Oxford Revise AQA GCSE Physics | Chapter P4 answers

P4: Supplying energy

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Mark

AO / Specification reference

01.1

the potential difference of the mains electricity in the UK is: about 230 V

the frequency of mains electricity in the UK is: 50 Hz

the mains supply in the UK produces a current that is: alternating

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AO1

4.2.3.1

01.2

earth or neutral

earth; neutral in any order

live; neutral in any order

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AO1

4.2.3.1

01.3

if the casing on an appliance becomes live, the earth wire conducts the current safely to earth

do not accept ‘for safety’

do not accept ‘to protect the user’

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AO1

4.2.3.1

02.1

power = potential difference × current or P = I × V

= 6 V × 1.5 A

(= 9 W)

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AO1

AO2

4.2.4.1

02.2

\(\textrm{power} = \frac{{{\rm{energy}}}}{{{\rm{time}}}}\)

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AO1

4.1.1.4

02.3

\(9 = \frac{{{\rm{energy}}}}{30}\)

energy = 9 × 30

= 270 (J)

accept 270 with no working for three marks

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AO2

4.2.4.2

02.4

both devices transfer the same amount of energy

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AO2

4.2.4.2

03.1

a fault can be caused by the live wire touching the case of the fan

the earth wire is connected to the case

providing a (low resistance) path to ‘earth’

so the current flows through the earth wire and not through the person touching the case

a large current flows, so the fuse melts so the current stops flowing

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AO2

4.2.3.2

03.2

make the case of the fan out of plastic/non-conducting material

if the live wire touches the case the current will not travel through the case to the person

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AO2

4.2.3.2

03.3

power = potential difference × current

= 230 × 4.5

= 1035 W

= 1000 W (to two significant figures)

accept 1000 W with no working for two marks

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AO1

AO2

4.2.4.1

03.4

power = current2 × resistance

energy = power × time

energy = current2 × resistance × time

5.4 = 52 × resistance × 0.63

\(\textrm{resistance} = \frac{{{5}{\rm{.4}}}}{{{\rm{(}}{{5}^{2}} \times {0}{\rm{.63)}}}}\)

= 0.3428

= 0.34 Ω (to two significant figures)

or

P = \(\frac{{\rm{E}}}{{\rm{t}}}\) and P = I2 × R

\(\left( {{\rm{P}} = \frac{{{5}{\rm{.4}}}}{{{0}{\rm{.63}}}}} \right)\) = 8.57 W

\(\left( {{\rm{R}} = \frac{{\rm{P}}}{{{{\rm{I}}^{2}}}}} \right)\) = \(\frac{{{8}{\rm{.57}}}}{5}\) = 0.3428

R = 0.34 Ω to two significant figures)

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or

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AO2

4.2.4.1

04.1

a transformer changes the potential difference/steps a potential difference up or down

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4.2.4.3

04.2

Level 3: Detailed explanation of why the National Grid uses a higher potential difference. Calculations of the current in each wire and of power loss in each wire

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AO3

4.2.4.3

Level 2: Explanation of why the National Grid uses a higher potential difference.

Calculation of the current in each wire or an attempt at calculation of power loss in each wire

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Level 1: Explanation of why the National Grid uses a higher potential difference. Calculation of the current in each wire or an attempt at calculation of power loss in each wire.

1-2

No relevant comment.

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Indicative content:

  • transmitting power at a higher potential difference means that the current is smaller
  • the wires have a resistance, so they will get hot
  • so there is less energy transferred to the thermal store of the surroundings
  • at a power of 80 × 106 W and a potential difference of 400 000V, the current in the wire is: current = \(\frac{{{\rm{power}}}}{{{\rm{potential\ difference}}}}\) = \(\frac{{{80} \times {1}{{0}^{6}}}}{{{\rm{400\ 000}}}}\) = 200 A
    • power loss is: P = I2 × R = 2002 × 4 = 1.6 × 105 W
  • at a power of 80 × 106 W and a potential difference of 4000 V the current in the wire is: current = \(\frac{{{\rm{power}}}}{{{\rm{potential\ difference}}}}\) = \(\frac{{{80} \times {1}{{0}^{6}}}}{4000}\) = 20 000 A

power loss is: P = I2 × R = (20 000)2 × 4 = 1.6 × 109 W

05.1

power = current × potential difference

accept symbol equation P = V × I

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4.1.1.4

05.2

potential difference of the mains = 230 V

power = 2 kW = 2000 W

2000 = current × 230

current = \(\frac{2000}{230}\)

= 8.69 A

= 8.7 A (to two significant figures)

accept 8.2 with no working for the four calculation marks

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AO2

4.2.4.2

05.3

potential difference = current × resistance

accept V = IR

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4.2.1.3

05.4

230 = 8.69 × resistance

resistance = \(\frac{230}{{{8}{\rm{.69}}}}\)

= 26.47 Ω

= 26 Ω(to two significant figures)

or

power = current2 × resistance

2000 = (8.69)2 × resistance

resistance = \(\frac{2000}{{{{{\rm{(8}}{\rm{.69)}}}^{2}}}}\)

= 26.48 Ω

= 26 Ω (to two significant figures)

allow 26.4 = 26 Ω if 8.7A used

allow 26.4 = 26 Ω if 8.7A used

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or

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AO2

4.2.1.3

4.2.4.1

05.5

energy transferred by kettle = power time

accept E = Pt

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4.2.4.2

05.6

E = 2000 × 2 × 60

= 240 000 J

for toaster:

240 000 = 1200 × time

time = \(\frac{{{\rm{240\ 000}}}}{1200}\)

= 200 seconds

= 3 minutes 20 seconds.

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05.7

yes

they both transfer energy from a chemical/nuclear energy store (in the power station) to the thermal energy store (of the surroundings)

by an electric current

no marks for ‘yes’

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AO2

4.2.4.2

06.1

2000 W means 2000 joules of energy are transferred per second/unit time

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4.2.4.2

06.2

230 V means 230 joules of energy are transferred by each coulomb of charge (that flows in the circuit)

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4.2.4.2

06.3

energy = power × time

accept symbol equation E = P × t

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4.2.4.2

06.4

E = 2000 × 5 × 60

= 600 000 J

accept 600 000 with no working for two marks

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AO2

4.2.4.2

06.5

energy = charge × potential difference

accept symbol equation E = Q × V

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4.2.4.2

06.6

600 000 = charge × 230

charge = \(\frac{{{\rm{600\ 000}}}}{230}\)

= 2608.7 C

accept 2609 with no working for three marks

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AO2

4.2.4.2

07.1

Metal rod

Time for nail to fall off in s

Time for nail to fall off in s

Time for nail to fall off in s

Mean time for nail to fall off in s

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evidence of metal rod as independent variable in the first column

evidence of repeat readings

evidence of calculation of mean

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AO2

AO3

4.1.3

07.2

two from:

  • distance of a nail from end of rod
  • Bunsen burner, type/air hole position/position on rod
  • amount of wax on nail
  • size/material/mass of nail
  • initial temperature of nail

one mark for each correct point up to a maximum of two marks

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AO3

4.1.2.1

07.3

the control variables are difficult to control

leading to a big uncertainty in the data produced

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AO3

4.1.2.1

07.4

use a different method for working out when the end of the rod has gotten hot e.g., thermal paint, temperature sensor attached to rod

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AO3

4.1.2.1

08.1

230 V for all appliances

this is the potential difference of the mains

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4.2.3.1

4.2.4.1

08.2

thermal (energy store)

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4.2.3.1

4.2.4.1

08.3

iron→ hairdryer → toaster

The current is proportional to the power if the potential difference is constant or I = \(\frac{{\rm{P}}}{{\rm{V}}}\)

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AO2

4.2.4.1

08.4

2000 = 8.72 resistance

resistance = \(\frac{2000}{{{{8.7}^2}}}\)

= 26.4

= 26 Ω (to two significant figures)

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AO2

4.2.4.1

09.1

the energy from the Sun will not run out (in the immediate future)

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4.1.3

09.2

4 × 1026 W

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4.2.4.2

09.3

energy per year = power × time

= 500 × 3.1 × 107

= 1.55 × 1010 J

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AO2

4.2.4.2

09.4

\(\textrm{area needed} = \frac{{{7} \times {1}{{0}^{18}}}}{{{1}{\rm{.55}} \times {1}{{0}^{10}}}}\)

= 4.5 × 108 m2

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4.1.3

10.1

(Ee = 0.5ke2)

= 0.5 × 500 × 0.012

= 0.025 J

accept 0.025 (J) with no working

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AO2

4.1.1.2

10.2

at the top of the first bounce there is more energy in the gravitational potential energy store

at the top of the second bounce the energy has been transferred to a gravitational potential energy store and the thermal energy store of the surroundings

there is less energy in the gravitational potential energy store, so the second bounce is not so high

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AO2

4.1.1.1

4.1.1.2

10.3

energy is transferred by forces/mechanically

and by heating

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AO2

4.1.1.1

11.1

the National Grid

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4.2.4.3

11.2

both transformers change the potential difference

the transformer near the power station/transformer 1 is a step-up transformer/increases the potential difference

the transformer near the house/transformer 2 is a step-down transformer/decreases the potential difference

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4.2.4.3

11.3

(the energy is transferred at a high potential difference so) the current is small

so the energy/power/heat lost is small

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4.2.4.3

12.1

power = potential difference current

9000 = 230 current

oven current = \(\frac{9000}{230}\)

= 39 A

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AO2

4.2.4.1

12.2

power = potential difference × current

2000 = 230 × current

toaster current = \(\frac{2000}{230}\)

= 8.7 A

difference between oven and toaster = \(\frac{39}{{{8}{\rm{.7}}}}\)

= 4.48

the current in the oven is over 4 times bigger

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AO3

4.2.4.1

12.3

the current is very large, so the heating effect is very big

the wire needs to be thicker so that there is less resistance and so less heating in the wire/the wire does not melt

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4.2.4.1

12.4

there is an earth wire connected to the casing of an appliance

through which current flows if the casing becomes live/connected to the live wire

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AO2

4.2.3.2

13.1

as the light intensity increases, the resistance decreases

at a decreasing rate

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4.2.1.4

13.2

the resistance of B also decreases with increased light intensity

but after light intensity reaches 30 lux, resistance of B is constant

whereas the resistance of A continues to go down

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AO3

4.2.1.4

13.3

when the light level is 13 lux, the resistance of A is about 40 k Ω

either:

the total resistance = (40 × 103) + (100 × 103) = 140 × 103 Ω

current in circuit = \(\frac{{\rm{V}}}{{\rm{R}}}\) = \(\frac{6}{{{140} \times {1}{{0}^{8}}}}\) = 4.28 × 10–5 A

so potential difference across the light dependent resistor

V = IR = (4.28 × 10–5) × (40 × 103) = 1.7 V

so the potential difference across the resistor Vout = 6 – 1.7 = 4.3 V

or:

V = \(\left( {\frac{{{40} \times {1}{{0}^{8}}}}{{\left[ {{100} \times {1}{{0}^{8}} + {40} \times {1}{{0}^{8}}} \right]}}} \right)\) × 6

= 1.7 V

so the potential difference across the resistor Vout = 6 – 1.7 = 4.3 V

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or

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A02

4.2.1.3

4.2.1.4

4.2.2

13.4

light dependent resistor A

for light dependent resistor B, the lowest value of the resistance of light dependent resistor B is 40 kΩ

so the lowest potential difference across the light dependent resistor is 1.7 V

so the maximum potential difference of Vout is 4.3 V (because they add up to 6 V)

to get a potential difference that is higher than 5 V, you need the potential difference across the light dependent resistor to be lower (than 1 V)

so you need a light dependent resistor that has a resistance that can go lower than 40 kΩ

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4.2.2

4.2.1.4

14

Level 3: Calculations of power generated in each bulb in each circuit.

The link between power and brightness is explicit and the statement about brightness is correct.

synoptic question involving ideas from the previous chapter

5-6

AO1

AO2

4.2.2

4.2.4.1

Level 2: Calculations of currents in circuits and a comment linking current and resistance to power. Statements about brightness that follow from previous reasoning.

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Level 1: Recognition that power depends on current and potential difference/resistance. A general statement that bulbs are brighter in parallel/less bright in series

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No relevant comment.

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Indicative content:

  • the brightness of the bulb depends on the power generated in it, P = I2R
  • in series, the current through each bulb is the same
    • current = \(\frac{{{\rm{potential\ difference}}}}{{{\rm{total\ resistance}}}}\) =\(\frac{12}{15}\) = 0.8 A
    • the 10 Ω lamp will have a power of 0.82 × 10 = 6.4 W
    • the 5 Ω lamp will have a power of 0.82 × 5 = 3.2 W
    • the 10 Ω lamp will be brighter.
  • in parallel the potential difference across each bulb is the same, 12 V
    • current through the 5 Ω lamp = \(\frac{{{\rm{potential\ difference}}}}{{{\rm{total\ resistance}}}}\) =\(\frac{12}{5}\) = 2.4 A
    • current through the 10 Ωlamp will be 1.2 A (as the resistance is double, current will be half)
    • the 10 Ω lamp will have a power of 1.22 × 10 = 14.4 W
    • the 5 Ω lamp will have a power of 2.42 × 5 = 28.8 W
    • the 5 Ω lamp will be brighter
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