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Deriving Q (t)=Qmax (1-e^(-t/tau)). Stuck. Please help.

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  1. Mar 27, 2016 #1
    1. The problem statement, all variables and given/known data
    I am trying to derive the equation for the charge of a capacitor as a function of time: Q (t) = Qmax (1-e^(-t/tau).


    2. Relevant equations
    Kirchhoff's loop law
    I (t) = dQ(t)/dt
    Delta Vbat= Epsilon
    Delta Vresistor= -I*R = [-dQ(t)/dt]*R
    Delta Vcapacitor= Q/C
    Qmax = C*Epsilon
    Tau = R*C


    3. The attempt at a solution
    Kirchhoff's loop law
    Epsilon - I*R - (Q/C) = 0
    Epsilon - [dQ(t)/dt]*R -(Q/C) =0
    [dQ(t)/dt] = [Epsilon -(Q/C) ]/R
    = (C*Epsilon -Q)/RC
    = (Qmax -Q) / tau

    Multiply both sides by dt and divide by Q
    dQ(t)/Q = [{(Qmax/Q) - 1}*{1/tau}]dt

    Integrate both sides
    lnQ(t) ={(Qmax/Q) - 1}*{1/tau}

    Take the inverse natural logarithm on both sides
    Q(t) = e^[(Qmax/Q) - 1}*{1/tau}]

    Voilà. What am I doing wrong. I've been trying to get the right equation for hours. Is what I found equivalent to the equation in the post title? Thanks in advance for any help.
     
  2. jcsd
  3. Mar 27, 2016 #2

    TSny

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    You didn't integrate the right side correctly. Note that Q on the right hand side is a function of t.

    Hint: Go back where you divided both sides by Q and divide both sides by Qmax - Q instead.
     
  4. Mar 27, 2016 #3
    Si I did as you suggested and got:

    [dQ(t)]/(Qmax - Q(t)) = (1/tau)dt
    ([dQ(t)]/Qmax) - ([dQ(t)]/Q(t)) = (1/tau)dt

    Integrate both sides
    (Q(t)/Qmax) - lnQ(t) = 1/tau
    Q(t) = [lnQ(t) + (1/tau)]*Qmax

    And then I'm stuck again...:(
     
  5. Mar 27, 2016 #4

    TSny

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    The second line does not follow from the first line.
    Note that ##\frac{a}{b-c} \neq \frac{a}{b} - \frac{a}{c}##. For example ##\frac{1}{3-1} \neq \frac{1}{3} - \frac{1}{1}##

    You should be able to integrate dQ(t)]/(Qmax - Q(t)) directly. If you need to, make the substitution u = Qmax - Q(t)
     
  6. Mar 27, 2016 #5
    I see. Sorry, my math skills are lacking...
    So the integral of dx/(a-x) is ln (a-x), hence the integral of dQ(t) / (Qmax - Q(t)) is ln ( Qmax - Q(t). (Right?)

    Thus Qmax - Q(t) = e^(1/tau)
    Q(t)= Qmax - e^(1/tau) --- which is incorrect....another hint please?
     
  7. Mar 28, 2016 #6

    cnh1995

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    It is -ln(a-x)..
    KVL equation becomes
    R(dq/dt)+q/C=V
    ∴Rdq/dt=(CV-q)/C
    dq/(Qmax-q)=dt/RC
    (CV=Qmax)..
    Integrate this equation using proper limits and you'll get the answer.
     
  8. Mar 28, 2016 #7

    TSny

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    As cnh1995 pointed out, when you integrated the left side of your equation, you missed a negative sign.

    Also, ##\int \frac{dt}{\tau} \neq \frac{1}{\tau}##. Recall that ##\int dx = x + k##, where ##k## is an arbitrary constant of integration. The constant of integration will be important.
     
  9. Mar 28, 2016 #8
    Almost there......

    dQ(t) /( Qmax - Q(t)) = (1/tau) dt

    Integrate both sides ( indefinite integral for the right side; upper limit = Q ( t ) & lower limit = 0 for the left side )

    This gives:
    -ln (|Q(t) - Qmax | ) + ln ( | 0 - Qmax | ) = t / tau

    Multiply both sides by -1
    ln (Q(t) - Qmax ) - ln ( Qmax ) = -t/tau

    ln(a) - ln(b) = ln(a/b) so...
    ln [{Q(t) -Qmax }/Qmax ] = -t/tau

    ln(x) = y ==> x = e^y
    (Q(t)/Qmax) - (Qmax/Qmax) = e^(-t/tau)
    Q(t) = [(e^(-t/tau)) + 1]*Qmax

    Still not correct. I don't know why I get +1 instead of -1 in the equation. ....
     
  10. Mar 28, 2016 #9

    TSny

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    Careful with removing the absolute value. Recall |x| = x if x ≥ 0 and |x| = -x if x < 0.
     
  11. Mar 28, 2016 #10
    Really? I thought an absolute value is always positive......
     
  12. Mar 28, 2016 #11

    TSny

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    Yes, that's absolutely :oldsmile: true, except for |0|.

    So, consider |x| when x = -2. Does |x| = x or does |x| = -x?
     
    Last edited: Mar 28, 2016
  13. Mar 29, 2016 #12

    Ok, I shall not give up!

    -ln (|Q(t) - Qmax | ) + ln ( | 0 - Qmax | ) = t / tau
    -ln ( | -(-Q (t) + Qmax) |) + ln ( Qmax ) = t/tau
    ln ( Qmax ) - ln ( Qmax -Q (t)) = t/tau
    Qmax/(Qmax - Q (t)) = e^(1/tau)
    Qmax - Q (t) = Qmax/(e^(1/tau))
    - Q (t) = Qmax*(e^(-1/tau)) - Qmax

    Multiply both sides by -1
    Q(t) = Qmax -Qmax*(e^(-1/tau))

    Q (t) = Qmax * ( 1 - e^(-1/tau))

    Is it too soon to celebrate?
     
  14. Mar 29, 2016 #13

    TSny

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    I believe that's correct. Good work!
     
  15. Mar 29, 2016 #14
    Thank you all so much! Hard, collaborative work does pay off!
     
  16. Mar 29, 2016 #15


    Sorry, distraction error. Please replace all e^(-1/tau) with e^(-t/tau).
     
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