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Coupled RC circuits with AC current source

  1. Dec 31, 2014 #1
    1. The problem statement, all variables and given/known data
    Hi, I already asked a question close to this, but now I have different conditions.
    This is the circuit

    proxy.php?image=http%3A%2F%2Fi59.tinypic.com%2Fjr8ev9.png




    [itex] C_1 = C_2 \\
    R_1= R_2[/itex]

    The current is an AC , and I would like to know the voltage at R_1 and at R_2
    I made some progress but I do not really know hot to continue.




    3. The attempt at a solution


    VR1 = VC1
    VR1 + VRc - VC2 = 0
    VR2 = VC2
    IR1 + IC1 + VRC = Iinj
    IRc = IR2 + VC2
    [tex] V_1 = V_{C1} = V_{R1} \\ V_2 = V_{R2} = V_{C2} [/tex]

    [tex] \frac{V_1}{R_1} + \frac{dV_1}{dt}*C_1 + \frac{V_{Rc}}{R_c} = I_{inj} \\
    \frac{V_{Rc}}{R_c} = \frac{V_2}{R_2} + C_2*\frac{dV_2}{dt}\\
    [/tex]


    [tex] \frac{dV_1}{dt}*C_1 = I_{\omega} -\frac{V_1-V_2}{R_c} - \frac{V_1}{R_1} \\
    \frac{dV_2}{dt}*C_2 = -\frac{V_2}{R_2} +\frac{V_1-V_2}{R_c} \\
    --> C_1 == C_2 , R_1 == R_2 \\
    \frac{dV_1}{dt} = \frac{I_{\omega}}{C_1} -\frac{V_1-V_2}{R_c*C_1} - \frac{V_1}{R_1*C_1} \\
    \frac{dV_2}{dt} = -\frac{V_2}{R_1*C_1} +\frac{V_1-V_2}{R_c*C_1} \\

    [/tex]
     
  2. jcsd
  3. Dec 31, 2014 #2

    gneill

    User Avatar

    Staff: Mentor

    With AC sources you can obtain steady-state solutions for such circuits using impedances for the components and phasor values for the sources. No need to write or solve differential equations, the usual DC circuit methods and techniques of analysis will work fine.
     
  4. Jan 4, 2015 #3

    rude man

    User Avatar
    Homework Helper
    Gold Member

    You can't have voltage and current terms in the same equation. Straightening this out is your first task.

    It is perectly OK to use differential equations as you have done, especially if you never had phasors..
     
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