Circuit with Two Sources and Switch

Discussion in 'Homework Help' started by mjakov, Jan 2, 2015.

  1. mjakov

    Thread Starter New Member

    Feb 13, 2014

    Hi all!

    Here is a circuit with two sources and a switch. The right source should be a dependent source according to the notation used in my book, although I fail to see any dependance. Could this be a typing error in the book?

    The task is to find  v=?

    The initial conditions before closing the switch would be:
     i_1 (0) = 0
     i_2 (0) = - \frac{k}{2}
     v(0) = k , this assumes that the right source is active starting from  - \infty . Could this assumption be wrong?

    Starting with loop analysis:
    (1):  -2 + 2i_1 + 2 (i_1 - i_2) = 0
    (2):  2 (i_2 - i_1) + \frac{di_2}{dt} + k = 0

    Eliminating  i_1 gives:
     i_2 + \frac{di_2}{dt} = 1 - k
    Solving this equation:
     i_2^h = A e^{-t}
     i_2^p = 1 - k
     i_2 = i_2^h + i_2^p = Ae^{-t} + 1 - k
    Inserting the initial condition:
     i_2(0) = A + 1 - k = - \frac{k}{2}
     A = \frac{k}{2} - 1
    It follows:
     i_2 = (\frac{k}{2} - 1) e^{-t} + 1 - k
    From Eq. (1):
     i_1 = \frac{i_2+1}{2}
     i_1 = \frac{1}{2} (\frac{k}{2} - 1 ) e^{-t} + 1 - \frac{1}{2}k
    Observe that  i_1 does not satisfy the initial condition for 0 in the last equation, only for 0+, after the switch is closed. So it would probably be better to multiply the result of  i_1 with the unit step function? I am asking this, because we can derive the formula for  i_1 without the step function, by simply inserting the initial condition in the diff. eq.
    Now to calculate v:
     v = 2 (i_1 - i_2)
     v = (1 - \frac{k}{2}) e^{-t} + k
    For example, for  k = 0.5 we obtain  v = \frac{3}{4} e^{-t} + \frac{1}{2} . The solution according to my book should be:  v = e^{-\frac{t}{2}} . For   k = 0 the solution is correct:  v = e^{-t}.

    From this I surmise that the exponential part should look like  e^{-kt} , but I do not see how this could be obtained from the differential equation above. Any help or hints would be appreciated.
    Last edited: Jan 3, 2015
  2. MikeML

    AAC Fanatic!

    Oct 2, 2009
    This is really an interesting circuit. I simed it for k=0 (green), 0.5 (red), 0.9 (blue), 1 (yellow) and slightly more than 1 (violet). I plot the voltage at node V and the current through the inductor. I close the switch at 2s. Before that, the voltage at V(v) and E2 = 0, and the inductor current is zero. At 2s, the switch is closed in 1ps, and V(v) jumps to 1V.

    For k=1, the voltage at both ends of the inductor is the same, so no current flows through the inductor, so V(v) is just half the step.
    For k=0, the voltage at the right end of the inductor is zero, so the circuit is just two 2Ω resistors in parallel (1Ω) fed by half a two volt step (1V) into 1H, so a simple RL time constant.
    For k=0.5 and o.9, the time constant is lengthened.
    For k>1. the current is ever increasing...

  3. mjakov

    Thread Starter New Member

    Feb 13, 2014
    Hi! Thanks for the answer. I assume that the right hand side source is dependent on small v in your simulation. This would make sense and the obtained graph is compatible with the solutions in my book. For yellow we should have v = 1 like in the simulation. Violet looks like a rising exponential. According to the solutions, for k = 2, v should be  v = e^t . I suppose you chose for yellow k slightly more than 1, because otherwise v would grow too fast to plot together with the other values. For, k = 2, would it be possible in practice, since that should case unbounded growth of v? P.S. It looks like the simulation approach is quite effective when troubleshooting the algebraic analysis :)
    Last edited: Jan 3, 2015
  4. MikeML

    AAC Fanatic!

    Oct 2, 2009
    It doesn't produce the equation, but it sure makes it easy to check the equation that you derive...

    The growth is unbounded for any value of k>1

    Here it is again with k=2. I plot the two voltages on a log plot to show the exponential growth, and that V(e1) is always 2x V(v), as you would expect. Since there is a voltage difference between the two ends of the inductor, as you would expect, the inductor current grows forever...