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AS Daily A Level Physics question

2026-07-01 OCR A DC Circuits (M4) 4.2.3 Electrical circuits — potential divider 4.2.2 Energy, power and resistance — resistance and temperature

A student builds a simple temperature probe using a 9.0 V supply and a potential divider: a 6.0 kΩ fixed resistor in series with an NTC thermistor. The output Vout is measured across the thermistor to ground. At 20 °C the thermistor is 6.0 kΩ. When warmed so its resistance halves, what happens to Vout? Assume the supply is ideal.

  1. A It decreases from 4.5 V to 3.0 V. (correct)
  2. B It increases from 4.5 V to 6.0 V.
  3. C It stays at 4.5 V (each gets half of 9.0 V).
  4. D It decreases from 4.5 V to 2.25 V.

Answer

The correct answer is A.

Correct: A — It decreases from 4.5 V to 3.0 V. A uses the correct ratio: initially equal resistances split 9.0 V to 4.5 V; halving the thermistor to 3.0 kΩ gives Vout = 9 × 3/(6 + 3) = 3.0 V. B assumes the smaller resistor gets a larger share; 6.0 V is actually the drop across the 6.0 kΩ fixed resistor after warming. C assumes half the supply regardless of resistance; equal split only occurs when resistances are equal. D assumes the current is unchanged so V drops in proportion to R; in fact total resistance falls, current rises, and the correct drop across the thermistor is 3.0 V, not 2.25 V.

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