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State Variable Method (Circuit Analysis)

  1. Jul 13, 2012 #1
    1. The problem statement, all variables and given/known data

    http://img12.imageshack.us/img12/1176/img20120713190506.jpg [Broken]

    2. Relevant equations
    V[itex]_{C}[/itex]' = dv/dt
    I[itex]_{L}[/itex]' = di/dt

    3. The attempt at a solution

    I did KCL at the centre node:

    I[itex]_{L}[/itex] + i[itex]_{2}[/itex] = [itex]\frac{V_{C}'}{5}[/itex]
    (Vc'/5 is basically the current coming from the capacitor line to the centre node (from equation ic = Cdv/dt)

    Re-arranging you get i2 = [itex]\frac{V_{C}'}{5}[/itex] - I[itex]_{L}[/itex]

    I then look at the inductor voltage next, basically the inductor voltage is the voltage across the 10ohm resistor at the top minus the voltage across the capacitor, so:

    V[itex]_{L}[/itex] = 10i[itex]_{1}[/itex] - V[itex]_{C}[/itex]

    Now I substitute this into the standard equation for an inductor:

    I[itex]_{L}[/itex]' = [itex]\frac{V_{L}}{L}[/itex]

    Which gives us 10i[itex]_{1}[/itex] - V[itex]_{C}[/itex] = 2.5I[itex]_{L}[/itex]'

    or i[itex]_{1}[/itex] = [itex]\frac{I_{L}'}{4}[/itex] + [itex]\frac{V_{C}}{10}[/itex]

    Altogether now in state-space representation:

    http://img824.imageshack.us/img824/3282/img20120713191406.jpg [Broken]


    I'm not sure I've done this correctly because there is no Ia (Current source) or Va(voltage source) from the circuit diagram involved, also I'm not sure if my final expression in matrix form is correct.

    Any help is appreciated.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jul 13, 2012 #2
    I got this class this fall and I'm starting to get rather scared. How much physics 2 did you need for this class?
     
  4. Jul 13, 2012 #3
    Physics 2? I'm not sure what that is. I'm not in the USA.

    You don't really need much pure physics for circuit analysis, it's all about employing the various methodologies taught, it's more like an algorithm for problem solving rather than "physics".
     
  5. Jul 13, 2012 #4
    Okay that's what someone else said too. Hopefully someone can help you with this problem!
     
  6. Jul 14, 2012 #5
    Anyone got any idea??
     
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