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RC circuit

  1. May 8, 2006 #1
    I came across the following problem recently:

    [​IMG]

    [itex]R_1 = R_2 = 4700 \Omega[/itex]
    [itex]C = 0.060 F[/itex]
    [itex]V = 12 V[/itex]


    S1 has been closed for a long time. At t = 0, S2 is closed. Sketch a graph of the currents [itex]I_1[/itex] through [itex]R_1[/itex] and [itex]I_2[/itex] through [itex]R_2[/itex].



    I wasn't really sure where to begin with this. I ended up with the following graph:

    [​IMG]

    I reasoned that at t=0, the entire voltage (12 V) appears across the capacitor (since S1 had been closed for a long time), so when [itex]R_2[/itex] is placed in parallel with it, there must be a 12 V potential difference across it, too. The current [itex]I_2[/itex] must be [itex]\frac{12}{4700}[/itex] then. The current through [itex]R_1[/itex] is 0 at t = 0 because S1 had been closed for a long time. At a long time after S2 is closed, however, the resistors are effectively in series with one another. Therefore the currents through each must be the same, and equal to the voltage of the battery over the sum of the resistances, or [itex]\frac{12}{9400}[/itex].

    I can't figure how exactly the graphs would look between the two points, but this is my best guess.


    Can anyone help explain to me exactly what happens in between? Much appriciated.
     
  2. jcsd
  3. May 9, 2006 #2

    Curious3141

    User Avatar
    Homework Helper

    Your answer is essentially correct, and the shape of your graph is the correct exponential one.

    To appreciate what is actually going on, you need to set up a system of equations involving Kirchoff's laws, and solve them (this will involve solving a first order differential equation).

    The equations are [tex]C\frac{dV_C}{dt} = I_1 - I_2[/tex]
    [tex]V_C - I_2R_2 = 0[/tex] and
    [tex]I_1R_1 - V + I_2R_2 = 0[/tex]

    which are basically Kirchoff's first and second laws applied to the loops and junctions of this circuit. [tex]V_C[/tex] is the potential difference across the capacitor.

    After solving that system, you will find that :

    [tex]I_2 = \frac{V}{R_1 + R_2}(1 + \frac{R_1}{R_2}e^{-\frac{R_1 + R_2}{CR_1R_2}t})[/tex]

    and [tex]I_1 = \frac{V - I_2R_2}{R_1}[/tex]

    which will correspond to your graph.

    In simple terms, you can visualise the capacitor as starting out with a charge of V volts, then starting to discharge across R2 when S2 is thrown. The discharge current will be exponentially decreasing with time. It will not discharge completely, instead, the system reaches steady state when the voltage across the capacitor is [tex]V_C = \frac{VR_2}{R_1 + R_2}[/tex], which is also the final potential drop across the second resistor.
     
    Last edited: May 9, 2006
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