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IGCSE · 4 · practical

Electricity and magnetism · Question 10

Electricity and magnetism · Original GioPhysics question with a detailed, mark-by-mark answer guide.

Demand
demanding
Marks
8
Topics
1
Answer
Complete
practical8 marks

A student investigates how the current in a filament lamp depends on the potential difference across it. She connects the lamp to a variable power supply, with an ammeter in series and a voltmeter across the lamp.

The student's results
V / V0.01.02.03.04.05.06.0
I / A0.000.170.280.360.420.470.51
  1. (a)

    State State how the ammeter and the voltmeter must each be connected in the circuit.

    2 marks
  2. (b)

    Determine Determine the resistance of the lamp when the potential difference across it is 1.0 V, and again when it is 6.0 V.

    3 marks
  3. (c)

    Explain Explain, in terms of the filament, why the resistance changes in the way your answers to (b) show.

    3 marks
Ready to self-mark?Reveal the detailed answer guide
Answer overviewKey answer: the ammeter is connected in series with the lamp the voltmeter is connected in parallel with (across) the lamp R = V / I used at both points
01

(a)

2 marks

State State how the ammeter and the voltmeter must each be connected in the circuit.

How to approach it

Turn the task into a measurable method: name the independent, dependent and controlled quantities; describe apparatus and repeats; then explain how the evidence will be processed and how uncertainty or safety will be managed.

  1. 1

    the ammeter is connected in series with the lamp

  2. 2

    the voltmeter is connected in parallel with (across) the lamp

02

(b)

3 marks

Determine Determine the resistance of the lamp when the potential difference across it is 1.0 V, and again when it is 6.0 V.

How to approach it

Turn the task into a measurable method: name the independent, dependent and controlled quantities; describe apparatus and repeats; then explain how the evidence will be processed and how uncertainty or safety will be managed.

  1. 1

    R = V / I used at both points

  2. 2

    at 1.0 V: R = 1.0 / 0.17 = 5.9 Ω

  3. 3

    at 6.0 V: R = 6.0 / 0.51 = 11.8 Ω (accept 12 Ω)

03

(c)

3 marks

Explain Explain, in terms of the filament, why the resistance changes in the way your answers to (b) show.

How to approach it

Turn the task into a measurable method: name the independent, dependent and controlled quantities; describe apparatus and repeats; then explain how the evidence will be processed and how uncertainty or safety will be managed.

  1. 1

    as the current increases the filament gets hotter

  2. 2

    the metal ions in the filament vibrate with greater amplitude

  3. 3

    so the electrons collide with them more often and the resistance increases

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