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A student investigated how the current in a filament lamp varied with the potential difference across the filament lamp - AQA - GCSE Physics - Question 1 - 2020 - Paper 1

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A student investigated how the current in a filament lamp varied with the potential difference across the filament lamp. Figure 1 shows part of the circuit used. F... show full transcript

Worked Solution & Example Answer:A student investigated how the current in a filament lamp varied with the potential difference across the filament lamp - AQA - GCSE Physics - Question 1 - 2020 - Paper 1

Step 1

Complete Figure 1 by adding an ammeter and a voltmeter.

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Answer

In Figure 1, add the ammeter in series with the filament lamp (denoted by a circle symbol) and the voltmeter in parallel with the filament lamp. Use the correct symbols:

  • Ammeter: a circle with 'A' inside, connected in series.
  • Voltmeter: two horizontal lines with 'V' inside, connected in parallel.

Step 2

Draw a line on Figure 2 to show the relationship between the negative values of current and potential difference.

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Answer

On Figure 2, draw a smooth, curved line that starts from the top left quadrant and ends in the bottom right quadrant. The line should pass through points that reflect the negative values of current and potential difference observed (for example, at -4.0 V, -0.2 A and -6.0 V, -0.23 A). Ensure the line does not become horizontal.

Step 3

Write down the equation which links current (I), potential difference (V) and resistance (R).

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Answer

The equation that links current (I), potential difference (V), and resistance (R) is:

V=I×RV = I \times R

Step 4

Determine the resistance of the filament lamp when the potential difference across it is 1.0 V.

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Answer

To determine the resistance of the filament lamp, use the formula:

R=VIR = \frac{V}{I} Given that the potential difference (V) is 1.0 V, we need the corresponding current (I) from Figure 2. Assume from the graph that at 1.0 V the value of current is 0.08 A. Therefore:

R=1.0V0.08A=12.5ΩR = \frac{1.0 \, \text{V}}{0.08 \, \text{A}} = 12.5 \, \Omega.

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