Oxford Revise AQA GCSE Physics | Chapter P5 answers

P5: Electric circuits

Question

Answers

Extra information

Mark

AO / Specification reference

01.1

A should be in the uppermost circle

V should be in the right hand circle

The resistor is between the cell and the voltmeter

The variable resistor is the lower-most circuit component

one mark for ‘A’ in correct place

one mark for ‘V’ in correct place

one mark for resistor labelled correctly

one mark for variable resistor labelled correctly

1

1

1

1

AO1

4.2.1.1

01.2

the rate of flow of charge/charge flowing per second

1

AO1

4.2.1.2

01.3

resistance = \(\frac{{{\rm{potential\ difference}}}}{{{\rm{current}}}}\) or potential difference = current × resistance

allow V = IR or R = \(\frac{{\rm{V}}}{{\rm{I}}}\)

1

AO1

4.2.1.3

01.4

R = \(\frac{6}{{{0}{\rm{.3}}}}\)

= 20

unit = Ω

accept 20 with no working for two calculation marks

1

1

1

AO2

4.2.1.3

02.1

circuits A and C are parallel circuits

no marks if more than one box ticked

1

AO1

4.2.2

02.2

A, C

B, D

B, D

in all cases, both letters required for the mark

letters can be in either order

1

1

1

AO2

4.2.2

02.3

no

the bulbs will be the same brightness

because they are in a series circuit

1

1

1

AO2

4.2.2

03

Level 3: Correct diagrams with description of measurements (total current and potential difference) to be taken in each circuit. Rearrangement of equation to give an equation for resistance. Correct statement about relative magnitudes of equivalent resistances.

5-6

AO1

4.2.2

Level 2: Diagrams or description of measurements (total current and potential difference) to be taken in each circuit lacking one or two details

Evidence of use of equation involving current, potential difference, and resistance. Correct statement about relative magnitudes of equivalent resistances.

3-4

Level 1: One correct diagram. Either potential difference or current measurement mentioned. Little or no evidence of use of equation. Little or no statement about the relative magnitudes of the equivalent resistances.

1-2

No relevant comment.

0

Indicative content:

  • two correct circuits drawn with an ammeter in each circuit in an appropriate place to measure the total current, with or without a voltmeter across the battery
  • one circuit should be a parallel circuit with a bulb in each circuit, with an ammeter in the circuit closest to the cell
  • the other circuit should be a single circuit with two bulbs and one ammeter
  • you need to measure the total current in each circuit, and the potential difference of the supply.
  • you use the equation potential difference = current × resistance
  • rearrange to give resistance = \(\frac{{{\rm{potential\ difference}}}}{{{\rm{current}}}}\) to calculate the equivalent resistance of each circuit

the equivalent resistance of the series circuit is bigger than the equivalent resistance of the parallel circuit

04.1

as the temperature increases, the resistance decreases

1

AO1

4.2.1.4

04.2

the reading on the voltmeter will not change

it is connected directly across the battery/there is only one component (other than the battery)

1

1

AO2

4.2.2

04.3

the potential differences across resistors in a series circuit are in the same proportion as the size of the resistances

there is 8 V out of the total 12 V across the thermistor, which is \(\frac{2}{3}\) of the total

so the resistance of the thermistor is 2 × 10 = 20 kΩ

this happens when the resistance is very cold

there is 3 V out of the total 12 V across the thermistor, which is \(\frac{1}{4}\)

so the resistance of the thermistor is 3.3 kΩ

this happens when the resistance is very hot

1

1

1

1

1

1

1

AO2

4.2.2

05.1

potential difference = 6 – 4

= 2 V

accept 2 V with no working for two marks

1

1

AO2

4.2.2

05.2

potential difference = current × resistance or V = IR

4 = 0.2 × resistance

resistance = \(\frac{4}{{{0}{\rm{.2}}}}\)

= 20 Ω

accept 20 with no working shown for three marks

1

1

1

AO1

AO2

4.2.1.2

05.3

R = \(\frac{2}{{{0}{\rm{.2}}}}\)

= 10 Ω

or

total resistance = \(\frac{6}{{{0}{\rm{.2}}}}\) = 30 Ω

so resistance of resistor = 30 – 20 = 10 Ω

1

1

or

1

1

AO2

4.2.2

05.4

decrease

adding a resistor in series increases the total resistance of the circuit so less current will flow

1

1

AO2

4.2.2

06.1

component A

1

AO2

4.2.1.4

06.2

evidence of reading current and potential difference from the graph e.g., current = 0.5 A, potential difference = 5.0 V

potential difference = current × resistance

resistance = \(\frac{{{\rm{potential\ difference}}}}{{{\rm{current}}}}\)

= \(\frac{{{5}{\rm{.0\ V}}}}{{{0}{\rm{.5\ A}}}}\)

= 10 (Ω)

accept any correct pair of readings

accept ten with no working shown for three marks

1

1

1

AO1

AO2

4.2.1.2

06.3

as the potential difference increases the resistance of component A stays the same

the ratio of all the potential difference and current readings stays the same = 10 (Ω)

as the potential difference increases, the resistance of component B increases

the ratio of the potential difference: current readings increases, from \(\frac{{{1}{\rm{.0}}}}{{{0}{\rm{.2}}}}\) = 5 Ω to \(\frac{{{3}{\rm{.0}}}}{{{0}{\rm{.4}}}}\) = 7.5 Ω

do not accept any answer involving gradients of the lines

1

1

1

1

AO2

4.2.1.4

06.4

at 3.0 V, total current = 0.3 A + 0.4 A

= 0.7 A

1

1

AO2

4.2.2

06.5

resistance = \(\frac{{\rm{V}}}{{\rm{I}}}\)

= \(\frac{3}{{{0}{\rm{.7}}}}\)

= 4.3 (Ω) [= 4.286(Ω)]

accept 4.3 with no working shown for two marks

1

1

07.1

negative

1

AO2

4.2.5.2

07.2

an arrow in a direction away from the sphere

1

AO2

4.2.5.2

07.3

(the force) decreases/gets smaller

1

AO2

4.2.5.1

07.4

charge = current × time or Q = It

4 × 10–6 = current × 0.2

\(\textrm{current} = \frac{4 \times 10^{- 6}}{0.2}\)

= 2 × 10–5 A

student must show working for full marks

1

1

1

1

AO1

AO2

4.2.1.2

08.1

electron(s)

1

AO1

4.2.5.1

08.2

negatively charged particles (electrons) are transferred from the jumper to the balloon

which leave a (net) negative charge on the balloon

1

1

AO2

4.2.5.1

08.3

yes

because opposite charges attract

and an attractive force keeps the balloon on the wall

or

no

the wall is neutral

but when the balloon is close to the wall it repels the electrons which leaves a net positive charge on the surface

all marks for this question are awarded for explanations, rather than for the ‘yes’/’no’ answer

1

1

or

1

1

AO2

4.2.5.2

08.4

the balloons will repel

they have the same charge on them/both are negatively charged

like/similar charges repel

1

1

1

AO1

AO2

4.2.5.2

09.1

charge = current × time or Q = It

0.6 = 15 × 10–3 × time

\(\textrm{time} = \frac{0.6}{15 \times 10^{-3}}\)

= 40 s

accept 40 with no working shown for three marks

accept 0.04 (not converting from mA) for one mark

1

1

1

AO1

AO2

4.2.1.2

09.2

graph should show a smooth curve with resistance increasing sharply initially before beginning to gradually plateau

One mark for a graph labelled mass on x-axis, resistance on y-axis

one mark for a curved shape with decreasing gradient

2

AO3

4.2.1.2

09.3

as the mass of salt increases the current increases

the potential difference is constant (6 V)

R = \(\frac{{\rm{V}}}{{\rm{I}}}\), so the resistance will decrease (as mass increases)

the current increases at a decreasing rate, so the resistance will decrease at a decreasing rate

from 25 g to 30 g the resistance decreases from 400 Ω to 240 Ω

from 75 g to 80 g the resistance decreases from 150 Ω to 146 Ω

1

1

1

1

1

1

AO2

AO3

4.2.1.2

10.1

graph should show that resistance initially drops sharply, but begins to gradually plateau as light intensity increases

the curve should be a smooth arch showing a negative correlation

one mark for plotting light intensity on x-axis and resistance on y-axis

one mark for correct units

one mark for correct shape of graph

3

AO2

4.2.1.4

10.2

sensible suggestion e.g., street lights, security lights

1

AO2

4.2.1.4

10.3

description of how it is used

appropriate circuit diagram

explanation of why it is needed

for example:

turning the lights on in a house when it gets dark outside

connect up the light dependent resistor in a circuit with an LDR, a resistor and a battery

circuit diagram with labelled components

use the output potential difference across the LDR or the resistor to switch on the lights

as the light level changes, the changing resistance produces a changing potential difference

which can be used to turn on the lights when the potential difference reaches a certain level

2

1

2

AO2

11.1

points should be plotted at the following co-ordinates:

(5, 1.5), (10, 3.8), (15, 4.6), (20, 5.9), (25, 7.8)

line of best fit should be straight

three or four points plotted correctly for one mark

all points plotted correctly for two marks

one mark for appropriate scales on correctly labelled axes

one mark for plotting an appropriate line of best fit

4

AO2

4.2.1.3

11.2

independent variable = length

dependent variable = resistance

control variable = type of metal/diameter of wire/temperature of wire

1

1

1

AO3

4.2.1.3

11.3

6.8 (Ω)

accept values between 6.5 and 7.3 (Ω)

1

AO2

4.2.1.3

11.4

take repeat measurements and calculate/plot the average/mean of repeat measurements

1

AO3

4.2.1.3

11.5

both students A and B are correct

the line is straight (so is linear)

and foes through (0,0) (so it is directly proportional)

1

1

1

AO3

4.2.1.3

12.1

energy in the kinetic energy store

is transferred to a gravitational potential energy store as the ball moves upwards.

as the ball moves through the air some energy is transferred to the thermal energy store of the surroundings

as the ball moves down energy is transferred from a gravitational potential energy store to a kinetic energy store

some of this energy is transferred to the thermal energy store of the hand

1

1

1

1

AO2

4.1.1.1

12.2

gravitational potential energy = mass × gravity × (change in) height

allow gpe = mgh

1

AO1

4.1.1.2

12.3

0.4 = 0.1 × 9.8 × change in height

change in height = \(\frac{{0.4}}{{{0}{\rm{.98}}}}\)

= 0.41 m

accept 0.4 with no working shown for three marks

1

1

1

AO2

4.1.1.2

13.1

\(\textrm{efficiency} = \frac{{{\rm{useful\ energy\ transferred}}}}{{{\rm{total\ energy\ transferred}}}}\) × 100

= \(\frac{{{\rm{4000\ }} \times {\rm{\ 1000}} \times {100}}}{{{\rm{10\ 000\ }} \times {\rm{\ 1000}}}}\) × 100

= 40 %

accept 40% with no working shown for three marks

accept 0.4 for 1 mark

1

1

1

AO1

AO2

4.1.2.2

13.2

efficiency of vehicle B is 2 × 40% = 80% or 0.8

0.8 = \(\frac{{{6000} \times {1000}}}{{{\rm{total\ energy\ transferred}}}}\)

total energy transferred = \(\frac{{{6000} \times {1000}}}{{{0}{\rm{.8}}}}\)

= 7 500 000 J = 7500 kJ

accept 7 500 000 J with no working for three marks

1

1

1

AO2

4.1.2.2

13.3

power = \(\frac{{{\rm{useful\ energy\ transferred}}}}{{{\rm{time\ taken}}}}\)

= \(\frac{6000 \times 1000}{2 \times 3600}\)

= 833 W

= 830 W to two significant figures

accept 833 with no working shown for two marks

1

1

1

1

AO1

AO2

4.1.1.4

14.1

A – the graph for a resistor with a small resistance

B – the graph for a resistor with a large resistance

C – the graph for a filament lamp

one mark for two lines correct

two marks for all lines correct

2

AO1

4.2.1.4

14.2

while potential difference is negative, current should be 0

once potential difference becomes positive, current should increase sharply

one mark for current with positive potential difference

one mark for approximately zero current with negative potential difference

2

AO1

4.2.1.4

14.3

none of them will light up

light emitting diode X is in the reverse direction/cannot allow current from positive to negative terminals of the battery.

1

1

AO2

4.2.1.4

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