AS Daily A Level Physics question
A designer uses a 6.0 V potential divider to feed a microcontroller input. The upper leg is an LDR connected to +6.0 V; the lower leg is a fixed 4.0 kΩ resistor to 0 V. The output voltage is taken at their junction. In bright lab light the LDR is 2.0 kΩ. The light level halves; assume the LDR’s resistance doubles when illumination halves. Which statement must be true?
Answer
The correct answer is B.
Correct: B — The output decreases from 4.0 V to 3.0 V (a 25% drop); in very dark conditions it tends towards 0 V. Bright: Vout = 6.0 × 4.0/(2.0 + 4.0) = 4.0 V; after halving light the LDR doubles to 4.0 kΩ so Vout = 6.0 × 4.0/(4.0 + 4.0) = 3.0 V, and as the LDR becomes very large in the dark the output tends to 0 V. A is wrong because it reverses the direction (the junction moves closer to 0 V as the top resistance grows) and its limiting behaviour incorrectly goes to the supply. B is correct as above. C is wrong because it assumes the output halves when the LDR resistance doubles; the divider ratio depends on both resistances, giving 3.0 V not 2.0 V (though its dark limit is right). D is wrong because the ratio does change (not just the total), so the output does not stay at 4.0 V, and its dark-limit trend is also reversed.