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Second order RLC circuit

  1. Nov 18, 2012 #1
    http://imgur.com/6aAMV

    So I need to find the current labeled as a function of time. THe switch opens at t=0 and I drew the circuit after the switch opens.

    I found the initial current to be -4 and the voltage on the capacitor to be 8.


    I'm having trouble trying to find [itex]di_L(0)/dt[/itex]. I know you just need to do a simple loop around the whole circuit, [itex]v_c+v_l+v_r=0[/itex]. The problem is I don't know the signs of the voltages, or rather I don't know what they should be. I get either [itex]di_L(0)/dt = -40/10, -24/10[/itex].
    Same for finding the second order eq.
    [itex]1/c\int i_c+ L di_L/dt + 8 i_L = 0[/itex] Simple loop around the outside. Are signs correct?
    and
    [itex]i_c - i_L - V_2/2 = 0[/itex] Node equation at the top node, V is equal to the voltage across the inductor and 8 ohm resistor so substitute

    [itex] i_c - i_L - 1/2 (Ldi_L/dt + 8i_L)=0[/itex]
    Substitute the third into the first, do algebra, and find answer.
    That part should be easy. What I need to know is whether the signs are correct. That is, when doing kichoff mesh and nodes how do you know what is positive and negative [or rather what side the element should be placed on]?

    Thanks
     
  2. jcsd
  3. Nov 18, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    The signs of the potential changes are determined by the assumed direction of the current. The components of capacitance and inductance have their equations defined assuming a current directed into the component and the "+" reference at the terminal where the current enters.

    attachment.php?attachmentid=53095&stc=1&d=1353273440.gif

    Note, for example, that in the above figure for the inductor that a small positive going change in current "dI" results in a positive change in the voltage across the inductor, thus giving the equation V = L dI/dt. Similarly for the capacitor, a positive current flowing into the capacitor increases the potential across the capacitor over time.

    For this problem I would probably solve it using a state variable (Laplace Transform) approach, since it handles all the initial condition details easily. Having a good table of Laplace transforms helps :smile:
     

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  4. Nov 18, 2012 #3
    I think I finally understand this + and -.
    I picked the current going to the right around the loop. Labeled each element with a plus and minus, current entered the plus so each one is positive. Blah blah balh, it all worked out.
     
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