao
(b) Answer: 2.0 A
(c) M1 recognise that the number per second n = I / e, oe, ft from (b)
(c) M1 correct substitution: n = 2.0 / (1.60 x 10-19)
(c) A1 1.25 x 1019 electrons per second, awrt, cao
(c) Answer: 1.25 x 1019 electrons per second
Question 3
(a) M1 convert area to m2: 2.0 mm2 = 2.0 x 10-6 m2
(a) M1 correct rearrangement v = I / (n A q) and substitution
(a) A1 v = 1.1 x 10-4 m/s, awrt 2 sf, cao
(a) Answer: 1.1 x 10-4 m/s
(b) B1 a thinner wire has a smaller cross-sectional area A, with n and q unchanged (same material, same charge carrier)
(b) B1 since I = n A v q and I is the same, a smaller A means v must be larger to keep I constant, oe
(b) Answer: The drift velocity increases because A is smaller; since I = nAvq is fixed and n and q are unchanged, v must increase to compensate.
(c) M1 correct area from diameter: A = π(d/2)2 = 2.8 x 10-7 m2
(c) M1 correct substitution into R = resistivity x L / A
(c) A1 R = 0.30 ohm, awrt 2 sf, cao
(c) Answer: 0.30 ohm
Question 4
(a) M1 correct rearrangement I = P/V and substitution
(a) A1 13.0 A, awrt, cao
(a) Answer: 13.0 A
(b) M1 convert time to seconds (240 s) and substitute into E = Pt
(b) A1 7.2 x 105 J (720 kJ), cao
(b) Answer: 7.2 x 105 J (720 kJ)
(c) M1 energy in kWh = power (kW) x time (h) = 3.0 x (4.0/60)
(c) M1 = 0.20 kWh
(c) A1 cso: cost = 0.20 x 28 = 5.6p
(c) Answer: 5.6p (shown)
(d) B1 less water means less energy is needed to heat it, so less energy is transferred/wasted and the cost is lower, oe
(d) Answer: Boiling less water needs less energy, so the cost of heating it is lower.
Question 5
(a) B1 moving the wiper of the potential divider varies the pd applied to the lamp smoothly from zero up to the supply value, oe
(a) B1 at each setting, record the ammeter reading (current) and the voltmeter reading (pd across the lamp)
(a) B1 reverse the connections to the lamp (or swap the battery terminals) to obtain readings with current in the opposite direction
(a) B1 repeat each reading (or take readings at closely spaced intervals) and/or allow the lamp to cool between readings to improve reliability
(a) Answer: Vary the pd smoothly using the potential divider, recording I and V at each setting; reverse the lamp connections to obtain negative pd/current readings; repeat readings for reliability.
(b) B1 R = V/I at 2.0 V: R = 5.0 ohm, cao
(b) B1 R = V/I at 6.0 V: R = 10.0 ohm, cao
(b) B1 correctly states resistance increases as the pd/current increases, oe
(b) Answer: 5.0 ohm at 2.0 V; 10.0 ohm at 6.0 V; resistance increases as pd (and current) increase.
(c) B1 as current increases, the temperature of the filament increases
(c) B1 the (lattice) ions vibrate with greater amplitude, so drifting electrons collide with them more frequently, increasing resistance, oe
(c) Answer: Higher current heats the filament; the hotter lattice ions vibrate more, causing more frequent collisions with drifting electrons, which increases resistance.
(d) B1 once forward biased beyond its threshold voltage, the diode's resistance becomes very small, so current would rise very rapidly for a small increase in pd
(d) B1 a protective resistor limits the current, preventing damage to the diode (or exceeding the range of the ammeter), oe
(d) B1 precaution: avoid leaving a large forward current flowing for long periods, to prevent the diode overheating (or use a low-voltage supply), oe
(d) Answer: A protective resistor limits the rapidly rising current once the diode conducts, preventing damage; the student should also avoid prolonged large currents that could overheat the diode.
Question 6
(a) B1 measure the diameter of the wire using a micrometer (or digital calliper) at several points along the wire and at different orientations, then calculate a mean, oe
(a) B1 clamp the wire straight along a metre ruler so it is taut but not stretched, to allow accurate length measurement
(a) B1 use a crocodile clip to select each length L of wire in the circuit, with an ammeter (in series) and voltmeter (in parallel across the length in circuit) to find R at each L
(a) B1 keep the current low, or switch off the circuit between readings, so the wire does not heat up (which would change its resistance/resistivity)
(a) B1 repeat each measurement of R (at each L) and calculate a mean to reduce random error / improve reliability
(a) Answer: Measure the wire's diameter with a micrometer (mean of several readings); clamp it taut against a ruler; use a crocodile clip and ammeter/voltmeter to find R at each L; keep current low to avoid heating; repeat readings for reliability.
(c) M1 correct area from diameter: A = π(0.16 x 10-3)2 = 8.0 x 10-8 m2
(c) M1 resistivity = gradient x A, ft from (b)
(c) A1 1.1 x 10-6 ohm metre, awrt 2 sf, cao (ft)
(c) Answer: 1.1 x 10-6 ohm metre
(d) B1 using the gradient removes the effect of a constant systematic error, such as contact resistance at the crocodile clip, because this appears as a y-intercept rather than affecting the gradient, oe
(d) B1 plotting several points and drawing a line of best fit averages out random errors in individual readings, oe
(d) Answer: The gradient method removes constant systematic errors (e.g. contact resistance, which shows up as an intercept), and averaging several points via a line of best fit reduces the effect of random error.
Question 7
(a) M1 correct substitution: 6.0 = I(2.5 + 0.50)
(a) M1 correct rearrangement I = 6.0/3.0
(a) A1 I = 2.0 A, cao
(a) Answer: 2.0 A
(b) M1 correct method: V = IR (or V = EMF - Ir), ft from (a)
(b) A1 5.0 V, cao (ft)
(b) Answer: 5.0 V
(c) M1 correct substitution into P = I2 r, ft from (a)
(c) A1 2.0 W, cao (ft)
(c) Answer: 2.0 W
(d) B1 a smaller external resistance R means a smaller total resistance (R + r), so the current I increases, oe
(d) B1 the pd across the internal resistance (the 'lost volts', Ir) therefore increases
(d) B1 since terminal pd = EMF - Ir, the terminal pd decreases, approaching zero as R approaches zero (short circuit), oe
(d) Answer: As R decreases, current increases, so the lost volts (Ir) increase; since terminal pd = EMF - Ir, the terminal pd falls, tending to zero in a short circuit.
Question 8
(a) M1 correct substitution: 1/Rp = 1/6.0 + 1/12
(a) A1 Rp = 4.0 ohm, cao
(a) Answer: 4.0 ohm
(b) B1 8.0 ohm, cao (ft from (a): R1 + Rp)
(b) Answer: 8.0 ohm
(c) M1 correct substitution I = V/Rtotal, ft from (b)
(c) A1 1.5 A, cao (ft)
(c) Answer: 1.5 A
(d) M1 correct substitution V = I Rp, ft from (a) and (c)
(d) A1 6.0 V, cao (ft)
(d) Answer: 6.0 V
(e) M1 correct substitution I3 = V/R3, ft from (d)
(e) A1 0.50 A, cao (ft)
(e) Answer: 0.50 A
(f) M1 I2 = V/R2 = 6.0/6.0, ft from (d)
(f) A1 cso: I2 = 1.0 A
(f) B1 1.0 + 0.50 = 1.5 A, equal to the total current in (c), confirming Kirchhoff's first law (current entering a junction equals current leaving it)
(f) Answer: I2 = 1.0 A; 1.0 + 0.50 = 1.5 A, matching the total current in part (c).
Question 9
(a) M1 total resistance R + RLDR = 3000 + 500 = 3500 ohm
(a) M1 correct substitution into Vout = 9.0 x 500/3500
(a) A1 1.3 V, awrt 2 sf, cao
(a) Answer: 1.3 V
(b) M1 correct substitution into Vout = 9.0 x 12000/15000
(b) A1 7.2 V, cao
(b) Answer: 7.2 V
(c) B1 as it gets dark, RLDR increases, so Vout (taken across the LDR) increases, as shown in parts (a) and (b), oe
(c) B1 when Vout rises above the transistor's base-emitter threshold voltage (about 0.7 V), the transistor switches on/starts to conduct
(c) B1 the transistor then allows current to flow in the output circuit (e.g. via a relay), switching on the lamp
(c) Answer: As darkness falls, RLDR (and so Vout) increases; once Vout exceeds the transistor's threshold voltage, the transistor conducts and switches on the lamp circuit.
(d) B1 Vout would now be small in darkness (rather than large), because it is taken across R rather than the LDR, oe
(d) B1 reasoning: as RLDR increases, the larger share of the supply pd is now across the LDR, leaving a smaller share across R (Vout), oe
(d) B1 correct supporting calculation: Vout = 9.0 x 3000/15000 = 1.8 V (ecf from correct method)
(d) Answer: Vout would fall to about 1.8 V in darkness (rather than rising), because Vout is now across the fixed resistor R, and R's share of the total pd decreases as RLDR increases.
Question 10
Level 3 (5-6): A full, coherent comparison of all three components, correctly describing each I-V characteristic (straight line through the origin for the resistor; curve of decreasing gradient for the lamp; negligible current until a threshold voltage then a rapid rise for the diode) and explaining the underlying physics for each (constant lattice vibration/collision rate at constant temperature; increased lattice vibration and collision rate with the drift electrons as the filament heats up; charge carriers only crossing the p-n junction in significant numbers once forward biased beyond the threshold voltage), using accurate specialist terminology throughout.
Level 2 (3-4): A correct description of the I-V behaviour of at least two of the three components, with a partial physical explanation (for example, linking temperature change to resistance change for the lamp, or describing the diode's threshold behaviour) for at least one. The answer has some logical structure and is mostly clear.
Level 1 (1-2): Basic, largely descriptive statements about the shape of one or two of the I-V graphs, with little or no physical explanation in terms of charge carriers or lattice vibration. The answer may lack clarity.
Level 0 (0): No relevant content, or the answer does not relate to the question.
Indicative content:
A metal wire resistor kept at constant temperature gives a straight-line I-V graph through the origin, showing that current is directly proportional to pd (Ohm's law); resistance is constant because temperature is constant, so the lattice ions vibrate with the same amplitude and the collision rate with drifting electrons does not change.
A filament lamp gives a curve whose gradient decreases as I and V increase, so resistance increases with current; as current increases, the filament's temperature rises, the lattice ions vibrate with greater amplitude, and this increases the rate of collision between the ions and the drifting electrons, increasing resistance.
A silicon diode carries almost no current in reverse bias, and negligible current in forward bias until a threshold (turn-on) voltage of around 0.6-0.7 V is reached; beyond this threshold, current increases very rapidly for a small further increase in pd.
The diode's behaviour arises because charge carriers only cross the p-n junction in significant numbers once the forward bias is large enough to overcome the potential barrier at the junction; below threshold, effectively no charge carriers have enough energy to cross, so almost no current flows.
Unlike the resistor and the lamp, the diode's I-V characteristic is not symmetrical about the origin, reflecting the fact that it allows current to flow easily in only one direction.
Question 11
(a) M1 correct substitution: Rp = (8.0 x 8.0)/(8.0 + 8.0), oe
(a) A1 4.0 ohm, cao
(a) Answer: 4.0 ohm
(b) M1 Rtotal = Rp + R3 + r, ft from (a)
(b) A1 7.5 ohm, cao (ft)
(b) Answer: 7.5 ohm
(c) M1 correct substitution I = EMF/Rtotal, ft from (b)
(c) A1 1.6 A, cao (ft)
(c) Answer: 1.6 A
(d) M1 correct method: terminal pd = EMF - I r, ft from (c)
(d) A1 9.6 V, cao (ft)
(d) Answer: 9.6 V
(e) M1 identify external resistance = Rp + R3 = 4.0 + 2.0 = 6.0 ohm, ft from (a)
(e) M1 efficiency = 6.0/7.5, ft from (b)
(e) A1 cso: = 0.80 = 80%
(e) Answer: 80% (shown)
(f) M1 pd across parallel section = I x Rp = 1.6 x 4.0, ft from (a) and (c)
(f) A1 I1 = 6.4/8.0 = 0.80 A, cao (ft)
(f) Answer: 0.80 A
Question 12
(a) B1 the temperature below which a material's electrical resistance becomes (suddenly) zero, oe
(a) Answer: The temperature below which a material's resistance suddenly becomes zero.
(b) B1 any one valid application, e.g. superconducting magnets in MRI scanners (produce very strong magnetic fields without resistive energy loss)
(b) B1 any second valid application, e.g. powerful electromagnets in particle accelerators, or superconducting cables for (loss-free) power transmission, oe
(b) Answer: Any two of: MRI scanner magnets; particle accelerator electromagnets; superconducting power transmission cables.
(c) M1 correct substitution into P = I2 R = (150)2 x 0.40
(c) A1 9.0 x 103 W (9.0 kW), cao
(c) Answer: 9.0 x 103 W (9.0 kW)
(d) B1 below its critical temperature the superconducting cable has zero resistance, so by P = I2 R no power is dissipated as heat in the cable, unlike the copper cable which continuously loses energy (e.g. 9.0 kW per km, as in part (c)), oe
(d) B1 so a greater proportion of the generated electrical power reaches the consumer / less energy is wasted as heat, oe
(d) B1 practical difficulty: the cable must be cooled continuously to below its critical temperature (e.g. using liquid nitrogen or liquid helium), which requires energy and specialist equipment, adding cost, oe
(d) Answer: The superconducting cable has zero resistance so no power is lost as heat (unlike the 9.0 kW/km lost in the copper cable), meaning more power reaches the consumer; however, the cable must be continuously cooled below its critical temperature, which is costly and requires specialist cryogenic equipment.