higher order circuit help

Discussion in 'Homework Help' started by dbhakta, Feb 20, 2012.

  1. dbhakta

    Thread Starter New Member

    Sep 23, 2009
    2
    0
    [​IMG]

    Can someone please help me on this problem

    C1 is initially charged to 1 volt.
    The switch closes at t=0. Find and plot all the currents.

    Since the switch is initially i assume all initial conditions are:
    i1(0) = i2(0) = i3(0) = i4(0) = i5(0) = 0

    so I write a KVL loop equation for each loop and get :

    ' - denotes derivative
    int - denotes integral

    Loop 1 : [1/c1 * int(i1*dt) + 1] + L1*i1' + R1*i2 = 0
    ==> L1*i1'' + R1i2' + i1/c1 = -1

    Loop 2: R1*i2 + 1/c2 *int(i3*dt) + R2*i3 + L2*i4' = 0
    ==> L2*i4'' + R2*i3' + R1*i2' +i3/c2 = 0

    Loop 3: L2*i4' = R3*i5

    KCL equations give:
    i1 = i2 + i3
    i3 = i4 + i5

    I solved for i1''(from loop1 equation) and i4''(from loop 2 equation).
    I'm kind of stuck at this point. I don't know how to setup the equations from here so that they can be solved. I tried solving for the first derivatives but that didn't really give me anything.
    Does anyone have any suggestions or tips?
     
  2. Vahe

    Member

    Mar 3, 2011
    75
    9
    I think the best way to approach this problem is to use the Laplace transform technique -- that way you will be able to work with algebraic equations rather than differential equations. You will then use the inverse Laplace transform to get the time domain expressions. Do you know how to do this?

    --Vahe
     
  3. dbhakta

    Thread Starter New Member

    Sep 23, 2009
    2
    0
    so the impedance of inductors go to sL and the impedance of capacitors become 1/sC correct? I think I would be able to solve this if it has a constant voltage source but since that is not the case I'm a little confused.

    I combine all elements in laplace domain except c1 and get Zeq = (80s^3 +538s^2 +181s+4)/(40s^2 + 249s+6).

    Im not sure how to solve for current i1 since the initial charge is on c1
     
    Last edited: Feb 20, 2012
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