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AS Daily A Level Physics question

2026-03-01 OCR A DC Circuits (M4) OCR-A 4.3.2 Electrical circuits — Potential divider and sensing circuits (thermistors/LDRs)

A portable incubator uses an NTC thermistor in a potential divider powered by 9.0 V. The fixed 3.0 kΩ resistor is connected to +9.0 V, the thermistor is connected to 0 V, and the output V_out is the voltage across the thermistor (from the junction to 0 V). At 20°C the thermistor is 3.0 kΩ. When warmed so its resistance halves, which statement best describes the change in V_out?

  1. A V_out decreases from 4.5 V to 3.0 V (about a 33% drop), because the thermistor's share of the total resistance falls. (correct)
  2. B V_out increases from 4.5 V to 6.0 V, because the larger current produces a larger voltage across the thermistor.
  3. C V_out remains about 4.5 V, because the supply voltage is unchanged.
  4. D V_out decreases from 4.5 V to 2.25 V, because halving the resistance halves its voltage.

Answer

The correct answer is A.

Correct: A — V_out decreases from 4.5 V to 3.0 V (about a 33% drop), because the thermistor's share of the total resistance falls. Initially the two 3.0 kΩ resistors split 9.0 V equally (4.5 V); halving the thermistor to 1.5 kΩ makes its fraction 1.5/(3.0+1.5) = 1/3, so V_out = 9.0 × 1/3 = 3.0 V. B is incorrect because although the current rises, the thermistor is a smaller fraction of the series total, so it gets a smaller share of the supply, not a larger one. C is incorrect because a potential divider’s output depends on the resistance ratio, which changes when the thermistor resistance changes. D is incorrect because the voltage across one series component does not simply halve when its resistance halves; it depends on its fraction of the total, which gives 3.0 V here, not 2.25 V.