Name: Electricity - ASTON ACADEMY GCSE PHYSICS · 2019-12-05 · Aston Academy Page 19 of 71...

ElectricityHigh Demand Questions

Name: ________________________

Class: ________________________

Date: ________________________

Time: 197 minutes

Marks: 192 marks

Comments:

Page 1 of 71Aston Academy

The photograph below shows a coffee machine. The coffee machine uses an electric element toheat water.1

(a) The coffee machine has a metal case.

Why would it be dangerous for the live wire of the electric cable to touch the metal case?

___________________________________________________________________

___________________________________________________________________

(1)

(b) The power output of the coffee machine is 2.53 kW.

The mains potential difference is 230 V.

Calculate the current in the coffee machine.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Current = ____________________ A

(3)


(c) The coffee machine heats water from 20 °C to 90 °C.

The power output of the coffee machine is 2.53 kW.

The specific heat capacity of water is 4200 J/kg °C.

Calculate the mass of water that the coffee machine can heat in 14 seconds.

___________________________________________________________________

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___________________________________________________________________

___________________________________________________________________

Mass = ____________________ kg

(5)

(Total 9 marks)

(a) Complete the sentence. Choose answers from the box.

charge potential difference power temperature time

The current through an ohmic conductor is directly proportional to the

_______________________ across the component, provided

that the _______________________ remains constant.

(2)

2


(b) Figure 1 shows a current − potential difference graph for a filament lamp.

Figure 1

Explain how the resistance of a filament lamp changes as the potential difference across itincreases.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(3)

(c) Many householders are replacing their filament lamps with LED lamps which are moreenergy efficient.

What does more energy efficient mean?

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(1)


A Light Dependent Resistor (LDR) is used to turn on an outside lamp when it gets dark.

Part of the circuit is shown in Figure 2.

Figure 2

(d) The light intensity decreases.

What happens to the potential difference across the LDR and the current in the LDR?

Potential difference ___________________________________________________

Current ____________________________________________________________

(2)

(e) What is the resistance of the LDR when the potential difference across it is 4 V?

Give a reason for your answer.

Explain your answer.

Resistance = _______________________ Ω

Reason ____________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)


(f) Calculate the current through the LDR when the resistance of the LDR is 5000 Ω.

Give your answer to 2 significant figures.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Current = _______________________ A

(4)

(Total 14 marks)

The picture shows an electric bicycle. The bicycle is usually powered using a combination of therider pedalling and an electric motor.

3

(a) A 36 volt battery powers the electric motor. The battery is made using individual1.2 volt cells.

(i) Explain how a 36 volt battery can be produced using individual 1.2 volt cells.

To gain full marks, you must include a calculation in your answer.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

(2)

(ii) The battery supplies a direct current (d.c.).

What is a direct current (d.c.)?

______________________________________________________________

______________________________________________________________

(1)


(iii) When fully charged, the battery can deliver a current of 5 A for 2 hours. The battery isthen fully discharged.

Calculate the maximum charge that the battery stores.

Show clearly how you work out your answer and give the unit.

______________________________________________________________

______________________________________________________________

Charge stored = ______________________________

(3)

(b) When powered only by the electric motor, the bicycle can carry a 90 kg rider at a maximumspeed of 6 m/s. Under these conditions, the maximum distance that the bicycle can coverbefore the battery needs recharging is 32 km.

The bicycle has a mass of 30 kg.

(i) Calculate the maximum kinetic energy of the bicycle and rider when the rider is notpedalling.

Show clearly how you work out your answer.

______________________________________________________________

______________________________________________________________

Kinetic energy = ______________________________ J

(2)

(ii) The bicycle can be fitted with panniers (bags) to carry a small amount of luggage.

What effect would fitting panniers and carrying luggage have on the distance thebicycle can cover before the battery needs recharging?

______________________________________________________________

Give a reason for your answer.

______________________________________________________________

______________________________________________________________

(2)

(Total 10 marks)


A homeowner had a new gas boiler installed.

(a) The following information is an extract from the information booklet supplied with the boiler.

Fuel Natural Gas

Water temperature 60 °C

Energy supplied to gas boiler 8.0 kJ/s (8.0 kW)

Efficiency 0.95

4

(i) Calculate the energy transferred each second by the gas boiler to the water insidethe boiler.


______________________________________________________________

______________________________________________________________

Energy transferred by the gas boiler each second = ___________ kJ

(2)

(ii) The energy value of the gas used in a home is measured in kilowatt-hours (kWh).

The homeowner has a pre-payment meter and pays £30 into his account. With apre-payment meter, gas costs 15p per kilowatt-hour.

Calculate the total number of hours that the gas boiler would operate for £30.


______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

Number of hours = _________________________

(2)


(b) Although the gas boiler is very efficient, some energy is wasted.

Explain what happens to the waste energy.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

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(2)

(Total 6 marks)

(a) The picture shows a person using a set of electronic ‘Body Fat Scales’. When the personstands on the scales, a small, harmless, electric current passes through the person’s body.The scales then calculate the resistance of the person’s body and convert the resistanceinto a prediction of body fat content.

(i) The scales contain two 3 V cells joined in series.

Calculate the resistance of a person’s body, if when he stands on the scales, acurrent of 0.12 mA passes through his body.

1000 mA = 1 A

Show clearly how you work out your answer and give the unit.

______________________________________________________________

______________________________________________________________

______________________________________________________________

Resistance = _______________________________

(3)

5


(ii) The scales can only produce a prediction of body fat content and not an accuratemeasurement.

Suggest why.

______________________________________________________________

______________________________________________________________

______________________________________________________________

(1)

(iii) It is recommended that the scales are not used immediately after a person has drunka large amount of water.

Suggest why.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

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(2)


(b) The diagram shows how someone could get an electric shock from accidentally cutting intoan electric cable. If this happens, and a Residual Current Circuit Breaker (RCCB) is beingused, the circuit will switch off automatically.

(i) A faulty appliance or circuit can be switched off by a RCCB or a fuse.

Compare the action of a RCCB with the action of a fuse.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

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(2)

(ii) The graph shows how the severity of an electric shock depends on the size of thecurrent and the time that the current flows through the body.


Using the RCCB helps prevent an electric shock seriously injuring the person usingthe hedge trimmers.

Using information from both the diagram and the graph explain how.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

(2)

(Total 10 marks)


The graph shows how the electric current through a 12 V filament bulb varies with the potentialdifference across the bulb.

(a) What is the meaning of the following terms?

electric current

___________________________________________________________________

___________________________________________________________________

potential difference

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(2)

6


(b) The resistance of the metal filament inside the bulb increases as the potential differenceacross the bulb increases.

Explain why.

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(3)

(c) Use data from the graph to calculate the rate at which the filament bulb transfers energy,when the potential difference across the bulb is 6 V.


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Rate of energy transfer = __________________ W

(2)

(Total 7 marks)

The table gives data about two types of low energy bulb.

Type of bulb Power inputin watts

Efficiency Lifetimein hours

Cost ofone bulb

Compact FluorescentLamp (CFL)

8 20% 10 000 £3.10

Light Emitting Diode(LED)

5 50 000 £29.85

7


(a) Both types of bulb produce the same useful power output.

(i) Calculate the useful power output of the CFL.


______________________________________________________________

______________________________________________________________

______________________________________________________________

Useful power output = _________________________ W

(2)

(ii) Calculate the efficiency of the LED bulb.


______________________________________________________________

______________________________________________________________

______________________________________________________________

Efficiency = ___________________________

(1)

(b) LED bulbs are expensive. This is because of the large number of individual electronic LEDchips needed to produce sufficient light from each bulb.

(i) Use the data in the table to evaluate the cost-effectiveness of an LED bulb comparedto a CFL.

______________________________________________________________

______________________________________________________________

______________________________________________________________

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(2)

(ii) Scientists are developing brighter and more efficient LED chips than those currentlyused in LED bulbs.

Suggest one benefit of developing brighter and more efficient LED chips.

______________________________________________________________

______________________________________________________________

(1)

(Total 6 marks)


(a) The resistance of a 24 W, 12 V filament lamp depends on the current flowing through thelamp. For currents up to 0.8 A, the resistance has a constant value of 2.5 Ω.

(i) Use the equation in the box to calculate the potential difference across the lamp whena current of 0.8 A flows through the lamp.

potential difference = current × resistance


______________________________________________________________

______________________________________________________________

Potential difference = ______________________________ V

(2)

8

(ii) When the potential difference across the lamp is 12 V, the current through the lamp is2 A.

On the axes below, draw a current–potential difference graph for the filament lampover the range of potential difference from 0 to 12 volts.

(2)


(iii) Why does the resistance of the lamp change when the current through the lampexceeds 0.8 A?

______________________________________________________________

______________________________________________________________

(1)

(b) The lamp is now included in a circuit. The circuit is switched on for 2 minutes. During thistime, 72 coulombs of charge pass through the lamp.

Use the equation in the box to calculate the energy transformed by the lamp while thecircuit is switched on.

energy transformed = potential difference × charge


___________________________________________________________________

___________________________________________________________________

Energy transformed = ______________________________ J

(2)

(Total 7 marks)


(a) The figure below shows a fridge with a freezer compartment.

The temperature of the air inside the freezer compartment is –5 °C.

The air inside the fridge forms a convection current when the fridge door is closed.

Explain why.

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(4)

9


(b) The table below shows information about four fridges.

Fridge Volume in litresEnergy used inone year in kWh

A 250 300

B 375 480

C 500 630

D 750 750

A householder concludes that the energy used in one year is directly proportional to thevolume of the fridge.

Explain why her conclusion is not correct.

Use data from the table in your answer.

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___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(c) New fridges are more efficient than fridges made twenty years ago.

Give one advantage and one disadvantage of replacing an old fridge with a new fridge.

Ignore the cost of buying a new fridge.

Advantage __________________________________________________________

___________________________________________________________________

Disadvantage _______________________________________________________

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(2)

(Total 8 marks)


(a) A student set up the circuit shown in the diagram. The student uses the circuit to obtain thedata needed to plot a current - potential difference graph for a diode.

(i) Draw, in the boxes, the circuit symbol for a diode and the circuit symbol for a variableresistor.

Diode Variable resistor

(2)

10

(ii) The student made two mistakes when setting up the circuit.

What two mistakes did the student make?

1. ____________________________________________________________

______________________________________________________________

2. ____________________________________________________________

______________________________________________________________

(2)


(b) After correcting the circuit, the student obtained a set of data and plotted the graph below.

Potential difference in volts

(i) At what potential difference did the diode start to conduct an electric current?

___________________________________ V

(1)

(ii) Use data from the graph to calculate the resistance of the diode when the potentialdifference across the diode is 0.3 V.

______________________________________________________________

______________________________________________________________

______________________________________________________________

Resistance = _____________ ohms

(3)


(c) The diagram shows the trace produced by an alternating current (a.c.) supply on anoscilloscope.

Each horizontal division on the oscilloscope screen represents a time of 0.01s.

(i) Calculate the frequency of the a.c. supply.

______________________________________________________________

______________________________________________________________

______________________________________________________________

Frequency = _______________________ hertz

(2)

(ii) A diode is now connected in series with the a.c. power supply.

Why does the diode cause the trace on the oscilloscope screen to change?

______________________________________________________________

______________________________________________________________

______________________________________________________________

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(2)

(Total 12 marks)


The picture shows an electric cooker hob. The simplified circuit diagram shows how the fourheating elements connect to the mains electricity supply. The heating elements are identical.

When all four heating elements are switched on at full power the hob draws a current of 26 Afrom the 230 V mains electricity supply.

(a) Calculate the resistance of one heating element when the hob is switched on at full power.

Give your answer to 2 significant figures.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Resistance = _______________ Ω(3)

11


(b) The table gives the maximum current that can safely pass through copper wires of differentcross-sectional area.

Cross-sectionalarea in mm2

Maximum safecurrent in amps

1.0 11.5

2.5 20.0

4.0 27.0

6.0 34.0

The power sockets in a home are wired to the mains electricity supply using cablescontaining 2.5 mm2 copper wires. Most electrical appliances are connected to the mainselectricity supply by plugging them into a standard power socket.

It would not be safe to connect the electric cooker hob to the mains electricity supply byplugging it into a standard power socket.

Why?

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(c) Mains electricity is an alternating current supply. Batteries supply a direct current.

What is the difference between an alternating current and a direct current?

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(2)

(Total 7 marks)


Solar panels are often seen on the roofs of houses.

(a) Describe the action and purpose of a solar panel.

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(2)

12

(b) Photovoltaic cells transfer light energy to electrical energy.

In the UK, some householders have fitted modules containing photovoltaic cells on theroofs of their houses.

Four modules are shown in the diagram.

The electricity company pays the householder for the energy transferred.

The maximum power available from the photovoltaic cells shown in the diagram is 1.4 ×103 W.

How long, in minutes, does it take to transfer 168 kJ of energy?

___________________________________________________________________

___________________________________________________________________

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___________________________ Time = ____________________ minutes

(3)


(c) When the modules are fitted on a roof, the householder gets an extra electricity meter tomeasure the amount of energy transferred by the photovoltaic cells.

(i) The diagram shows two readings of this electricity meter taken three months apart.The readings are in kilowatt-hours (kWh).

21 November 0 0 0 4 4

21 February 0 0 1 9 4

Calculate the energy transferred by the photovoltaic cells during this time period.

______________________________________________________________

Energy transferred = ____________________ kWh

(1)

(ii) The electricity company pays 40p for each kWh of energy transferred.

Calculate the money the electricity company would pay the householder.

______________________________________________________________

______________________________________________________________

Money paid = ____________________

(2)

(iii) The cost of the four modules is £6000.

Calculate the payback time in years for the modules.

______________________________________________________________

______________________________________________________________

Payback time = ____________________ years

(3)

(iv) State an assumption you have made in your calculation in part (iii).

______________________________________________________________

______________________________________________________________

(1)


(d) In the northern hemisphere, the modules should always face south for the maximumtransfer of energy.

State one other factor that would affect the amount of energy transferred during daylighthours.

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___________________________________________________________________

(1)

(Total 13 marks)

The diagram shows an a.c. generator.

The coil rotates about the axis shown and cuts through the magnetic field produced by themagnets.

13

(a) (i) A potential difference is induced between X and Y.

Use the correct answer from the box to complete the sentence.

electric generator motor transformer

This effect is called the ___________________________ effect.

(1)

(ii) What do the letters a.c. stand for?

______________________________________________________________

(1)

(iii) Name an instrument that could be used to measure the potential difference betweenX and Y.

______________________________________________________________

(1)


(b) Graph 1 shows the output from the a.c. generator.

Graph 1

(i) One of the axes on Graph 1 has been labelled ‘Potential difference’.

What should the other axis be labelled?

______________________________________________________________

(1)

(ii) The direction of the magnetic field is reversed.

On Graph 1, draw the output from the a.c. generator if everything else remains thesame.

(2)

(c) The number of turns of wire on the coil is increased. This increases the maximum inducedpotential difference.

State two other ways in which the maximum induced potential difference could beincreased.

1. _________________________________________________________________

___________________________________________________________________

2. _________________________________________________________________

___________________________________________________________________

(2)

(Total 8 marks)

The current in a circuit depends on the potential difference (p.d.) provided by the cells and thetotal resistance of the circuit.

(a) Using the correct circuit symbols, draw a diagram to show how you would connect 1.5 Vcells together to give a p.d. of 6 V.

(2)

14


(b) Figure 1 shows a circuit containing an 18 V battery.

Two resistors, X and Y, are connected in series.

• X has a resistance of 3 Ω.

• There is a current of 2 A in X.

Figure 1

(i) Calculate the p.d. across X.

______________________________________________________________

______________________________________________________________

P.d. across X = ______________________ V

(2)

(ii) Calculate the p.d. across Y.

______________________________________________________________

______________________________________________________________

______________________________________________________________

P.d. across Y = ______________________ V

(2)

(iii) Calculate the total resistance of X and Y.

______________________________________________________________

______________________________________________________________

______________________________________________________________

Total resistance of X and Y = ______________________ Ω(2)


(c) Figure 2 shows a transformer.

Figure 2

(i) An 18 V battery could not be used as the input of a transformer.

Explain why.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

(2)

(ii) The transformer is 100% efficient.

Calculate the output current for the transformer shown in Figure 2.

______________________________________________________________

______________________________________________________________

______________________________________________________________

Output current = ______________________ A

(2)

(Total 12 marks)


The diagram shows a temperature sensing circuit used to control a heating system in a house.15

(a) What quantity does the ammeter measure?

___________________________________________________________________

(1)

(b) The current in the circuit is 3.5 mA when the potential difference across the thermistor is4.2 V

Calculate the resistance of the thermistor.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Resistance = ____________________ Ω(3)

(c) Calculate the charge that flows through the thermistor in 5 minutes when the current is 3.5mA.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Charge = ____________________ C

(3)


(d) Explain why the potential difference across the thermistor changes as the temperature inthe house decreases.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(e) The circuit shown in the diagram can be modified to turn lights on and off by replacing thethermistor with a Light Dependent Resistor (LDR).

Draw the circuit symbol for an LDR in the space below.

(1)

(Total 10 marks)


A student investigated how current varies with potential difference for two different lamps.

Her results are shown in the figure below.

(a) Complete the circuit diagram for the circuit that the student could have used to obtain theresults shown in the figure above.

(3)

16

(b) Which lamp will be brighter at any potential difference?


Use the figure above to aid your explanation

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___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)


(c) Lamp B has the higher resistance at any potential difference.

Explain how the figure above shows this.

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___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(d) Both lamps behave like ohmic conductors through a range of values of potential difference.

Use the figure above to determine the range for these lamps.


___________________________________________________________________

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___________________________________________________________________

(3)

(Total 10 marks)


An electrician is replacing an old electric shower with a new one.

The inside of the old shower is shown in Figure 1.

Figure 1

© Michael Priest

(a) If the electrician touches the live wire he will receive an electric shock.

Explain why.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(4)

17


(b) Different electrical wires need to have a cross-sectional area that is suitable for the poweroutput.

Figure 2 shows the recommended maximum power input to wires of different cross-sectional areas.

Figure 2

The new electric shower has a power input of 13.8 kW.

Determine the minimum diameter of wire that should be used for the new shower.

The diameter, d, can be calculated using the equation:

A is the cross-sectional area of the wire.

___________________________________________________________________

___________________________________________________________________

Minimum diameter = _________________________ mm

(2)


(c) The charge that flows through the new shower in 300 seconds is 18 000 C.The new electric shower has a power of 13.8 kW.

Calculate the resistance of the heating element in the new shower.

Write down any equations you use.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Resistance = _______________________ Ω(5)

(Total 11 marks)


Figure 1 shows a graph of current against potential difference for a solar cell when light ofintensity 450 W/m2 is incident on it.

Figure 1

18

(a) Determine the power output of the solar cell when the potential difference is 0.5 V

Use data from Figure 1.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Power = ____________________ W

(3)


(b) Draw a sketch graph on Figure 2 to show how the power output of the solar cell varies withpotential difference between 0.1 V and 0.5 V

No values need to be included on the vertical axis.

Figure 2

(2)

(c) The maximum power output of this solar cell is 0.52 W

When the light intensity is 450 W/m2 the cell has an efficiency of 0.15 at the maximumpower output.

Calculate the area of the solar cell.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Area = ____________________ m2

(4)


(d) A householder has four solar cells.

Each of the solar cells has a resistance of 0.78 Ω

Explain how the solar cells should be connected so that the total resistance is as low aspossible.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(Total 11 marks)

Figure 1 shows a Van de Graaff generator that is used to investigate static electricity.

Before it is switched on, the metal dome has no net charge.

After it is switched on, the metal dome becomes positively charged.

Figure 1

© Michael Priest

19


(a) Explain how an uncharged object may become positively charged.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(3)

(b) Figure 2 shows a plan view of the positively charged metal dome of a Van de Graaffgenerator.

Draw the electric field pattern around the metal dome when it is isolated from itssurroundings.

Use arrows to show the direction of the electric field.

Figure 2

(2)


(c) Another positively charged object is placed in the electric field.

Look at Figure 3.

Figure 3

In which position would the object experience the greatest force?

Tick one box.

P

Q

R

S

(1)

(Total 6 marks)

The image shows a battery-powered drone.20


(a) The battery in the drone can store 97.5 kJ of energy.

When the drone is hovering, the power output of the battery is 65.0 W

Calculate the time for which the drone can hover.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Time = ____________________ seconds

(3)

(b) The battery powers 4 motors in the drone.

Each motor has a resistance of 1.60 Ω when the power input to each motor is 19.6 W

The 4 motors are connected in parallel with the battery.

Calculate the current through the battery.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Current = ____________________ A

(4)

(Total 7 marks)


A 12 V filament bulb is connected to a 12 V power supply.The graph shows how the current changes after the bulb is switched on.

21

(a) (i) After 0.10 seconds, the bulb works at its normal brightness.

What is the current through the bulb when it is working at normal brightness?

Current = _____________________ A

(1)

(ii) The bulb works at normal brightness for 30 seconds before it is switched off.

Calculate the charge that flows through the bulb in the 30 seconds before it isswitched off. Give the unit.

______________________________________________________________

______________________________________________________________

______________________________________________________________

Charge = _________________ unit _________________

(3)

(iii) Calculate the energy transferred by the 12 V bulb when it is working at normalbrightness for 30 seconds.

______________________________________________________________

______________________________________________________________

Energy transferred = _____________________ J

(2)


(b) Between 0.02 seconds and 0.08 seconds, there is an increase in both the resistance andthe temperature of the metal filament inside the bulb.

Explain, in terms of the electrons and ions inside the filament, why both the temperatureand the resistance increase.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(Total 8 marks)


Mark schemes

(a) risk of electric shock (if someone touched the case)allow risk of electrocution (if someone touched the case)

1

1

(b) 2530 = I × 230this mark may be awarded if P is incorrectly / not converted

1

this mark may be awarded if P is incorrectly / not converted1

I = 11 (A)

this answer only

an answer of 0.011 (A) scores 2 marks1

an answer of 11 (A) scores 3 marks

(c) E = 2530 × 14this mark may be awarded if P is incorrectly / not converted

1

E = 35 420 (J)

this answer only1

35 420 = m × 4200 × 70

allow their calculated E = m × 4200 × 701

1

m = 0.12 (kg)

allow an answer that is consistent with their calculated value of E1

[9]

(a) potential differenceallow p.d.allow voltage

1

temperature1

in this order only

2


(b) the current increases (when the potential difference increases)1

(which) causes the temperature of the filament to increase1

(so) the resistance increases

do not accept resistance increases and then levels off1

(c) a higher proportion / percentage of the (total) power / energy input is usefullytransferred

wastes less energy is insufficient

orhigher (useful) power / energy output for the same (total) power / energy input

1

(d) potential difference increases1

current decreases1

(e) 1000 (Ω)

reason only scores if R = 1000 (Ω)1

potential difference is shared in proportion to the resistance

allow a justification using a correct calculation1


(f) 12 = I × 70001

1

I = 1.71 × 10−3 (A)

an answer that rounds to 1.7 × 10−3 (A) scores 3 marks1

I = 1.7 × 10−3 (A)

this answer only

orI = 0.0017 (A)

an answer of 2.4 × 10−3 (A) scores 2 marks

if no other marks scored allow 1 mark for calculation of total resistance

(7000 Ω)1

an answer of 1.7 × 10−3 (A) scores 4 marks

[14]

(a) (i) (connect) 30 (cells)13

in series1

(ii) current always flows in the same directionorcurrent only flows one way

1

(iii) 36 000

allow 1 mark for correctly converting 2 hours to 7200 seconds

answers 10 or 600 score 1 mark2

coulombs / C

do not accept c1

(b) (i) 2160

allow 1 mark for correct substitution, ie ½ × 120 × 62

answers of 1620 or 540 score 1 mark2


(ii) reduce it1

any one from:

• draws a larger current (from battery)

• motor draws greater power (from battery)

accept energy per second for power

accept more energy needed to move the bicycle

• greater resistance force (to motion) / air resistance / drag / friction

accept less streamlined

more mass to carry is insufficient1

[10]

(a) (i) 7.6

allow 1 mark for correct substitution and / or transformation

95 × 8.02

4

ie

(ii) 25 (hours)

allow 1 mark for obtaining number of kWh = 200

an answer of 26(.3) gains both marks2

(b) any two from

• transferred to the surroundings / air / atmosphere

• becomes spread out

• shared between (many) molecules

• (wasted as) heat / sound2

[6]

(a) (i) 50 000

allow 1 mark for correct substitution, ie

6 = 0.00012 × R

or 6 = 0.12 × R

or answers of 25 000 or 50 gain 1 mark

or allow 1 mark for an incorrect answer caused by one error only ieusing 3V or an incorrect conversion of current

2

5


ohm / Ωan answer 50kΩ gains 3 marks

1

(ii) (body) resistance changes

or

body fat/resistance affected by (many) factors

accept named factor, eg age, gender, height, fitness, bonestructure, muscle, drinking water related to body fat / resistance

1

(iii) gives misleading / wrong/inaccurate value

do not credit if specifically linked to a change in mass / weight1

(because) high water content changes body resistance

accept a specific change to resistance

water changes body mass is insufficient1

(b) (i) RCCB – detects difference between current in live and neutral (wires)

accept RCCB can be reset1

fuse – (overheats and) melts

accept blows for melts1

(ii) switches the circuit / hedge trimmers off within 60 milliseconds

allow for 1 mark the RCCB / it is (very) fast.

do not accept the bigger the current the faster the RCCBswitches off

2

[10]

(a) electric current(rate of) flow of (electric) charge / electrons

accept

with Q and t correctly named1

6

potential differencework done / energy transferred per coulomb of charge(that passes between two points in a circuit)

accept

with W and Q correctly named1


(b) metals contain free electrons (and ions)

accept mobile for free1

as temperature of filament increases ions vibrate faster /with a bigger amplitude

accept atoms for ions

accept ions/atoms gain energy

accept vibrate more for vibrate faster

do not accept start to vibrate1

electrons collide more (frequently) with the ionsor(drift) velocity of electrons decreases

do not accept start to collide

accept increasing the p.d. increases the temperature (1 mark)

and

(and) resistance increases with temperature (1 mark) if no othermarks scored

1

(c) 7.8

allow 1 mark for obtaining value 1.3 from graph

or allow 1 mark for a correct calculation using an incorrect currentin the range 1.2-1.6 inclusive

2

[7]

(a) (i)

1.6 (W)

allow 1 mark for correct substitution ie

2

7

(ii)

32 (%) / 0.32ortheir (a)(i) ÷ 5 correctly calculated

ignore any units1


(b) (i) any two from:

• comparison over same period of time of relative numbers of bulbsrequired eg over 50 000 hours 5 CFL’s required to 1 LED

accept an LED lasts 5 times longer

• link number of bulbs to cost eg 5 CFL’s cheaper than 1 LED

an answer in terms of over a period of 50 000 hours CFLs cost£15.50 (to buy), LED costs £29.85 (to buy) so CFLs are cheaperscores both marks

an answer in terms of the cost per hour (of lifetime) being cheaperfor CFL scores 1 mark if then correctly calculated scores bothmarks

• over the same period of time LEDs cost less to operate (than CFLs)2

(ii) any one from:

• price of LED bulbs will drop

do not accept they become cheaper

• less electricity needs to be generated

accept we will use less electricity

• less CO 2 produced

• fewer chips needed (for each LED bulb)

• fewer bulbs required (for same brightness / light)

• less energy wasted

do not accept electricity for energy1

[6]

(a) (i) 2

allow 1 mark for correct substitution i.e. 0.8 × 2.5 provided nofurther step shown

2

8

(ii) straight line drawn from origin to 2, 0.8ortheir (a)(i), 0.8

1


curve from 2, 0.8 to 12,2ortheir (a)(i) 0.8 to 12,2

accept curve from 2, 0.9 to 12,2

or

their (a)(i) 0.9 to 12,2

‘convex’ curve required

accept a curve that flattens between 10 and 12V1

(iii) filament / lamp gets hot

accept temperature increases1

(b) 108

allow 1 mark for correct substitution i.e. 1.5 × 72 provided no furtherstep shown

2

[7]

(a) air near freezer compartment is cooled or loses energy

accept air at the top is cold1

9

cool air is (more) dense or particles close(r) together (than warmer air)

do not allow the particles get smaller / condense1

so (cooler) air falls1

air (at bottom) is displaced / moves upwards / rises

do not allow heat rises

accept warm air (at the bottom) rises1

(b) if volume is doubled, energy use is not doubledorvolume ÷ energy not a constant ratio

1

correct reference to data, eg 500 is 2×250 but 630 not 2×3001


(c) accept suitable examples, eg

advantage:

• reduces emissions into atmosphere• lower input power or uses less energy or wastes less energy• costs less to run

cost of buying or installing new fridge is insufficient

ignore reference to size of fridge1

disadvantage:

• land fill• energy waste in production• cost or difficulty of disposal• transport costs

1

[8]

(a) (i) symbol for a diode

accept 1

10

1

symbol for a variable resistor

(ii) voltmeter is in series or voltmeter is not in parallel1

ammeter is in parallel or ammeter is not in series

accept an answer in terms of how the circuit should be corrected

voltmeter and ammeter are wrong way around is insufficient1

(b) (i) 0.2 (V)

accept any value between 0.20 and 0.21 inclusive1

(ii) 37.5

allow 1 mark for I = 0.008orallow 2 marks for correct substitution, ie 0.3 = 0.008 × Rorallow 1 mark for a correct substitution using I = 0.8 or I = 0.08or I = 0.009orallow 2 marks for answers of 0.375 or 3.75 or 33(.3)

3


(c) (i) 25

allow 1 mark for obtaining period = 0.04(s)2

(ii) diode has large resistance in reverse / one direction1

so stops current flow in that / one direction

allow diodes only let current flow one way / direction

allow 1 mark for the diode has half-rectified the (a.c. power) supply1

[12]

(a) 35

an answer with more than 2 sig figs that rounds to 35 gains 2 marks

an answer 8.8 gains 2 marks

an answer with more than 2 sig figs that rounds to 8.8 gains 1 mark3

11

allow 2 marks for correct method, ie

allow 1 mark for I = 6.5 (A) or R =

(b) (maximum) current exceeds maximum safe current for a 2.5 mm2 wire

accept power exceeds maximum safe power for a 2.5 mm2 wire

or(maximum) current exceeds 20 (A)

(maximum) current = 26 (A) is insufficient1

a 2.5 mm2 wire would overheat / melt

accept socket for wire

do not accept plug for wire1

(c) a.c. is constantly changing direction

accept a.c. flows in two directions

accept a.c. changes direction

a.c. travels in different directions is insufficient1

d.c. flows in one direction only1

[7]

(a) water heated by radiation (from the Sun)

accept IR / energy for radiation1

12


water used to heat buildings / provide hot water

allow for 1 mark heat from the Sun heats water if no other marksgiven

references to photovoltaic cells / electricity scores 0 marks1

(b) 2 (minutes)

1.4 × 103 =

gains 1 mark

calculation of time of 120 (seconds) scores 2 marks3

(c) (i) 150 (kWh)1

(ii) £60(.00) or 6000 (p)

an answer of £6000 gains 1 mark

allow 1 mark for 150 × 0.4(0) 150 × 40

allow ecf from (c)(i)2

(iii) 25 (years)

an answer of 6000 / 240or6000 / their (c)(ii) × 4gains 2 marks

an answer of 6000 / 60or6000 / their (c)(ii) gains 1 mark, ignore any other multiplier of (c)(ii)

3

(iv) any one from:

• will get £240 per year

accept value consistent with calculated value in (c)(iii)• amount of light is constant throughout the year• price per unit stays the same• condition of cells does not deteriorate

1

(d) any one from:

• angle of tilt of cells• cloud cover• season / shade by trees• amount of dirt

1

[13]

(a) (i) generator113


(ii) alternating current1

(iii) voltmeter / CRO / oscilloscope / cathode ray oscilloscope1

(b) (i) time1

(ii) peaks and troughs in opposite directions1

amplitude remains constant

dependent on first marking point1

(c) any two from:

• increase speed of coil• strengthen magnetic field• increase area of coil

do not accept larger2

[8]

(a) attempt to draw four cells in series114

correct circuit symbols

circuit symbol should show a long line and a short line, correctlyjoined together

example of correct circuit symbol:

1

(b) (i) 6 (V)


V = 3 × 2 scores 1 mark

provided no subsequent step2

(ii) 12 (V)

ecf from part (b)(i)

18 – 6

or

18 – their part (b)(i) scores 1 mark2


(iii) 9 (Ω)ecf from part (b)(ii) correctly calculated

3 + their part (b)(ii) / 2

or

18 / 2 scores 1 mark

provided no subsequent step2

(c) (i) need a.c.1

battery is d.c.1

(ii) 3 (A)


18 × 2 = 12 × Is scores 1 mark2

[12]

(a) current115

(b) 4.2 = 3.5 × 10–3 × R1

R = 4.2 / 3.5 × 10–3

1

R = 1200 (Ω)an answer of 1200 (Ω) scores 3 marks

an answer of 1.2 scores 2 marks1

(c) conversion from minutes to seconds (300 s)1

Q = 0.0035 × (5 × 60)1

Q = 1.05 C

an answer of 1.05 (C) scores 3 marks

an answer of 17.5 scores 1 mark

an answer of 1050 or 0.0175 scores 2 marks1

(d) (potential difference) increases1

(because thermistor) resistance increases

2nd mark dependent on scoring 1st mark1


(e)

1

[10]

(a)

battery in series with bulb and ammeter1

16

voltmeter in parallel with bulb1

variable resistor

or

variable power pack

or

potentiometer1

(b) A is brighter because it has a higher current (than lamp B at any p.d.)1

(therefore A has a) higher power output (than bulb B)

accept higher energy output per second1

(c) lower current (than lamp A) for the same potential difference

accept answer in terms of R = V / I1

lower gradient (than lamp A)1

(d) 0 – 2 Volts

allow a range from 0 V up to any value between 1 and 2 V.1

(for an ohmic conductor) current is directly proportional to potential difference

allow lines (of best fit) are straight and pass through the origin1

(so) resistance is constant1

[10]


(a) (because the) potential of the live wire is 230 V117

(and the) potential of the electrician is 0 V1

(so there is a) large potential difference between live wire and electrician1

charge / current passes through his body

allow voltage for potential difference1

(b) diameter between 3.50 and 3.55 (mm)

allow correct use of value of cross-sectional area of 9.5 to 9.9(mm2) with no final answer given for 1 mark

2

(c) 18000 = I × 3001

I = 18000 / 300 = 601

13 800 = (602) × R1

R = 13 800 / 602

1

3.83 (Ω)1

allow 3.83(Ω) with no working shown for 5 marks

answer may also be correctly calculated using P = IV and V = IR if230 V is used.

[11]

(a) current at 0.5 V = 0.91 (A)1

P = 0.91 × 0.51

P = 0.455 (W)

an answer of 0.455 (W) scores 3 marks1

18

(b) straight line with positive gradient

allow for 1 mark a straight line that passes through (0.1, 0)1

positive y-axis intercept

ignore any values on y-axis1


(c)

1

total P = 3.47 (W)1

1

area = 7.7 × 10–3 (m2)

an answer of 7.7 × 10–3 (m2) scores 4 marks

allow use of student’s calculated incorrect total power for last 2marking points

1

(d) connect the solar cells in parallel1

(so that) the current has multiple paths it can take

or

the total resistance is less than the resistance of one solar cell1

[11]

(a) negatively charged119

electrons are transferred1

from the (neutral) object1

(b) minimum of four lines drawn perpendicular to surface of sphere

judge by eye1

minimum of one arrow shown pointing away from sphere

do not accept any arrow pointing inwards.1

(c) Q1

[6]


(a) 97 500 = 65.0 × t1

1

t = 1500 (s)

an answer of 1500 (s) scores 3 marks

an answer of 1.5 scores 2 marks1

20

(b) 19.6 = I2 × 1.601

1

I = 3.5 (A)

allow 1 mark for a correct value for I correctly multiplied by 41

current through battery = 14 (A)

an answer of 14 (A) scores 4 marks1

[7]

(a) (i) 1.71

(ii) 51or30 × their (i) correctly calculated

allow 1 mark for correct substitution i.e. 1.7

or their (i)2

coulomb / C

do not accept c1

(iii) 612ortheir (ii) × 12 correctly calculatedortheir (i) × 360 correctly calculated

allow 1 mark for correct substitution i.e. E = 12 × 51

or 12 × their (ii)

or their (i) × 3602

21


(b) ions vibrate fasterorions vibrate with a bigger amplitude

accept atoms for ions throughout

accept ions gain energy

accept ions vibrate more

ions start to vibrate is insufficient1

electrons collide more (frequently) with the ionsor(drift) velocity of electrons decreases

electrons start to collide is insufficient

there are more collisions is insufficient, unless both electrons andions are implied

1

[8]


Examiner reports

(a) 48% of students answered this question correctly. The danger would be, if a persontouched the case, they would receive an electric shock or be electrocuted. The casebecoming live was insufficient for the mark.

1

(b) 73% of students scored 3 marks. Most students who didn’t score 3 marks either didn’tconvert the power or converted it incorrectly, scoring 2 marks if the rest of their calculationswere correct.

(c) This question was a challenging two-step calculation. 50% of the students scored all 5marks. Many students failed to calculate the energy transferred using E=Pt, but if theyattempted to calculate an energy, even if incorrect, they could score the last 3 markingpoints in the calculation. Some students calculated the mass of water that could be heatedper second and then multiplied by 14 (seconds).

(a) 57% of students scored 2 marks for this question, with 32% scoring 1 mark. The mostcommon mistake was not stating which quantity should remain constant, ‘charge’ being acommonly seen incorrect answer.

2

(b) This question discriminated well between students with 13% of students scoring 3 marks,30% scoring 2 marks and 40% scoring 1 mark. Some students incorrectly stated that thecurrent decreased or the resistance decreased. Contradictory statements often negatedmarks, e.g. pd increases causing current to increase but then decrease.

(c) 24% of students answered this question correctly. Students needed to answer in terms ofamount / proportion / percentage input compared to useful output in order to score themark.

(d) This question differentiated well, with 26% of students scoring 2 marks, while 39% ofstudents scored 1 mark.

(e) 13% of students scored 2 marks on this question. Very few correct answers of 1000 ohmswere seen. When 1000 ohms was given, the reason usually justified the award of 2 marks,so very few 1 mark answers were seen.

(f) 12% of students scored 4 marks on this question. Many students failed to add theresistance of the two series resistors together. The most common incorrect answer was0.0024 (A), which scored 2 marks. Most students who calculated the correct current alsocorrectly showed their answer to 2 significant figures.


(a) (i) Only a few students scored both marks. The first mark for calculating 30 cells wasmost frequently scored however the mark for series circuit was rarely given.Appearances of ‘parallel circuit’ were fortunately rare. Those students who didattempt to describe the connection needed for the cells to make a battery said‘positive to negative’ without realising that this could be achieved both in series andparallel.

(ii) About three fifths of students could describe a direct current however there were a lotof common misconceptions: a current which goes straight to where it is needed, itflows through all of the circuit, a current which travels in a straight line. A few studentstried to describe d.c. only in terms of a CRO display which was insufficient for thisquestion.

(iii) About a quarter of students scored all three marks. The most common mistake wasthe failure to convert the time of two hours to seconds. Of the third of students whogained two marks, the majority gave the ‘fall back answer’ of 10 or 600 with thecorrect unit, those who gained one mark achieved this mostly with answers of 10 or600 with very few gaining 1 mark from a correct unit without a correct calculation. Anumber of students had been well trained to put a really large C but unfortunately anumber of small c’s and q’s were also in evidence. A few students scored no mark.

(b) (i) This question was well done compared to recent years, with nearly three quarters ofstudents scoring bothmarks. Common mistakes were the failure to square ‘6’, anderroneously changing 120 kg to 120000 g.

(ii) Most students correctly stated that the distance was reduced (in their own words) butsome mistakenly gave an answer of ‘time being less between recharge’ thus notanswering the question. A common misconception is that more KE is needed tomove the bicycle with a greater load rather than more energy from the battery beingtransferred as more KE for a bicycle of greater mass moving at the same speed. Veryfew students who mentioned KE in their answer managed to do so in an appropriateway to score the mark. The two fifths of students who scored the reason markgenerally did so by simply stating that ‘more energy is needed’.

3

(a) (i) The vast majority of students attempted this calculation, with over three quartersgetting the correct answer. The most common errors were in transposing theequation incorrectly, or including the water temperature of 60°C in the calculation.

(ii) Around half of the students were able to use the total cost of £30 and the cost perkWh of 15p to calculate a figure of 200; however that is the point at which moststopped, quoting an answer of 200 hours. A few then attempted to involve the power,with only a minority ending up with the correct answer. A few students did not attemptto answer this question.

(b) Most students gave at least one correct point in answer to this question, with around halfbeing able to give two points correctly.

4


(a) (i) This question was poorly answered with only a third of students completing thecalculation correctly. Many students missed the information that the scales containedtwo cells and so used 3 V in the calculation. A large number of students were unableto convert mA to A. Some students chose the correct equation, but were unable totranspose it correctly. It was extremely disappointing that only a tenth of studentswere able to give the unit for resistance.

(ii) This question was poorly answered by most students. Many students made referenceto other tissues, such as muscle, but were unable to relate this to changes in body fator body resistance. Common errors included responses relating to clothing andresponses in terms of the inaccuracy of the apparatus or lack of precision.

(iii) A small number of students correctly identified the water would cause a change inbody resistance and so give misleading values. Many students however, took thisidea further and suggested a risk of electrocution or internal burns because waterwas a good conductor. A significant number of students simply stated that you wouldweigh more because you had drunk the water.

(b) (i) Very few students were able to write that the RCCB works by detecting a differencebetween the current in the live wire and the current in the neutral wire. However,many students were able to score this mark by giving the alternative answer that theRCCB could be reset. A small number of students were under the impression that anRCCB is a type of fuse or that it contains a fuse inside. Over half of students wereable to score a mark by stating that the fuse melts / blows. However, students shouldbe aware that other descriptions such as, breaks, explodes, snaps and bursts are notacceptable.

(ii) There were few responses where the student had calculated from the graph that theRCCB had to switch off within 60 milliseconds. Many students seemed to havemissed the instruction to use information from the diagram or did not understand thesignificance of the difference in the current in the live and neutral wires. A number ofstudents did gain a mark for the idea that the RCCB was fast, quick or actedimmediately. Some students thought that ‘automatically’ was the same as fast. A fewstudents misunderstood the graph and thought incorrectly that the RCCB switchedthe circuit off faster if the current was higher.

5


(a) Only just over half of students were able to give an acceptable meaning for electric current.There were many vague answers such as ‘the flow of electricity‘. Very few students, wereable to give the meaning of potential difference. Common errors often included the idea ofa force pushing the electricity around.

(b) This question was poorly answered. Very few students referred to free electrons at all. Moststudents gained one mark for an answer linking the increasing p.d. to the temperaturehowever, few went on to then link this to the resistance increase. A significant number ofstudents thought that the resistance caused the p.d. Those students who tried to explainresistance in terms of electrons colliding more often as the ions vibrate faster found itdifficult to express themselves clearly. Many students had ‘start’ to vibrate and ‘start’ tocollide, implying this was not happening before. Others had ions moving, rather thanvibrating. Some students linked resistance to it being harder for electrons to pass, withoutgiving any explanation why.

(c) About half the students scored both marks. The most common error was to misread thegraph scale when obtaining the current at 6 volts. However, students that did misread thegraph could still score one mark by showing a correct calculation using their incorrectvalue.

6

(a) (i) Almost a half of students gained both marks. The common errors were to multiply 8by 20 without then dividing by 100 or divide 20 by 8 giving an answer of 2.5.

(ii) A significant number of students thought this question could not be answered due tolack of data. These students had failed to read the question stem that told them bothbulbs had the same useful power output. Students scoring both marks in part (a)(i)usually also scored this mark. However, there were a significant number of studentsgiving answers in excess of 100 %.

(b) (i) Students were asked to use the data and not simply to repeat it. However, there weremany good answers that compared the cost over the same period of time and gainedboth marks. Different, but still valid answers, which used correct calculations in termsof cost per hour or hours per pound spent, were regularly seen.

(ii) There was a wide variety of answers, many unfortunately, too vague to credit. Themost popular correct answer was in terms of ‘waste less energy’.

7


(a) (i) The majority of students, over 90%, scored 2 marks. Very few gained 1 mark.

(ii) Very few students produced a correct graph to gain 2 marks. A large percentage ofstudents did not appreciate the importance of using the answer to part (a)(i).Common “wrong” answers included a straight line from the origin to (12,2), a straightline from (12,2) through (2,0.8) and continued back to the y axis, and two straightlines, one from origin to (2,0.8) and then from that point to (12,2). Otherwise, all sortsof curves, including s-shaped and convex all the way from the origin.

(iii) There were only a minority of correct answers, less that 25%. Many answers referredto how the resistance changed as the current increased but not why. There were alsolots of answers given in terms of the resistance changing to allow or stop the currentflowing and for safety to keep the “fuse” from blowing.

(b) The majority of students gave correct answers. The most common incorrect response was“144”, by multiplying the charge by the time in minutes. An answer scoring one mark wasrare – and seemingly from students who did not possess a calculator.

8

(a) This question was well answered on the whole, with around half of students scoring at leastthree of the four marks. Many answers started with warmer air rising, rather than with thecooler air falling. Whilst many students made reference to changes in density, they oftenincorrectly referred to the ‘particles becoming denser’.

(b) Around a fifth of students achieved both marks. Many answers indicated that ‘directlyproportional’ meant that the two values had to be the same, as in fridge D. Some studentsworked out the difference in volume between each fridge and the previous one, and alsothe difference in energy used. As these were not the same, they stated that the data did notshow proportionality.

To check if results are directly proportional, either the ratio of the volume to energy usedneeds to be a constant or the volume and energy used needs to change by the samemultiplier.

(c) Nearly two-thirds scored at least one mark, but only around a fifth scored both. Manystudents seem to have overlooked the instruction to ignore the cost of buying a new fridge.Many answers indicated that ‘more efficient’ meant that the new fridge was colder, or keptfood fresher for longer.

9


(a) (i) Considering this question was asking for recall of standard symbols it is surprisingand disappointing that over half of the students scored zero. Only a small minority ofthe students gained both marks. The variable resistor caused fewer problems thanthe diode.

(ii) Again poorly answered with only about a third of the students gaining both marks andjust less than half of the students scoring zero. Many of the students mentioned theswitch and the direction of the diode as a reason for the circuit being incorrect. Othersshowed even less understanding of the circuit.

(b) (i) The majority of the students gained this mark.

(ii) Just over half of the students got this question correct gaining the full three marks. ASignificant number of students made errors by taking the numbers 0.08, 0.8 or 0.009from the graph.

(c) (i) Just less than a quarter of students managed to achieve two marks on this question;with a further small number gaining one mark by obtaining the period, with asignificant number believing this was the final answer.

(ii) Very few students achieved both marks on this question – most made no mention ofhigh resistance in the reverse direction. Less than half of the students achieved thesecond marking point with reference to current flowing only in one direction.

10

(a) The correct answer of 35 was rarely given; 8.8 as the resistance for all 4 heating elementswas the most common answer. A number of students then incorrectly divided by 4 to findthe resistance of a single heating element. These students usually gained a single mark fordividing the p.d. of 230 V by the current of 26 A. The concept of significant figures is still notwell understood with very few students scoring full marks.

(b) Very few students scored both marks, however half of students did score one mark. Mostanswers lacked detail and some made reference to potential difference rather than current.Several students confused current and p.d, writing that 230 V was too high for the wire.Surprisingly few students scored the mark for overheating with many references to plugs /things blowing up and fuses melting rather than the wire.

(c) Nearly half of students gained both marks. For direct current, frequent incorrect answersincluded: ‘the current goes straight to the device’, or ‘straight to the source with no wires’,‘direct current travels in straight lines’. With regard to alternating current incorrect answersseen frequently included ‘alternating current goes up and down’, ‘only flows in a parallelcircuit’ or ‘goes in many directions.’

11


(a) Only a tenth of students correctly described the action of a solar panel. The remainderdescribed a photovoltaic cell which is not in the specification.

(b) The remainder of the question concerned photovoltaic cells which were introduced here.

The calculation to find the time to transfer a certain amount of energy, given a certain valueof power available gained full marks from a quarter of the students. The calculationinvolved the interpretation of data given in standard form, the conversion of kJ to J and thefinal answer given in minutes. Another quarter of the students only dropped one mark forleaving the answer in seconds.

(c) (i) Students were required to take the difference between two meter readings in this part

(ii) multiply the answer by 40p in this part

12

(iii) work out a payback time for some photovoltaic cells in this part.

The two readings were three months apart and many students had problems relatingthis time to the correct fraction of a whole year. Although nine-tenths of studentscorrectly completed the first two steps far less scored the marks in this part.

(iv) Almost two-thirds of students correctly stated the assumption behind the calculationof payback time.

(d) Most students knew that specific weather conditions such as cloud cover would affect theenergy transferred during daylight hours.

(a) Of the whole exam paper, this question had the highest percentage of students who did notattempt an answer. Around three-quarters of students correctly identified that four cellswould be needed and drew the correct symbols. However, these were often joined bydotted lines, or not joined at all.

(b) The calculations were very well answered with nearly all students scoring both marks forpart (i) and more than three-quarters scoring full marks for parts (ii) and (iii).

(c) (i) Around half of students had the correct idea. However, some failed to score bothmarks by just referring to either the fact that the transformer needs alternating currentto work, or that the battery supplies direct current, but not referring to both. Incorrectanswers commonly referred to the voltage being too high, or too low.

(ii) This calculation question was well answered, with around three-quarters of studentsscoring both marks.

14


(a) (i) Nearly all of the students scored this mark, with the most common errors being 1.6 or1.75.

(ii) About three quarters of the students correctly calculated the numerical answer,sometimes with an error carried forward from (i). However, only about half of thestudents could give the correct unit.

(iii) This was poorly answered with the most common response being 12 x 30 = 360. Justless than one quarter of the students scored these two marks.

(b) This was very poorly answered with two thirds of the students scoring zero and a furthertenth not attempting the question at all. A few answers suggested that the students hadsome notion of why the resistance increased, but the answers were too vague or impreciseto meet the second marking point.

21


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