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Basic RC circuit problem

  1. Nov 5, 2012 #1
    1. The problem statement, all variables and given/known data

    In the following transient circuit, assume at t<0, the circuit is at steady state.
    Find i0(t) for t>0
    http://sphotos-b.xx.fbcdn.net/hphotos-ash3/c0.0.241.241/p403x403/521743_510378412315012_79982992_n.jpg [Broken]

    2. Relevant equations



    3. The attempt at a solution
    First let's try to find Vc(t)
    capacitor acts as an open.
    the current of the whole circuit Is = 12V / [(4+2) || 10] = 6.4 mA
    current going through the 2KΩ = 6.4 mA x (10/16) = 4mA
    Vc(0+) = V2K = 4mA x 2k = 8V

    ok, right here I am not sure if Vc(∞) = 0 V
    to find time constant, we need the Thevenin resistance RTh = 4 || (2+10) =3k
    T = 200μF x 3K = 3/5

    so we got our vc(t) = 8e-5t/3

    This is where I got stuck, I really dont know how to relate io(t) with Vc(t)

    Please help.
    Thank you very much!
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 5, 2012 #2
    It's not. Any current that comes out of the + terminal of the battery must reenter the -ve terminal of the battery, which means it must go through the capacitor. At t→∞ no current flows through the capacitor so no current can flow out of the battery either. If no current flows out of the battery...

    Need to fix this up given the information above.

    The capacitor and the battery impose a voltage across the resistors and the current through the battery will equal the current through the capacitor. Knowing Vc(t) for all t is going to make this possible to solve without solving any differential equations.
     
    Last edited: Nov 5, 2012
  4. Nov 6, 2012 #3
    I understood that current from the source 24V wont be able to return to the (-) end of it. but I still dont know how to find Vc(∞).


    So basically we now have 2 sources of voltage the 24V and the capacitor with Vc(t), and we just need to add up the 2 values of current from each source, so I will just need to use superposition method for this?
     
  5. Nov 6, 2012 #4
    Ok.. the voltage Vc+24 appears across the 4kohm resistor and the series 10k+2k. The only way no current flows through those resistors is if the voltage across them is zero.

    The capacitor cannot act like a voltage source because its voltage changes with current passing through it. Ideal voltage sources look like short circuits to the rest of the circuit because any amount of current can be pumped into them, yet no change in voltage occurs. That is why an ideal voltage source can be treated as a short circuit in superposition but a capacitor voltage cannot.

    I had something more mundane in mind :-- KVL :) Vc+24 is the voltage across the other resistors. The current flowing through those resistors is just (Vc+24)/R . Also keep in mind the current through the capacitor is the same as the current through the 24V source.


    *** The initial condition of a capacitor can also be taken care of by treating the capacitor like an unenergized capacitor in series with a voltage source of Vc(0) volts. This comes from the Laplace transform: i = C dv/dt, I(s) = sC V(s) - Cv(0), V(s) = I(s) * (1/(sC)) + v(0)/s, v = 1/C ∫ i dt + v(0). If you haven't seen this yet, you shouldn't use it.
     
  6. Nov 6, 2012 #5

    gneill

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    Staff: Mentor

    Yes, the potential across the capacitor changes with time but at any given instant it has a particular value. As far as the rest of the circuit is concerned, in terms of the potentials that are driving currents at that instant the capacitor is indistinguishable from a voltage source with that instantaneous potential difference.

    Modeling capacitors as voltage sources at particular instants can be a handy technique for determining the currents in a circuit at that instant (such as at t = 0+ in this case). In fact, it is possible in general to model a capacitor which has an initial potential difference as a voltage source (with the same potential difference) in series with an uncharged capacitor of the same value; The circuit behavior remains the same and the potential across the series pair will match that of the original charged capacitor over time.
     
  7. Nov 7, 2012 #6
    Ah, but the poster wanted to use superposition by shorting out the capacitor in one superposition step. This cannot be done because the capacitor cannot absorb current without changing voltage, so it does not appear as a short circuit as an ideal source would to the rest of the circuit.
     
  8. Nov 8, 2012 #7
    i got the solution in class, my professor said that just add the voltage source with the voltage of capacitor and divide by the total resistance of 10k+2k.

    However, I am still not clear about the polarity of the voltage of capacitor.
     
  9. Nov 8, 2012 #8
    plus 4k in parallel. The answer is above too :)

    Choose a polarity; I would choose vc is higher at the curved plate so that capacitor voltage has the same orientation as the battery. This means your initial capacitor voltage is -8V, as you found earlier, because that had the other side at positive voltage.

    After the switch is open, the current out of the cap/source branch is (vc+24)/((10k+2k)||4k). The current through the capacitor is i=C dvc/dt. If positive current leaves the branch with the cap in it, this will cause the cap voltage to drop or rise? This will select the sign of dvc/dt for you.


    The method you were pursuing had initial voltage on the cap at -8v. The final voltage on the cap will be such that no current flows in the resistors, which means (vc+24)=0 (no voltage across the resistors). Then you could fit a first order exponential function to that.
     
    Last edited: Nov 8, 2012
  10. Nov 8, 2012 #9

    gneill

    User Avatar

    Staff: Mentor

    It works fine for any particular instant. In general a capacitor with an initial charge (and thus potential difference) can be modeled as a voltage source with that charge in series with an uncharged capacitor with the same capacitance. For that instant the you can treat the voltage source just as any other voltage source and treat the capacitor as a short in order to determine the currents at that particular instant. No potential change can occur on a capacitor in zero time (an "instant") which is why we can set Vc0+ = Vc0- for a capacitor potential when a switch is thrown at t=0.
     
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