A2 Daily A Level Physics question
A museum display uses a light sensor based on a potential divider: an LDR is connected to +9.0 V and a 4.0 kΩ fixed resistor to 0 V. The output voltage V_out is taken at their junction and measured by a high‑resistance data logger. In low light the LDR has 8.0 kΩ. A visitor shines a torch so the LDR resistance halves. Which statement must be true about V_out?
Answer
The correct answer is C.
Correct: C — It decreases from 6.0 V to 4.5 V because the share is R_LDR/(R_LDR + 4.0 kΩ). Using V_out = 9.0 × 8.0/(8.0+4.0) = 6.0 V initially, then 9.0 × 4.0/(4.0+4.0) = 4.5 V after the resistance halves. A assumes a falling sensor resistance makes the sensor take a larger share of the supply, but with the output across the top resistor, its share falls when its resistance falls. B assumes voltage is directly proportional to the sensor resistance alone; it ignores that the total resistance also changes, so the drop is not 50%. C matches the correct ratio reasoning for a potential divider with negligible loading. D treats the divider output as fixed by the supply, which is false—the junction voltage depends on the resistance ratio.