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. Answer is: S closed --> C charges up to Vs Instantly/very quickly S open: discharge starts Exponential discharge 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?