|Share this thread:|
Jan25-10, 06:13 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?
|Register to reply|
|Timing circuit (lm555)||Electrical Engineering||6|
|Equinox Timing||Astronomy & Astrophysics||4|
|Timing diagram||Engineering, Comp Sci, & Technology Homework||4|
|Logging out, Timing out||Forum Feedback & Announcements||1|
|Variable cam timing||General Engineering||2|