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Current prob

  1. Mar 8, 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 8, 2006 #2
    How many places are you going to post this question? It's been answered.

    -Dan
     
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