Jan25-10, 07:13 AM
In a timing circuit, an electronic processor operates so that the buzzer sounds when Vc is greater than Vs.
The switch S is normally open. Explain in detail what happens in the circuit after the switch S is closed for a moment then opened again. Your answer should include an appropriate calculation and a sketch graph.
S closed --> C charges
up to Vs
S open: discharge starts
Vc = Vs e^(-t/RC)
¾ Vs = Vs e^(-t/RC)
ln ¾ = -t/RC (1)
t = 29.7 s
Buzzer sounds for 29.7 s
However, shouldnt the answer be 2 x 29.7, as the buzzer also sounds during the time the capacitor is charged and Vc > 0.75Vs?
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