Oxford Revise AQA GCSE Physics | Chapter P18 answers

P18: Induced potential and transformers

Answers

Extra information

Mark

AO / Specification reference

01.1

an induced potential difference is produced by the generator effect/when a conductor moves in a magnetic field/when a magnetic field changes around a conductor

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AO1

4.7.3.1

01.2

a device that produces an alternating p.d. is a: dynamo

a device that changes the p.d. is a: transformer

a device that produces a direct p.d. is a: generator

a device that changes a sound wave to an electrical signal is a: microphone

one mark for one correct line

two marks for two correct lines

three marks for three/four correct lines

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AO1

4.7.3.2

4.7.3.3

4.7.3.4

01.3

alternator

step-up

primary

magnetic field

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AO1

4.7.3.2

4.7.3.3

4.7.3.4

02.1

move the magnet faster

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AO1

4.7.3.1

02.2

move the magnet in and out of the coil

the potential difference changes from positive to negative

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1

AO2

4.7.3.1

02.3

the direction of the magnetic field inside the coil is opposite to that of the magnet

or words to that effect

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AO1

4.7.3.1

03.1

diagram with downwards arrow labelled ‘force of gravity on student’/’weight’

upwards arrow labelled ‘force of plank on student’/’normal force’/’reaction force’

arrows of equal length

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AO2

4.5.1.4

03.2

the forces on the student are equal in magnitude and opposite in direction

so there is no net force/no resultant force

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1

AO2

4.5.6.2.1

03.3

one mark for diagram with one pivot

one mark for large arrow labelled W at 1.2 m

one mark for small arrow labelled S at 2 m

one mark for correct distances labelled

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AO1

AO2

4.5.4

03.4

total clockwise moment = total anticlockwise moment

1.2 × weight = 400 × 2.0

\(
\begin{array}{lr}
\rm{weight} = \frac{800}{1.2}\\
= 670\rm{N}\ (667\rm{N})
\end{array}
\)

answer given to two significant figures

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1

AO1

AO2

4.5.4

03.5

weight = mass × gravitational field strength

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AO1

03.6

\(
\begin{array}{lr}
667 = \rm{mass} \times 9.81\\
\rm{mass} = \frac{667}{9.81}\\
= 68\ \rm{kg}\ (67.99)
\end{array}
\)

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1

AO2

4.5.4

04.1

a changing potential difference in the primary coil

produces a changing magnetic field in the core

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1

AO2

4.7.3.1

4.7.3.4

04.2

number of turns on the primary coil

potential difference across the primary coil

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1

AO2

4.7.3.4

04.3

The induced potential difference is proportional to the number of turns on the secondary coil

e.g., if you double the number of turns from 10 to 20, the mean induced potential difference doubles from 2.5V to 5.0V

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1

AO3

4.7.3.4

04.4

\(
\begin{array}{lr}
\rm{V} = \rm{kN}, \rm{k} = \frac{\rm{V}}{\rm{N}} = \frac{2.5}{10} = 0.25\\
\rm{N} = \frac{\rm{V}}{\rm{k}} = \frac{3}{0.25}\\
= 12
\end{array}
\)

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1

AO3

4.7.3.4

05.1

two coils around an iron core

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AO1

4.7.3.4

05.2

\(
\begin{array}{lr}
\frac{\rm{V}_\rm{p}}{12} = \frac{2000}{100}\\
\rm{V}_\rm{p} = 12\ \times\ 20\\
= 240\ \rm{V}
\end{array}
\)

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AO2

4.7.3.4

05.3

no

it is a step down because Np is bigger than Ns

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1

AO1

AO2

4.7.3.4

05.4

iron is a magnetic material

the magnetic field in the core in stronger

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1

AO1

AO2

4.7.3.4

06.1

the wire is a moving conductor in a magnetic field

which is cutting the magnetic field (lines) causing a changing magnetic field

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1

AO1

AO2

4.7.3.1

06.2

the graph shows how the induced pd varies with the position of the coil

the induced potential difference is maximum when the coil is parallel to the magnetic field/cuts the field lines at 90°/perpendicularly

the coil could be at B or D

this is where the (plane of) the coil is parallel to the field

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1

AO1

AO2

4.7.3.2

06.3

rotate the coil through 90°/put the coil at 90° to its current position

put an alternating current through the turns of the electromagnet

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1

AO3

4.7.3.1

07.1

~ = a.c./alternating current

—- = d.c./direct current

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1

AO3

4.7.3.1

07.2

power = potential difference × current

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AO1

07.3

mains potential difference = 230 V

current = 1.8 A

power = 230 × 1.8

= 414

= 410 W

recalling mains potential difference

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AO1

AO2

4.7.3.4

07.4

power = potential difference × current

= 19.5 × 4.62

= 90.09

= 91 W

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AO2

4.7.3.4

07.5

the output power is much smaller than the input power

the transformer is not 100% efficient

some of the energy transferred by the current is transferred to the thermal energy store of the surroundings so the adapter gets hot

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1

1

AO3

4.7.3.4

08.1

one mark for general shape

one mark for potential difference in one direction only/all positive or all negative

one mark for labelled axes

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AO2

4.7.3.2

08.2

the potential difference is only one sign/only positive/only negative

which shows that it is in one direction only

which will produce a direct (not an alternating) current

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1

AO3

4.7.3.2

08.3

the potential difference would be negative/positive

must be opposite potential difference to that shown in graph of question 8.1

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AO2

4.7.3.2

08.4

there would be more cycles/oscillations

the magnitude of the potential difference is twice as big

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1

AO2

4.7.3.2

09.1

a transformer contains wire that heats up when a current flows through it

transferring energy to the thermal energy store of the surroundings

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AO2

4.7.3.4

09.2

power = potential difference × current

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AO1

4.7.3.4

09.3

\(
\begin{array}{lr}
\rm{current} = \frac{{{\rm{power}}}}{{{\rm{potential\ difference}}}}\\
120\ \rm{MW} = 1.2\ \times\ 10^8\ \rm{W}\\
400\ \rm{kV} = 400\;000\rm{V}\\
\frac{1.2 \times 10^8}{400\;000}\\
= 300\ \rm{A}
\end{array}
\)

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AO2

4.7.3.4

10

Level 3: Correctly links pressure to movement of diaphragm. Links movement to potential difference in positive and negative direction. Well organised answer.

5-6

AO1

AO2

4.7.3.3

Level 2: Links movement of air to movement of diaphragm. Links movement to potential difference in but not direction. Some organisation of answer.

3-4

Level 1: Some link between air movement and movement of coil. Answer shows poor organisation.

1-2

No relevant content.

0

Indicative content:

  • there are areas of high and low pressure in a sound wave
  • because air molecules in a sound wave move backwards and forwards as it is a longitudinal wave
  • when the high-pressure area hits the diaphragm, it pushes the diaphragm in
  • the coil moves through the magnetic field
  • which produces a potential difference in one direction
  • when the low-pressure area hits the diaphragm, it pulls the diaphragm out/ the diaphragm is not pushed in as far
  • which produces a potential difference in the other direction

11.1

20Hz – 20 000Hz/20 – 20kHz

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AO1

4.6.1.4

11.2

speed = frequency × wavelength

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AO1

4.6.1.2

11.3

5000 = 400 000 × wavelength

\(
\begin{array}{lr}
\rm{wavelength} = \frac{5000}{400\ 000}\\
= 0.013\ \rm{m}
\end{array}
\)

do not credit answers with three significant figures (0.0125)

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AO2

4.6.1.2

11.4

distance = speed × time

distance = 5000 × 8.4 × 10–7

= 0.0 042m

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AO1

AO2

4.5.6.1.2

11.5

\(
\begin{array}{lr}
\frac{0.0042}{2}\\
= 0.0 021\ \rm{m}
\end{array}
\)

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12.1

\(\frac{1.85 + 2 + 1.9}{3} = 1.92\)

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AO2

4.7.3.1

12.2

one mark for points correctly plotted

one mark for line of best fit

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AO2

AO3

4.7.3.1

12.3

0.8 V

allow 0.7 – 0.9 V

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AO2

4.7.3.1

12.4

random error

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AO3

4.7.3.1

13.1

the atmosphere is assumed to have a constant density

so the pressure at a point is due to the weight of air above that point

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1

AO1

4.5.5.2

13.2

as you come down a mountain the height/weight of air above you increases

so the pressure increases

which is the same when you dive in the ocean

but the liquid pressure is much bigger than the atmospheric pressure.

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AO1

AO2

4.5.5.2

4.5.5.1.2

13.3

\(
\begin{array}{lr}
\frac{\rm{density\ of\ water}}{\rm{density\ of\ air}} = \frac{84}{{{0}{.1}}}\\
=840
\end{array}
\)

water is 840 times denser than air

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AO2

4.5.5.2

4.5.5.1.2

14.1

diagram with 4 arrows:

vertical arrow down: weight/force of gravity on boat

vertical arrow up: upthrust/force of water on boat

horizontal arrow: thrust

opposing horizontal arrow: drag/force of water and air on boat

allow two separate arrows for air/water resistance

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AO1

AO2

4.5.1.2

4.5.1.4

14.2

force = mass × acceleration

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AO1

14.3

3000 = 850 × acceleration

\(
\begin{array}{lr}
\rm{acceleration} = \frac{300}{850}\\
= 3.5\ \rm{m/s}^2
\end{array}
\)

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AO2

4.5.6.2.2

14.4

force = mass × acceleration

= 2.7 × 850

= 2295 N (2300)

resultant force = engine force – drag force

2295 = 3000 – drag force

drag force = 3000 – 2295

= 705 N

= 710 N (two significant figures)

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AO1

AO2

4.5.6.2.2

14.5

final velocity2 – initial velocity2 = 2 × distance × acceleration

(14) 2 – (0) 2 = 2 × distance × 2.7

\(
\begin{array}{lr}
\rm{distance} = \frac{196}{2 \times 2.7}\\
= 36\ \rm{m}\ (36.3)
\end{array}
\)

allow v2 = u2 + 2as

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1

1

AO2

4.5.6.1.5

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