A2 Daily A Level Physics question
A security device uses an LDR in series with a 10 kΩ fixed resistor across a 5.0 V supply. The output Vout is the potential difference across the LDR and drives a comparator that turns a buzzer on when Vout > 3.0 V. In room light the LDR is 20 kΩ; when a torch shines on it, its resistance halves. Which outcome must be true?
Answer
The correct answer is A.
Correct: A — Vout decreases from about 3.3 V to about 2.5 V, so the buzzer switches off. A is correct because Vout = 5.0 × R_LDR/(R_LDR + 10 kΩ) gives about 3.3 V at 20 kΩ and 2.5 V at 10 kΩ, falling below the 3.0 V threshold so the buzzer turns off. B is wrong because those voltages are for the fixed resistor, not the LDR, and 2.5 V is still below 3.0 V so it would not switch on. C is wrong because an LDR’s resistance decreases in brighter light, so Vout across the LDR must fall, not rise to ~4.2 V. D is wrong because it again treats Vout as across the fixed resistor and also miscalculates the brighter-light value (it should rise to 2.5 V, not drop to ~1.3 V).