AS Daily A Level Physics question
A 5.0 V microcontroller reads V_out from a potential divider used as a light sensor. The divider is made from a fixed 1.0 kΩ resistor and an LDR that is about 0.50 kΩ in bright light and about 5.0 kΩ in dim light. V_out is the junction voltage measured relative to 0 V. The control algorithm requires V_out to be higher in dim light than in bright, and roughly 2.5 times larger. Which statement must be true about how the LDR is connected and how V_out changes?
Answer
The correct answer is A.
Correct: A — Connect the LDR to 0 V (fixed 1.0 kΩ to +5 V); then V_out in dim light is about 2.5 times that in bright, and V_out increases as it gets darker. A With the LDR at the lower leg, V_out is the share across the lower component, so as the LDR rises from 0.50 kΩ to 5.0 kΩ, V_out rises from about 1.67 V to about 4.17 V; the ratio (≈4.17/1.67) is ≈2.5, and in the limit of very dark (LDR ≫ 1.0 kΩ) V_out tends to the 5.0 V supply. B Putting the LDR at the upper leg makes the junction across the fixed 1.0 kΩ, so V_out falls with darkness (≈3.33 V to ≈0.83 V), not rising by a factor ≈2.5. C With the LDR at the lower leg the trend is the opposite of what is stated: V_out increases with darkness and the dim/bright ratio is ≈2.5, not ≈0.4. D With the LDR at the upper leg the direction (decreases with darkness) is correct, but the magnitude is not: the dim/bright ratio is about 0.25, far from ~1.5.