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Timing Circuit - Buzzer

  1. Jan 25, 2010 #1
    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?
     
  2. jcsd
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