A-Level Physics May/June 2024 Q5(b)(ii): The temperature of Y is changed to 190°C. The resistance of X remains unchanged. Determ…

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  From the graph, at a temperature of 190°C, the resistance of thermistor Y is $R_Y = 25 \, \Omega$.

  The circuit is a potential divider. The potential difference across Y, $V_Y$, can be found using the potential divider formula. The resistance of X is given as unchanged, so $R_X = 450 \, \Omega$. The total e.m.f. is $E = 9.0 \, \text{V}$.

  $V_Y = E \times \frac{R_Y}{R_X + R_Y}$

  $V_Y = 9.0 \times \frac{25}{450 + 25}$

  $V_Y = 9.0 \times \frac{25}{475}$

  $V_Y = 0.47368... \, \text{V}$

  To 2 significant figures, $V_Y = 0.47 \, \text{V}$.

  potential difference = 0.47 V

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  How the marks are awarded

  • C1 — Correctly reading the resistance of the thermistor Y from the graph at 190°C as 25 Ω.
  • C1 — Correctly substituting the known values (E, Rx, and the new Ry) into the potential divider equation, or an equivalent method such as finding the total current first.
  • A1 — Calculating the final correct answer for the potential difference as 0.47 V, or to 3 s.f. (0.474 V).

  Common mistakes

  • Calculating the potential difference across resistor X instead of thermistor Y, leading to an answer of 8.53 V.
  • Misreading the resistance from the graph, for example by reading from the wrong temperature or misinterpreting the axis scale.
  • Using the resistance of Y from the previous part of the question instead of finding the new resistance at 190°C.
  • Making a calculation error when adding the resistances in the denominator of the potential divider formula (e.g., 450 + 25 = 470).

  Examiner tip: For potential divider questions, always identify the total voltage and the total resistance of the series branch first, then apply the ratio of the component's resistance to the total resistance to find its share of the voltage.

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