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A2 Daily A Level Physics question

2026-07-15 OCR A DC circuits: potential dividers 4.3.3(b) 4.3.3(c)(i) 4.3.1(c)

In a lab temperature probe, a 6.0 V supply feeds a fixed 4.0 kΩ resistor in series with an NTC thermistor. V_out is taken across the thermistor. At 20 °C the thermistor is 4.0 kΩ; when warmed its resistance halves. Which statement must be true about V_out, and why?

  1. A A It increases from 3.0 V to about 4.0 V — lower resistance takes more of the supply.
  2. B B It decreases from 3.0 V to about 1.5 V — halving the resistance halves the voltage.
  3. C C It decreases from 3.0 V to about 2.0 V — the thermistor is 2 kΩ of a 6 kΩ series pair. (correct)
  4. D D It stays at about 3.0 V — the supply voltage is unchanged.

Answer

The correct answer is C.

Correct: C — C It decreases from 3.0 V to about 2.0 V — the thermistor is 2 kΩ of a 6 kΩ series pair. Initially V_out = 6.0 × (4.0 / (4.0 + 4.0)) = 3.0 V; after heating, V_out = 6.0 × (2.0 / (4.0 + 2.0)) = 2.0 V, because the share of total resistance becomes 2/6. A wrongly assumes a lower resistance gets a larger share of the supply; in a divider, a smaller fraction of total resistance gets a smaller fraction of the voltage. B applies a naïve V ∝ R for an isolated component and ignores that the total series resistance also changes, so the fraction is not simply halved. C is correct as shown by the ratio 2 kΩ out of 6 kΩ. D ignores the divider action; the supply is fixed but the voltage distribution between series components changes with their resistances.

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