Kirk Balk Academy

Physics Intervention – Topic 2 – Electricity (Resistance)Q1.(a) Electrical circuits often contain resistors.

The diagram shows two resistors joined in series.

Calculate the total resistance of the two resistors.

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Total resistance = .................................................. Ω (1)

(b) A circuit was set up as shown in the diagram. The three resistors are identical.

(i) Calculate the reading on the voltmeter..............................................................................

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Reading on voltmeter = .................................................. V (2)

(ii) The same circuit has now been set up with two ammeters.

Draw a ring around the correct answer in the box to complete the sentence.

smaller than

The reading on ammeter A2 will be equal to the reading on ammeter A1.

greater than

(1)

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(Total 4 marks)Q2.A student wants to investigate how the current through a filament lamp affects its resistance.

(a) Use the circuit symbols in the boxes to draw a circuit diagram that she could use.

12 V battery Variable resistor Filament lamp voltmeter ammeter

(2)

(b) Describe how the student could use her circuit to investigate how the current through a filament lamp affects its resistance.

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(c) The student’s results are shown in Figure 1.

Describe how the resistance of the filament lamp changes as the current through it increases.

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..........................................................................................(1)

(d) Use Figure 1 to estimate the resistance of the filament lamp when a current of 0.10 A passes through the lamp.

Resistance = .......................................... Ω(1)

(e) The current-potential difference graphs of three components are shown in Figure 2.

Use answers from the box to identify each component.

diode filament lamp light dependent resistor

resistor at constant temperature thermistor

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(3)(Total 11 marks)

Q3.A student set up the electrical circuit shown in the figure below.

(a) The ammeter displays a reading of 0.10 A.

Calculate the potential difference across the 45 Ω resistor.

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..........................................................................Potential difference = .................................... V(2)

(b) Calculate the resistance of the resistor labelled R.

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Resistance = ............................................. Ω (3)

(c) State what happens to the total resistance of the circuit and the current through the circuit when switch S is closed.

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(Total 7 marks)

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Q4.An electrician is replacing an old electric shower with a new one.

The inside of the old shower is shown in Figure 1.

(a) If the electrician touches the live wire he will receive an electric shock.

Explain why..............................................................................

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(b) Different electrical wires need to have a cross-sectional area that is suitable for the power output.

Figure 2 shows the recommended maximum power input to wires of different cross-sectional areas.

Figure 2

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The new electric shower has a power input of 13.8 kW.

Determine the minimum diameter of wire that should be used for the new shower.

The diameter, d, can be calculated using the equation:

A is the cross-sectional area of the wire.

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Minimum diameter = ................................................. mm(2)

(c) The charge that flows through the new shower in 300 seconds is 18 000 C.The new electric shower has a power of 13.8 kW.

Calculate the resistance of the heating element in the new shower.

Write down any equations you use.

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Resistance = .............................................. Ω(5)

(Total 11 marks)

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M1.(a) 25(Ω)1

(b) (i) 2(V)allow 1 mark for showing a correct method, ie 6 / 3

2

(ii) equal to1

[4]

M2.(a) battery, lamp and ammeter connected in series with variable resistor1

voltmeter in parallel with (filament) lamp1

(b) Level 2 (3–4 marks):A detailed and coherent description of a plan covering all the major steps is provided.The steps are set out in a logical manner that could be followed by another person toobtain valid results.

Level 1 (1–2 marks):Simple statements relating to relevant apparatus or steps are made but they maynot be in a logical order. The plan would not allow another person to obtain valid results.

0 marks:No relevant content

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 made4

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(c) (as current increases) resistance increases (at an increasing rate)1

(d) any value between 6.3 and 6.9 (Ω)1

(e) A: Filament lamp1

B: Resistor at constant temperature1

C: Diode1

[11]

M3.(a) V = 0.10 × 451

4.5 (V)1

(b) R = 12 / 0.101

total resistance = 120 (Ω)1

R = 120 – 105 = 15 (Ω)1

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(c) (total) resistance decreases1

(so) current increases1

[7]

M4.(a) (because the) potential of the live wire is 230 V1

(and the) potential of the electrician is 0 V1

(so there is a) large potential difference between live wire and electrician1

charge / current passes through his bodyallow voltage for potential difference

1

(b) diameter between 3.50 and 3.55 (mm)allow correct use of value of cross-sectional area of 9.5 to 9.9 (mm2) with no final answer given for 1 mark

2

(c) 18000 = I × 3001

I = 18000 / 300 = 601

13 800 = (602) × R1

R = 13 800 / 602

1

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3.83 (Ω)1

allow 3.83(Ω) with no working shown for 5 marksanswer may also be correctly calculated using P = IV and V = IR if 230 V is used.

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