A student sets up a simple series circuit containing a 6.0 V cell, a switch and a lamp. When the switch is closed, a current of 0.40 A flows through the lamp for 3.0 minutes.
(a)An ammeter is used to measure the current in the circuit. Which statement about an ideal ammeter is correct?(1)
A) It has zero resistance and is connected in series
B) It has infinite resistance and is connected in series
C) It has zero resistance and is connected in parallel
D) It has infinite resistance and is connected in parallel
(b)Use the equation: charge = current x time (Q = I x t). Calculate the charge that flows through the lamp in 3.0 minutes.(3)
(Total for Question 1 is 4 marks)
2
UK mains electricity is supplied to homes as alternating current (a.c.).
(a)State the frequency of the UK mains supply.(1)
(b)State the typical potential difference of the UK mains supply.(1)
(Total for Question 2 is 2 marks)
3
Figure 2 shows three identical lamps connected in series with a 12 V battery.
(a)The current through lamp 1 is 0.50 A. State the current through lamp 2.(1)
(b)The total resistance of the circuit is 24 ohm. Use the equation V = I x R to calculate the current supplied by the battery.(3)
(Total for Question 3 is 4 marks)
4
Figure 3 shows two resistors connected in parallel across a 9.0 V battery. The current in the first branch is 1.5 A and the current in the second branch is 0.75 A.
(a)State the potential difference across the second resistor.(1)
(b)Calculate the total current supplied by the battery.(2)
(Total for Question 4 is 3 marks)
5
A polythene rod is rubbed with a dry cloth. The rod becomes negatively charged.
(a)Explain, in terms of electrons, why the rod becomes negatively charged.(2)
(b)The charged rod is brought close to a thin stream of water and the stream bends towards the rod. Suggest why this happens.(2)
(Total for Question 5 is 4 marks)
6
A resistor is connected to a 6.0 V power supply. A current of 0.24 A flows through it.
(a)Use the equation V = I x R to calculate the resistance of the resistor.(2)
(b)The resistor is replaced with one of resistance 40 ohm, connected to the same 6.0 V supply. Calculate the new current, and state how the brightness of a lamp connected in series with this resistor would compare to using the 25 ohm resistor in part (a).(3)
(Total for Question 6 is 5 marks)
7
A hairdryer is rated at 230 V and draws a current of 8.7 A when in use.
(a)The hairdryer transfers electrical energy usefully to the surroundings by heating the air, and wastefully as sound. Which statement about this energy transfer is correct?(1)
A) Energy is destroyed as sound is produced
B) The total energy transferred equals the useful energy transferred only
C) The total energy transferred equals the sum of the useful and wasted energy transferred
D) Energy is created inside the heating element
(b)Use the equation P = V x I to calculate the power of the hairdryer.(3)
(Total for Question 7 is 4 marks)
8
A student carries out the required practical to investigate the I-V characteristics of a resistor and a filament lamp. The circuit contains a low-voltage power supply, a variable resistor, an ammeter, a voltmeter and the component under test.
(a)State the correct positions of the ammeter and the voltmeter in this circuit.(2)
(b)Describe the shape of the I-V graph obtained for a resistor at constant temperature.(2)
(c)The student then tests a filament lamp. Explain why the I-V graph for the filament lamp curves so that the current increases less rapidly as the potential difference increases.(3)
(Total for Question 8 is 7 marks)
9
A thermistor is connected into a control circuit used to switch on a greenhouse heater automatically.
(a)State how the resistance of a thermistor changes as its temperature increases.(1)
(b)Explain how the thermistor's changing resistance can be used to switch on the heater automatically when the greenhouse temperature falls.(3)
(Total for Question 9 is 4 marks)
10
The student repeats the I-V required practical using a diode in place of the resistor.
(a)Describe how the current through a diode changes as the potential difference in the forward direction is increased from 0 V.(2)
(b)The diode is then connected in reverse bias. Explain what happens to the current, and state one practical use of this property.(3)
(Total for Question 10 is 5 marks)
11
A student investigates how the length of a wire affects its resistance. For a 50 cm length of the wire, the graph of potential difference against current is a straight line through the origin. At one point on the line, the current is 0.20 A when the potential difference is 1.6 V.
(a)In this investigation, identify the independent variable and the dependent variable.(2)
(b)The gradient of a potential difference against current graph is equal to resistance. Show that the resistance of the 50 cm wire is 8.0 ohm.(3)
(c)The student repeats the experiment using a wire of the same material and cross-sectional area but double the length. Predict, with a reason, how the resistance of this wire compares with the 50 cm wire.(3)
(Total for Question 11 is 8 marks)
12
A kettle's heating element has resistance 20 ohm and carries a current of 11.5 A.
(a)Use the equation P = I2 x R to calculate the power output of the heating element.(3)
(b)Use the equation P = V2 / R, together with your answer to part (a), to show that this is consistent with a potential difference of about 230 V across the element.(3)
(Total for Question 12 is 6 marks)
13
An electric shower has a power rating of 9.5 kW and is used for 15 minutes each day.
(a)Use the equation E = P x t to calculate the energy transferred by the shower in one day. Give your answer in kWh.(2)
(b)Electricity costs 28p per kWh. Calculate the total cost, in pounds sterling, of using the shower every day for one week (7 days). Give your answer to 2 decimal places.(4)
(Total for Question 13 is 6 marks)
14
A washing machine is rated at 230 V, 2.8 kW and is connected to the mains supply using a three-core cable.
(a)State the function of the earth wire and the function of the live wire in a three-core cable.(2)
(b)Use the equation P = V x I to calculate the normal operating current of the washing machine, and hence select the most appropriate fuse rating from 3 A, 5 A and 13 A.(3)
(Total for Question 14 is 5 marks)
15
Figure 4 shows a circuit with a battery connected to a 4.0 ohm resistor (R1) in series with a section containing two resistors, R2 and R3, connected in parallel with each other.
(a)The current from the battery is 1.5 A. Use V = I x R to calculate the potential difference across R1.(3)
(b)The battery's potential difference is 12 V. Calculate the potential difference across the parallel section of the circuit.(3)
(c)The current through R2 is 0.90 A. Calculate the resistance of R2, then calculate the current through R3.(3)
(Total for Question 15 is 9 marks)
16
Fuel tankers can become charged with static electricity as fuel flows through the delivery pipe during refuelling. If enough charge builds up, a spark could occur and ignite flammable fuel vapour, causing an explosion. Explain how the tanker becomes charged, and describe two methods used to reduce the risk of a spark occurring during refuelling.
(Total for Question 16 is 6 marks)
17
A negatively charged sphere is isolated in space. A second, much smaller object with a positive charge is placed near the sphere.
(a)Describe the shape of the electric field around the isolated charged sphere.(2)
(b)Explain, in terms of electric fields, why the positively charged object experiences a force towards the sphere.(2)
(Total for Question 17 is 4 marks)
18
A step-down transformer at a local substation has 8000 turns on the primary coil and 400 turns on the secondary coil. The primary coil is connected to the National Grid at a potential difference of 22000 V.
(a)Use the equation Vp / Vs = Np / Ns to calculate the potential difference across the secondary coil.(3)
(b)The current in the secondary coil is 45 A. Use the equation Vp x Ip = Vs x Is to calculate the current in the primary coil, and explain one advantage of transmitting electrical energy through the National Grid at high voltage rather than low voltage.(4)
(Total for Question 18 is 7 marks)
19
Electrical power of 2.0 MW is transmitted along a cable of total resistance 5.0 ohm. Option A transmits this power using a current of 100 A. Option B transmits the same power using a current of 10 A (achieved by using a higher transmission voltage). Use the equation P = I2 x R to show that the power wasted as heat in the cable is much greater for Option A than for Option B, and state why the National Grid transmits power at high voltage.
(Total for Question 19 is 6 marks)
20
A technician measures the resistance of a wire sample at different temperatures as part of a quality check for a heating element, obtaining the results shown: at 20 C the resistance is 18.0 ohm; at 100 C the resistance is 24.3 ohm.
(a)Calculate the percentage increase in resistance between 20 C and 100 C.(3)
(b)The wire is connected to a 12 V supply at 100 C. Use I = V / R to calculate the current, then use P = V x I to calculate the power dissipated at this temperature.(4)
(Total for Question 20 is 7 marks)
Mark scheme · P2 Electricity
Question 1
(a) B1 A - zero resistance, connected in series
(a) Answer: A
(b) M1 converts time to seconds: t = 3.0 x 60 = 180 s
(b) M1 substitutes correctly: Q = 0.40 x 180
(b) A1 Q = 72 C cao
(b) Answer: 72 C
Question 2
(a) B1 50 Hz
(a) Answer: 50 Hz
(b) B1 230 V
(b) Answer: 230 V
Question 3
(a) B1 0.50 A (current is the same at all points in a series circuit)
(a) Answer: 0.50 A
(b) M1 rearranges to I = V / R
(b) M1 substitutes: I = 12 / 24
(b) A1 I = 0.50 A cao
(b) Answer: 0.50 A
Question 4
(a) B1 9.0 V (p.d. is the same across parallel branches, equal to the battery p.d.)
(a) Answer: 9.0 V
(b) M1 adds branch currents: 1.5 + 0.75
(b) A1 2.25 A cao
(b) Answer: 2.25 A
Question 5
(a) B1 friction between rod and cloth transfers electrons from the cloth to the rod
(a) B1 rod gains an excess of electrons, giving it an overall negative charge oe
(a) Answer: Electrons are transferred from the cloth to the rod by friction, giving the rod an excess of electrons (negative charge).
(b) B1 the charged rod induces an opposite charge on the near side of the water oe
(b) B1 unlike charges attract, so the water is pulled towards the rod
(b) Answer: The rod's electric field induces an opposite charge in the water, and since unlike charges attract, the stream bends towards the rod.
Question 6
(a) M1 rearranges and substitutes: R = 6.0 / 0.24
(a) A1 R = 25 ohm cao
(a) Answer: 25 ohm
(b) M1 I = 6.0 / 40
(b) A1 I = 0.15 A cao
(b) B1 ft the lamp would be dimmer, because the current is smaller than in part (a) (0.15 A < 0.24 A)
(b) Answer: 0.15 A; dimmer than with the 25 ohm resistor
Question 7
(a) B1 C - conservation of energy: total = useful + wasted
(a) Answer: C
(b) M1 substitutes: P = 230 x 8.7
(b) A1 P = 2001 W (awrt 2000 W or 2.0 kW)
(b) Answer: 2000 W (2.0 kW)
Question 8
(a) B1 ammeter connected in series with the component under test
(a) B1 voltmeter connected in parallel across the component under test
(a) Answer: Ammeter in series with the component; voltmeter in parallel across the component.
(b) B1 a straight line
(b) B1 through the origin, showing current is directly proportional to potential difference
(b) Answer: A straight line through the origin (current directly proportional to potential difference).
(c) B1 as potential difference increases, current increases and the filament heats up
(c) B1 resistance of the filament increases with temperature
(c) B1 higher resistance means a smaller increase in current for a given increase in p.d., so the graph gradient decreases oe
(c) Answer: The filament heats up as current increases, its resistance rises, so the current increases less for each extra volt, curving the graph.
Question 9
(a) B1 resistance decreases as temperature increases
(a) Answer: Resistance decreases.
(b) B1 as temperature falls, the thermistor's resistance increases
(b) B1 this increases the potential difference across the thermistor in the control circuit oe
(b) B1 the change in p.d./current is used to trigger the switching circuit, turning the heater on
(b) Answer: Falling temperature increases thermistor resistance, changing the p.d./current in the circuit enough to trigger the switch and turn on the heater.
Question 10
(a) B1 almost no current flows until a threshold (turn-on) voltage is reached
(a) B1 above the threshold voltage, current increases rapidly
(a) Answer: Almost no current below a threshold voltage; above it, current rises rapidly.
(b) B1 virtually no (negligible) current flows in reverse bias
(b) B1 because the diode has a very high resistance in the reverse direction
(b) B1 use: to allow current to flow in one direction only, e.g. in a rectifier converting a.c. to d.c. oe
(b) Answer: Negligible current flows (very high resistance in reverse); used in rectifier circuits to convert a.c. to d.c.
Question 11
(a) B1 independent variable: length of the wire
(a) B1 dependent variable: resistance of the wire (or current/p.d. measured)
(a) Answer: Independent: length of wire. Dependent: resistance of the wire.
(b) M1 correct method: R = V / I
(b) M1 substitutes: R = 1.6 / 0.20
(b) A1 R = 8.0 ohm cso (answer given)
(b) Answer: 8.0 ohm (shown)
(c) B1 resistance doubles, to 16 ohm ft
(c) B1 resistance is directly proportional to length oe
(c) B1 a longer wire gives charge carriers more collisions with the fixed ions, increasing resistance oe
(c) Answer: 16 ohm; resistance is directly proportional to length because charge carriers collide with more ions along a longer wire.
Question 12
(a) M1 substitutes: P = 11.52 x 20
(a) A1 P = 2645 W (awrt 2600-2650 W)
(a) Answer: 2645 W (approx. 2.6 kW)
(b) M1 rearranges to V = √P x R
(b) M1 substitutes (using ft power from a): V = √2645 x 20
(b) A1 V = √52900 = 230 V cso, consistent with the stated value
(b) Answer: 230 V (shown)
Question 13
(a) M1 converts time correctly: t = 15/60 = 0.25 h
(a) A1 E = 9.5 x 0.25 = 2.375 kWh (awrt 2.4 kWh)
(a) Answer: 2.375 kWh
(b) M1 ft weekly energy: 2.375 x 7 = 16.625 kWh
(b) M1 cost in pence: 16.625 x 28 = 465.5 p
(b) A1 converts to pounds: 465.5p = GBP 4.655
(b) A1 GBP 4.66 awrt (accept GBP 4.65-4.66)
(b) Answer: GBP 4.66
Question 14
(a) B1 earth wire: a safety wire that connects the metal casing to earth, carrying fault current safely away and preventing electric shock oe
(a) B1 live wire: carries the alternating potential difference (at about 230 V) from the supply oe
(a) Answer: Earth wire: safety wire preventing shock by carrying fault current to earth. Live wire: carries the alternating p.d. from the supply.
(b) M1 rearranges and substitutes: I = 2800 / 230
(b) A1 I = 12.2 A (awrt)
(b) B1 ft selects the 13 A fuse, because the fuse rating must exceed the operating current (5 A would blow immediately)
(b) Answer: 12.2 A; a 13 A fuse should be used
Question 15
(a) M1 identifies that the current through R1 equals the total battery current (series section)
(a) M1 substitutes: V = 1.5 x 4.0
(a) A1 V = 6.0 V cao
(a) Answer: 6.0 V
(b) M1 uses ft p.d. across R1: p.d.s in a series loop add up to the supply p.d.
(b) M1 12 - 6.0
(b) A1 6.0 V ft
(b) Answer: 6.0 V
(c) M1 ft uses V (parallel) / I: R2 = 6.0 / 0.90
(c) A1 R2 = 6.7 ohm (2 s.f.) awrt
(c) B1 ft current through R3 = total current - current through R2 = 1.5 - 0.90 = 0.60 A
(c) Answer: R2 = 6.7 ohm; current through R3 = 0.60 A
Question 16
Level 1 (1-2): Basic statement that friction/movement of fuel causes charging, or a brief mention of one risk-reduction idea, with little or no explanation. Answer may be a simple list.
Level 2 (3-4): Describes the charging mechanism with some detail (e.g. friction/electron transfer as fuel moves through the pipe) and describes at least one risk-reduction method, with some linking to the danger of sparks.
Level 3 (5-6): Clear, logically structured explanation of how friction between the moving fuel and the pipe transfers electrons and builds up static charge on the tanker, why this is dangerous (a spark could ionise the air and ignite flammable vapour), and describes two valid risk-reduction methods with clear reasoning for how each works.
Indicative content:
friction between the flowing fuel and the inside of the pipe transfers electrons
this causes a build-up of (static) charge on the tanker/fuel
if the charge (potential difference) becomes large enough, it can ionise the air and cause a spark
fuel vapour is flammable, so a spark risks igniting it and causing an explosion
an earthing/bonding strap or cable connects the tanker to the ground, allowing charge to flow away safely rather than building up
using conducting materials for hoses/pipes allows charge to disperse gradually instead of building up to a dangerous level
reducing the flow rate of the fuel reduces the rate at which charge builds up, keeping the potential difference lower
Question 17
(a) B1 a radial field, with field lines pointing towards the sphere (since it is negative)
(a) B1 field lines are more spread out further from the sphere, showing the field gets weaker with distance oe
(a) Answer: A radial field pointing towards the sphere, weakening (lines spreading out) with distance from the sphere.
(b) B1 the positive object lies within the electric field of the sphere, so a force acts on it
(b) B1 unlike charges attract, and the force increases the closer the object gets (field is stronger nearer the sphere) oe
(b) Answer: The object is within the sphere's field, and since unlike charges attract, it experiences an attractive force that grows stronger closer to the sphere.
Question 18
(a) M1 rearranges to Vs = Vp x Ns / Np
(a) M1 substitutes: Vs = 22000 x 400 / 8000
(a) A1 Vs = 1100 V cao
(a) Answer: 1100 V
(b) M1 rearranges and substitutes (ft Vs from a): Ip = (1100 x 45) / 22000
(b) A1 Ip = 2.25 A ft
(b) B1 transmitting at high voltage means a lower current is needed to transmit the same power
(b) B1 lower current means less energy is dissipated as heat in the cables (P = I2 x R), so transmission is more efficient oe
(b) Answer: Ip = 2.25 A; high-voltage transmission uses a lower current, reducing heat losses (P = I2R) in the cables and improving efficiency.
Question 19
M1 substitutes for Option A: P = 1002 x 5.0
A1 P(A) = 50000 W (50 kW)
M1 substitutes for Option B: P = 102 x 5.0
A1 P(B) = 500 W (0.5 kW)
A1 correct comparison: Option A wastes 100 times more power than Option B
B1 therefore transmitting at high voltage (low current) minimises energy wasted as heat, so the National Grid uses high voltage
Answer: P(A) = 50000 W, P(B) = 500 W; Option A wastes 100 times more power, so high-voltage (low-current) transmission is used to minimise losses.