AQA GCSE Physics Paper 1 Higher, June 2025: question 3

A PrepWise guide to a real paper, with our descriptions and answers. Not made or approved by AQA.

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PrepWise guide: AQA GCSE Physics Paper 1 HigherQuestion 33 of 10

A student measures the current through an LED at different values of potential difference. Figure 4 is the circuit, with its supply connected the first way. Figure 5 is the graph of the readings.

Figure 4
Circuit diagram: a battery, switch, ammeter, variable resistor and LED in a loop, with a voltmeter connected across the LED.
Figure 5
Graph of current against potential difference for an LED. Current is zero up to 2.0 volts, then rises in a curve.
03.1[1 mark]

Using the circuit in Figure 4, name one way to change the LED's potential difference.

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Adjust the variable resistor in the circuit.

1 mark

How the marks are earned

  1. Give one practical change that alters the LED's potential difference: use or adjust the variable resistor, add or remove cells, add a second battery, pick a battery with a different potential difference, or put extra parts in the loop.
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03.2[1 mark]

Which word equation is right? Tick one box.

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potential difference = current + resistance
potential difference = current × resistance
potential difference = current ÷ resistance
potential difference = resistance ÷ current
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potential difference = current × resistance. In symbols:. V = I × R

1 mark

How the marks are earned

  1. One mark: tick the equation where potential difference equals current multiplied by resistance.
Revise Resistance & Ohm's Law
03.3[4 marks]

Find the LED's resistance at 3.0 V, reading from Figure 5.

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Go up from 3.0 V on the horizontal axis to the curve, then across to the vertical axis. The current is 0.125 A.

Start from this equation:. V = I × R. Put in the values:. 0.125 × R = 3.0. Rearrange for the resistance:. R = 3.0 over 0.125. = 24 Ω

4 marks

How the marks are earned

  1. One mark: read the current from the graph at the stated potential difference.
  2. One mark: substitute into V = I × R.
  3. One mark: rearrange to R = V ÷ I.
  4. One mark: the correct resistance, 24. The unit, ohms, is not needed for this mark, but always write it. For the first mark, read between 0.12 A and 0.13 A. Later marks follow your own reading between 0.10 A and 0.15 A.

Why it scores

A careful graph reading is the first mark. A small reading error is allowed, and later marks follow your own value.

The substitution, the rearrangement and the answer each earn a mark, so show all three lines.

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03.4[2 marks]

The student swaps the supply connections in Figure 4. The ammeter now reads 0.0 A at every potential difference. Explain this reading.

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An LED has a very high resistance in the reverse direction. So the current is too small for the ammeter to show, and it reads 0.0 A.

2 marks

How the marks are earned

  1. One mark: the LED has a very high resistance.
  2. One mark: this very high resistance applies in the reverse direction. This mark depends on the first.
  3. If neither is scored, one mark is given for saying an LED only lets current through one way.

Why it scores

The first mark is the very high resistance, and the second links it to the reverse direction. The second mark depends on the first.

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AQA owns the copyright in its exam questions, so we cannot copy their wording. These descriptions and answers are ours. Not made or approved by AQA.[8 marks]