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Differential equations - circuits

  1. Feb 14, 2008 #1
    1. A circuit contains an inductance of 40 henrys, a capacitance of .005 farads and a resistance of 240 ohms and is driven by a constant voltage of 60 volts. Given that the initial charge in the system is 5000 Coulombs and the initial current is zero amperes, discuss the subsequent behavior of the current in the circuit.

    So
    R = 240
    L = 40 H
    E = 60 V
    Q = 5,000 Coulombs
    C = .005 F
    I = 0 amperes

    I am really not sure how to begin this problem. I can't tell if I am supposed to be coming up with an equation or just talking about the behavior? If I am supposed to find an equation, where does this come from? Most of the equations I have involve only two or three variables (Such as L dI/dt + RI = E). Am I able to write an equation and then graph it?
     
  2. jcsd
  3. Feb 14, 2008 #2

    Defennder

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    You didn't specify how the circuit elements are connected. Are they all in series, which I assume them to be? Just to double-check, the voltage source is DC 60 V right?

    If so, then all you need do is apply Kirchoff's voltage law to the single loop in the circuit. You have the expressions for the potential difference across the capacitor and inductor right? Once you get the equation, you can either differentiate it with respect to t to obtain a 2nd ODE in terms of I, or take it as it is and leave it as a 2nd ODE in terms of Q.

    After solving the 2nd ODE, plug in the initial values of current and charge. That should give you the answer.
     
  4. Feb 14, 2008 #3
    not sure...

    I am an education major and I think they allowed me into the wrong differential equations class! I have no idea what I am doing. In the book it says that they are all typical RLC circuits? That is all I know.

    Do I use the equations

    L dI/dt + 1/C Q + RI = E

    I = dQ/dt

    L d^2I/dt^2 + 1/C dQ/dt + R dI/dt = dE/dt
    OR
    L d^2I/dt^2 + R dI/dt + 1/C I = dE/dt

    I don't really understand what exactly I differentiate or what to plug in.

    do I do :

    40 dI/dt + (1/.005)(5,000) + 240I = 60
    40 dI/dt + 1.0x10^-6 + 240I = 60
    40 dI/dt + 240I = -999940

    and then differentiate? I'm very confused.
     
  5. Feb 14, 2008 #4

    Defennder

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    Don't mix the two variables together for Q and I. I = dQ/dt, so you can easily express any ODE you get in terms of either Q or just differentiate it with respect to t to get in terms of I only. Why are you treating E just like it is a function of time? You said E is constant 60V right? Which means E is just a constant. dE/dt = 0.

    Your first statement for the ODE is correct. Just that you should express dQ/dt as I so that you have a 2nd ODE in terms of I only.

    Your first numerical equation is incorrect. Q(t) is a function of t, so you can't write Q = 5000C as a constant. That is an initial value, which you have to use later once you solve the ODE.

    Have you learnt anything about solving 2nd order linear homogenous differential equations yet? If you haven't, I'm sure your textbook or notes would contain a general solution for 2nd ODE or at least one for RLC series circuits. Most 1st year physics majors are introduced to RLC circuits and and their solutions (involving solving the 2nd linear ODE) without having taken a DE class.
     
  6. Feb 14, 2008 #5
    We just started 2nd Order Linear Homogeneous Diff Eq. I just finished a bunch of problems such as : y'' + 2y' - 5y = 0
    I'm guessing I'm supposed to get these equations to look something like that.
    Is the numerical equation then:
    40 d^2I/dt^2 + 1/.005 I + 240 dI/dt = 0

    If so, writing it in similar for as my example would it be, using the variable I:

    40 I'' + 240 I' + 1/.005 I = 0

    then (40 d^2 + 240 d + 1/.005) I = 0


    use the quadratic equation to find the solutions
     
  7. Feb 14, 2008 #6

    Defennder

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    Yeah you've got it. Just solve it according to the method, then plug in the given intial values to determine the constants, and you're done.
     
  8. Feb 14, 2008 #7
    thank you so much. sometimes i just need to talk through it with someone!
     
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