RC Circuit

Discussion in 'Homework Help' started by Robert.Adams, Sep 20, 2010.

  1. Robert.Adams

    Thread Starter Active Member

    Feb 16, 2010
    112
    5
    I'm having trouble setting up the attached circuit. The dependent source is the main source of my troubles.

    I'm guessing you'd use KCL on this circuit, but how would you set up the left branch?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    A starting point is to relate the current ix(t) to the capacitor current.

    Assume the capacitor current ic(t) flows out of the the positively marked capacitor plate in a clockwise direction.

    The current ic(t) will split in the parallel resistor branch comprising the 80k and the 20k.

    Using the current divider rule you should be able to show that

    ix(t)=-0.2*ic(t) - eqn (1)

    The voltage equation for the loop would be something like

    Vc(t)=12x10^4*ix(t)-80k*ix(t) - eqn (2)

    Vc(t) is given by the integral term Vc(t)=(1/C)∫ic(t)dt - one would then substitute for ix(t) in eqn (2) {using eqn (1) above} to form a differential equation expressed solely in terms of ic(t).

    I've not convinced myself that the circuit as drawn will have a convergent steady-state solution. Are you sure you have drawn it correctly with respect to the direction of ix(t) say?
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    A simulation of your circuit as drawn confirms my suspicions about an unbounded solution. See attachment.
     
  4. Robert.Adams

    Thread Starter Active Member

    Feb 16, 2010
    112
    5
    The solution key says the answer is dVc/dt - 50Vc = 0 and the time constant is -0.2 seconds.

    I went through with the KVL method and got this equation for Ix:
    dIx/dt + 50Ix = 0.

    I got this from setting up a left loop and a right loop. I solved the right loop for i1 in terms of ix. Substituting this into the first loop, I get an equation for Vc in terms of Ix. Then I use Vc = 1/C int(-5ix, dt) to get the equation to only terms of Ix.

    Somehow the time constant came out backwards. This might be the non-convergence problem you're talking about.

    (The original attachment is straight from an old test, is not my redrawing).
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The key solution for the given circuit configuration is correct. So yes it is a case of an unbounded solution. In electrical circuits time constant is normally associated with transients which converge to a steady state condition.

    You can readily prove the key solution as follows.

    ix(t)=-0.2*ic(t) where ic(t) is the capacitor current out of the top plate.

    But ic(t)=-C*(dVc(t)/dt) where Vc(t) has the indicated polarity.

    So ix(t)=5x10^-7*(dVc(t)/dt)

    Also

    Vc(t)= 120,000*ix(t)-80,000*ix(t)=40,000*ix(t)

    and after substitution

    Vc(t)=0.02*(dVc(t)/dt)

    Hence

    Vc(t)-0.02*(dVc(t)/dt)=0

    or

    50*Vc(t)-dVc(t)/dt=0

    or as indicated in the key solution

    dVc(t)/dt-50*Vc(t)=0
     
Loading...