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time constant help?!

what is the time constant in the graph in page number 15?

http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/june2010-qp/6PH04_01_que_20100618.pdf

and please explain how you did it
(the answer should be between 14 and 20)
Reply 1
Avatar for Joinedup
Joinedup
20
You'd measure the time for the voltage to decreace by ~37% (1/e * 100)

here the Vpeak is ~ 5V and the Vminimum is ~3.5V

37% of 5V is 1.85V

so you'd measure (or attempt to extrapolate) the point where the voltage had fallen to 1.85V below 5V and measure the time to the preceding Vpeak.

it wouldn't be very accurate from a small graph, but you could put limits on reasonable valuse which is what I presume the questions asking for.
Reply 2
Avatar for ashboot
ashboot
Not sure if this is the correct answer?

RC = (-6x10^-3) / ln(3.5/5) = 16ms

Approx 6x10^-3 s to drop from 5v to 3.5v

RC = -t / ln ( v/v0 )
(edited 11 years ago)
Original post by ashboot
Not sure if this is the correct answer?

RC = (-6x10^-3) / ln(3.5/5) = 16ms

Approx 6x10^-3 s to drop from 5v to 3.5v

RC = -t / ln ( v/v0 )


how to did you get the number (-6x10^-3) ?
can you please give more details?
Reply 4
Avatar for ashboot
ashboot
Original post by >>MMM<<
how to did you get the number (-6x10^-3) ?
can you please give more details?


looking at the graph, it took roughly 6 ms for the capacitor to discharge... This is from looking at the peak voltage and the lowest point. Do you know your prefixes? m is x10^-3, and its -t so that's why it is -6x10^-3

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