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Homework Help: Current Problem

  1. Mar 7, 2006 #1
    The charge flowing through a circuit is

    q(t)=[3e^(-t) - 5e^(-2t)]--------(1)

    find the value of t and then current i.
    as i=dq/dt.
    i am doing it like this

    i=-3e^-t + 10e^-2t..................taking derivative

    let e^-t=u...........for eaze

    i=10[u^2-3u/10]

    let 1=0 then

    10[u^2-3u/10]=0

    10[u^2 - 2(u)(3/20) + (3/20)^2 - (3/20)^2]=0 .....using formula

    10[u-3/20]^2 - 10[3/20]^2=0

    10[u-3/20]^2=10[3/20]^2

    10[u-3/20]^2=9/40

    [u-3/20]^2=9/400

    u-3/20=sqrt[9/400]

    u=sqrt[9/400] + 3/20

    e^-t = 3/20 + 3/20.......as u=e^-t

    Taking log on both side

    ln[e^-t]= ln[3/10]

    -t= -1.204
    -----------------|
    t= 1.204 seconds |------ am i doing ok till here?
    -----------------|
    by putting this in equation 1 we will get the vale for charge,q.
     
  2. jcsd
  3. Mar 7, 2006 #2

    berkeman

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    Staff: Mentor

    What do you mean find the value of t? The value of t when what happens?
     
  4. Mar 8, 2006 #3
    i think!

    everyone knows that i= q/t in general but to find current i at some instant we have i= di/dt. In the given equation above if we can find the value of t we can put this value of t back in the equation and can find the charge q. by having charge q and time t, current i can be found by using formula i=q/t and the value of t can only be determined by taking derivative of the equation. at least i see only this possibility.
     
  5. Mar 8, 2006 #4

    Hootenanny

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    Staff Emeritus
    Science Advisor
    Gold Member

    So basically you have
    [tex]q(t) = 3e^{-t} - 5e^{-2t}[/tex]
    [tex]q '(t) = 10e^{-2t} - 3e^{-t}[/tex]
    You now can't solve any further without having more information. As berkeman said, when do you need to know t? Do you want to find t when q is maximum / minimum etc.
     
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