Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Can an integral that is a variable of itself be solved?

  1. Oct 7, 2011 #1
    Under the assumption that the voltage is [itex]V_f\left(1-e^{-\frac{t}{RC}}\right)[/itex], where [itex]V_f[/itex] is the final voltage, how would I determine the relationship between current [itex]I[/itex] and time [itex]t[/itex]?

    [tex]I = \int_0^T \frac{V_f\left(1-e^{-\frac{t}{RC}}\right) - RI}{L} \,dt \,[/tex]

    [itex]L[/itex] the magnetic inductance, [itex]R[/itex] the resistance, and [itex]C[/itex] the capacitance, are constants.

    How would I plot current [itex]I[/itex] as a function of time [itex]t[/itex]? (The only variables here are [itex]I[/itex] and [itex]t[/itex].) Let's assume initial conditions of [itex]I=0[/itex] and [itex]t=0[/itex]. My problem here is that the variable I am trying to calculate is a variable inside the integral that is used in deriving the variable itself! How are such problems handled? Any help is appreciated! :smile:
     
  2. jcsd
  3. Oct 7, 2011 #2
  4. Oct 7, 2011 #3

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    So, I take it that you mean that [itex]\displaystyle I(T\,) = \int_0^T \frac{V_f\left(1-e^{-\frac{t}{RC}}\right) - RI(t)}{L} \,dt\,.[/itex]

    Differentiate both sides w.r.t T and solve for I'(T). See if you can integrate the result to get I(τ).
     
  5. Dec 1, 2011 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    OP has a related question in this other post.

    note:
    [itex]I=\int RI.dt[/itex] - differentiate both sides gives

    [itex]dI/I = R.dt[/itex]

    ... what was the problem?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Can an integral that is a variable of itself be solved?
Loading...