Oxford Revise AQA A Level Physics | Chapter P29 answers

Question

Answers

Extra information

Mark

AO

1.1

Gate, drain, and source all correctly labelled

1

3.13.1.1

1.2

Has very high input impedance/resistance

No loading of logic gate output

1

3.13.1.1

1.3

MOSFET can easily become charged and switch on – the resistor prevents this/keeps gate at 0 when no input

1

3.13.1.1

1.4

Diode

An induced e.m.f. is created when a motor is switched on or off

Without diode this would destroy MOSFET

1

3.13.1.1

2.1

2.2 V is min p.d. across gate and source to form a channel between drain and source/gate voltage when the transistor is just switched off

1

3.13.1.1

2.2

Resistance increases as light intensity decreases

Potential difference across the LDR increases

Until threshold voltage achieved/current flows between drain and source and LED switched on

1

3.5.1.5

2.3

p.d. across R = 6 − 2.2 = 3.8 V

1

3.5.1.5

3.1

Max 4 marks from:

• Behaviour in forward bias direction same/same threshold voltage and then low resistance
• Diode has large breakdown voltage/zener diodes have low breakdown voltages
• Diode breaks if breakdown voltage exceeded/voltage across zener/zener voltage remains constant even when current exceeded
• Reverse current of zener remains constant/at least 5 mA until breakdown voltage applied

Must have similarities and differences for full marks

max 4

3.13.1.2

3.2

Correct symbol drawn

Arrow pointing up

1

3.13.1.2

3.3

1

3.5.1.4

3.4

V = 9 – 4.3 = 4.7 V

1

3.5.1.1

4.1

In parallel with zener diode so p.d. = 4.7 V

1

3.13.1.2

4.2

p.d. = 10 – 4.7 = 5.3 V

1

3.13.1.2

4.3

I_Z = 0.044 – 0.011 = 0.033 A

1

3.13.1.2

4.4

P = IV = 0.033 A × 4.7 = 0.16 W < 0.25 W

Use 250 mW zener diode

1

3.13.1.2

5.1

Reverse

1

3.13.1.3

5.2

Inverted parabola

Peak at 850 nm

Scale from 550 nm (min) to 1150 nm (max)

Ignore y-axis scale

1

3.13.1.3

5.3

Power = I × Area = 10 × 1 × 10⁻⁶ m²

Current = sensitivity × power = 0.62 × 10 × 1 × 10⁻⁶

Current = 6.2 × 10⁻⁶ A

1

3.13.1.3

5.4

Max 4 marks from:

• Photodiode in series with a resistor and a power supply
• V_out across the resistor
• When it is dark, the current is negligible and there is no p.d. across resistor. No V_out
• When it is light, there is a current and hence a p.d. across the resistor – this is V_out
• Resistor chosen to provide suitable p.d. to activate alarm

All marks can be awarded for suitably labelled diagram

max 4

3.13.1.3

6.1

Magnet placed on frame of set in line with hall effect sensor (allow vice versa)

When the magnet is close to the sensor, voltage output is high/far from it, voltage output low

Output fed to circuits for airbags/output amplified and input to circuits for airbags

1

3.13.1.4

6.2

3.5 × 10⁻⁴ T = 0.35 mT

V = 9 × 0.35 = 3.2 mV

0.014 T = 14 mT

V = 9 × 14 = 126 mV

1

3.13.1.4

6.3

0.065 mT

V = 0.59 mV

Too small to affect reading as this was max strength

20% of smallest value so would affect accuracy of output

1

3.13.1.4

7.1

As the magnet passes the sensor, it registers a voltage output

The number of voltage outputs can be counted

Using circumference of wheel, the distance can be measured/number of counts multiplied by distance of wheel

1

3.13.1.4

7.2

Distance = π × D = π × 0.05 m

1

3.13.1.4

7.3

Have more magnets on disc/have a smaller radius disc

Allow any sensible suggestion here

1

3.13.1.4

8.1

Reverse

1

3.13.1.3

8.2

Leakage current when there is no/zero light intensity

1

3.13.1.3

8.3

Current = 0.6 A W⁻¹ × 0.2 × 10⁻³ W

Current = 1.2 × 10⁻⁴ A

1

3.13.1.3

8.4

p.d. across R = IR = 1.2 × 10⁻⁴ A × 5000 Ω = 0.6 V

V_out = 9.0 – 0.6 = 8.4 V

1

3.13.1.3

Question

Answer

1

2

P = IV = 10 × 10⁻³ A × 2.7 V = 0.03 W

3