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6 - 2 Electricity – Trilogy
1 .0 Most domestic appliances are connected to the mains electricity.
1.1 What is the frequency of mains electricity?
Tick one box
[1 mark]
1.05 A
50 Hz
230 V
1.2 What is the potential difference of mains electricity?
Tick one box
[1 mark]
1.05 A
50 Hz
230 V
1.3 Most domestic appliances are connected to the mains electricity with a plug.
Explain why a plug needs a live and a neutral wire.
[2 marks]
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1.4 The law specifies the colour that mains wires should be for all domestic electrical
circuits.
It is important that the live wire is easy to identify to reduce the risk of an electric shock.
Explain how an electric shock can be caused by a person touching the live wire.
[2 marks]
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The live wire carries current to the appliance and the
neutral wire is needed to complete the circuit bringing
any current back to the plug.
The live wire has current flowing through it and has a
high potential difference at 230V compared to a human
at around 0v this means charge can flow from the live
wire to the human causing an electric shock.
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1.5 An iron is supplied with a current of 3 A from the mains. The resistance of the iron is
Calculate the power of the iron.
[2 marks]
Power = ______________ W
2 .0 A student wants to investigate how the current through a filament lamp affects its resistance.
2.1 Use the circuit symbols in Figure 1 to draw a circuit diagram that he could use.
[2 marks]
Figure 1
12 V battery variable
resistor
filament
lamp
voltmeter ammeter
2.2 Describe how the student could use her circuit to investigate how the current through a
filament lamp affects its resistance.
[4 marks]
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First, measure the initial current and potential
difference in the series circuit after you have the
reading take out the battery.Then increase the
resistance using the variable resistor to get positive
values and switch battery to get negative values for
it.Next, plot these values on a graph.The graph
should have a S shape.
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4. 0 A student investigated how current varies with potential difference for two different
lamps of the same type.
Her results are shown in the Figure 3 below.
Figure 3
4.1 Draw the circuit diagram for the circuit that the student could have used to obtain the
results shown in the figure above.
[3 marks]
4.2 The student made the following conclusion,
‘Lamp A is twice a bright as lamp B’.
Use data from Figure 3 to explain why the student’s conclusion is correct.
[3 marks]
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At all potential differences current was double in lamp
A compared to lamp B, for example at 6V current was
0.3 amps in lamp B compared to 0.6 in lamp A and so
a higher current means an increase in light intensity.
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MARK SCHEME
Qu No. Extra Information Marks 1.1 50 Hz 1 1.2 230 V 1 1.3 Live wire carries the (alternating) potential difference/voltage (from the supply) Neutral wire completes the circuit
1.4 connection is made to earth charge can flow through the body. or large potential difference across the body Accept answer in terms of a complete circuit or establishing a path (for charge to flow)
- 5 P = (3)^2 x 100 900 (W) Allow one mark for P=I^2 V if substitution incorrect. Allow 900 (W) with no working for 2 marks
Qu No. Extra Information Marks 2.1 battery, lamp and ammeter connected in series with variable resistor voltmeter in parallel with (filament) lamp
Level 2: (^) A detailed and coherent description of the experiment. The response provides a logical sequence.
Level 1: Simple description of the experiment with some steps missing. The response may not be in a logical sequence and may not lead to the collection of valid results.
Level 0: No relevant content. 0 Indicative content
- ammeter used to measure current
- voltmeter used to measure potential difference
- resistance of variable resistor altered to change current in circuit or change potential difference (across filament lamp)
- resistance (of filament lamp) calculated or R=V / I statement resistance calculated for a large enough range of different currents that would allow a valid conclusion about the relationship to be made
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Qu No. Extra Information Marks 3 .1 V = 0.025 × 75 1.9 (V) Allow 1.9 (V) with no working for 2 marks
3 .2 total resistance = 6 / 0. R = 240 - 225 = 15 (Ω)
3 .3 resistance decreases current increases
Qu No. Extra Information Marks 4 .1 battery in series with bulb and ammeter voltmeter in parallel with the bulb variable resistor or variable power supply
4 .2 correct pair of current readings at the same pd therefore current in lamp A is twice the current in lamp B so lamp A is twice as powerful and lamp B (hence is twice as bright) eg at 10 V, IA = 0.74A and IB = 0.37A must refer to power/rate of energy transfer
4 .3 R = V / I
lowest R = 0.6 / 0. R = 6 Ω Highest R = 10 / 0. R = 13.5 Ω Difference = 13.5 – 6 = 7.5 Ω allow R= 1.0 / 0. R = 6.25 Ω (other values may be acceptable but the values from the graph must be when V ≤ 1V and the lamp can reasonably be assumed to be ohmic) allow 7.25 Ω if consistent