In a timing circuit, an electronic processor operates so that the buzzer sounds when Vc is greater than Vs.(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Timing Circuit - Buzzer

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