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A2 Daily A Level Physics question

2026-02-08 OCR A Sensor circuits (M4) 4.3.2 Electric circuits: potential divider and sensor applications 4.2.2 Energy, power and resistance: thermistors and LDRs (qualitative behaviour)

A microcontroller reads the output of a 5.0 V potential divider used as an over‑temperature sensor on a heatsink. The divider has a fixed 10 kΩ resistor in series with an NTC thermistor; the output to the ADC is taken across the thermistor to ground. At 20 °C the thermistor is 10 kΩ. When the heatsink warms to 40 °C, the thermistor’s resistance halves. Which output change is most likely?

  1. A Increases to about 3.3 V because the divider gives more voltage to the thermistor.
  2. B Decreases from 2.5 V to about 1.7 V as the thermistor’s share of the divider falls. (correct)
  3. C Remains near 2.5 V because the divider is symmetrical.
  4. D Halves to about 1.25 V because halving resistance halves its voltage.

Answer

The correct answer is B.

Correct: B — Decreases from 2.5 V to about 1.7 V as the thermistor’s share of the divider falls. A NTC thermistor’s resistance halves, so V_out = 5.0 × R_T/(10 kΩ + R_T) changes from 5.0 × 10/(10+10) = 2.5 V to 5.0 × 5/(10+5) ≈ 1.7 V. A assumes the output is taken across the fixed resistor or that heating raises the sensor voltage; with output across the thermistor, its reduced fraction 5/(10+5) lowers V_out. B matches the ratio calculation: 2.5 V at 20 °C to ≈1.7 V at 40 °C. C wrongly assumes the 1:1 split persists; once the thermistor changes from 10 kΩ, the divider is no longer symmetrical. D assumes constant current so voltage scales directly with resistance; in a divider the current also changes, so the voltage does not simply halve.

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