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Analysis of RL circuit

  1. Jul 27, 2006 #1
    Hi guys, I'm currently deriving all the equations provided by the my physics professor on the RL, RC and LCR circuits. My trouble is not the physics but the mathematics.

    For example in the RL circuit the differential equation that needs to be solved is the following:

    [tex] L\frac{dI}{dt} +IR = V_0\cos{wt} [/tex]

    I reconize that you need a integrating factor, however the integral after that step seems ridiculous to solve, if you guys will just help me out with the right step to solve the RL circuit, I'm sure i can solve the rest with ease.
  2. jcsd
  3. Jul 27, 2006 #2


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    Since that's a linear equation with constant coefficients, the integrating factor is just [itex]e^{\frac{R}{L}t}[/itex]. Multiplying the entire equation by that,
    [tex]L\frac{de^{\frac{R}{L}t}I}{dt}= V_0e^{\frac{R}{L}t}cos(\omega t)[/tex]

    Is it [itex]\int e^x cos(x) dx[/itex] that you are concerned about? That's a fairly standard "integration by parts" problem.

    However, there's a fairly easy way to solve "linear equations with constant coefficients". First look at the "homogeneous part":
    [tex]L\frac{dI}{dt}+ RI= 0[/tex]
    That's separable:
    [tex]L \frac{dI}{I}= -R dt[/tex]
    [tex]L ln(I)= -Rt+ C[/tex]
    [tex]I= Ce^{-\frac{R}{L}t[/itex]
    Now just find any one function that satisfies the entire equation and add to that general solution of the homogeneous part. I would suggest trying
    [tex]I= A sin(\omega t)[/tex]
    and seeing what A must be in order to satisfy the equation.
  4. Jul 27, 2006 #3
    aren't you omitting a term proportional to [tex] Q [/tex] where Q is the charge of a system..in that case you would have a second-order differential equation of the form:

    [tex] a \frac{d^2 y}{dx^2 } +b \frac{dy}{dx}+cy=dCos(wx) [/tex] taking into account that..

    [tex] I=-\frac {dQ}{dt} [/tex]

    :tongue2: :grumpy: :cool: :rolleyes: that seems to have more "physical" meaning (the circuit current is a dumped oscillator so the charges is getting smaller and smaller as [tex] Q(t)=Q_0 (t)e^{-\gamma t}Sin(\omega t) [/tex]
  5. Jul 27, 2006 #4


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    No. Not in a pure LR circuit - in the absence of capacitance.
  6. Jul 27, 2006 #5
    Cheers hall, perfect help yet again. Yes the integral by parts is trivial, i guess i just let all those constants (R.L,V) confuse the trivial maths...
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