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Sunday, March 21, 2010

MP 21.62 part B (Assignment 4 #16) Practice Problem

The capacitor in an RC circuit (R = 140 \Omega, C = 40 \mu F) is initially uncharged.
PartB
Find the current in the circuit one time constant (\tau = RC) after the circuit is connected to a 9.0-{\rm V} battery.

I use this but didnt work

I = V/R (1-e^-t/RC)

When the problem said one time constant, does that always mean that t/RC = 1, or only the RC is equal to 1?

I think Prof gave the similar problem on the last quiz, where it asked the current after 2 time constant. And he just plugged the "2" into t/RC
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3 comments:

  1. mlmMETALLICAmlmMarch 21, 2010 at 7:41 PM

    so one time constant means that t = RC. What did you put for the time?

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  2. e.michaelaMarch 21, 2010 at 8:51 PM

    I = 9V / 140 ohms (1-e^-1)
    and I got 41mA but that's not the correct answer.

    Part A asks the charge, I did similar thing and it worked
    q = 40 * 9V (1-e^-1)
    = 230 microC, and that's right

    ReplyDelete
  3. e.michaelaMarch 22, 2010 at 2:06 AM

    Finally I got that.
    Didnt realize the equation was wrong
    it should be I= V/R (e^t/RC)
    instead of I= V/R (1-e^t/RC)

    ReplyDelete

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