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AS Daily A Level Physics question

2026-02-03 OCR A Electric circuits: potential dividers (Module 4) OCR-A Module 4.2.3 Electrical circuits: series/parallel rules and potential dividers

A technician uses a 12.0 V supply and a potential divider made from a 3.0 kΩ resistor in series with a 6.0 kΩ resistor. The output voltage is taken across the 6.0 kΩ resistor. Later, both resistors are replaced by ones that are 50% higher (4.5 kΩ and 9.0 kΩ). Which statement must be true about the output voltage and why?

  1. A It decreases to about 5.3 V, because the larger total resistance reduces the current and therefore the drop across that branch.
  2. B It increases to 12.0 V, because higher resistances use up more of the supply voltage.
  3. C It rises to 9.0 V, because increasing that resistor from 6.0 kΩ to 9.0 kΩ gives it a bigger share of 12.0 V.
  4. D It remains at 8.0 V, because the output is a fixed fraction of the supply set by the ratio of the two resistances, which is unchanged. (correct)

Answer

The correct answer is D.

Correct: D — It remains at 8.0 V, because the output is a fixed fraction of the supply set by the ratio of the two resistances, which is unchanged. Initially Vout = 12 × 6/(3+6) = 8.0 V; after scaling both to 4.5 kΩ and 9.0 kΩ, Vout = 12 × 9/(4.5+9) = 8.0 V. A confuses current reduction with voltage division: although current falls, V across a given leg depends on the ratio, not on the absolute current, so 5.3 V is incorrect. B assumes larger resistances automatically take the entire supply, which ignores that the split must still sum to 12 V and is set by the ratio. C assumes the absolute value of the lower resistor alone sets its share; the ratio 6:(3+6) and 9:(4.5+9) are both 2/3, so 9.0 V is not obtained.

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