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Parallel RC circuit with realistic battery

  1. Oct 25, 2011 #1
    Here's the problem:

    Consider the circuit in which a battery
    with internal resistance r can be connected via a switch to a resistor
    and capacitor in parallel.
    Suppose that the capacitor is initially uncharged and then at time
    t = 0 the switch is closed. After the switch is closed a current I_b(t)
    flows through the battery, a current I_R(t) flows through the resistor
    with resistance R, and the charge on the upper plate of the capacitor increases at a rate
    dq/dt .

    -What is dq/dt for times t ≥ 0 ?
    -What is dq/dt at t = 0 ?
    -What is I_R (t = 0)?
    -What is the voltage VC across the capacitor at t = 0 ?

    -What is dq/dt at very long times ( t → ∞ )?
    -What is the voltage VC across the capacitor at very long times?
    -Is VC at very long times less than, equal to, or greater than ε ?
    -What is I_R (t → ∞) ?

    So I'm been wrestling with this question fr a while. I know through loop rule that the current across the resistor is equal to Q/CR (from a loop of just the capacitor and resistor). I solved the differential for charge and found q(t)= EC/2 (1-e^(-2t/CR)). And I know how the whole system would act if the capacitor and resistor were in series but I'm really stuck. Any help would gladly be appreciated!!
     
  2. jcsd
  3. Oct 25, 2011 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    You can substitute the real battery with an ideal one in series with a resistor, [itex]r[/itex]. Then you analyse the circuit as any other quasistationary network problem.

    Let [itex]i[/itex] be the total current through the battery, [itex]i_1[/itex] the current through the external resistor and [itex]i_2[/itex] the current through the capacitor, and [itex]Q[/itex] the charge on the posive plate. Then you have from the loop built by the capacitor and the external resistor, [itex]R[/itex],

    [tex]i_2=\dot{Q}=C R \dot{i}_1[/tex]

    and from the loop with the battery and the two resistors

    [tex]r i + R i_1=U.[/tex]

    From the knot rule you have[itex]i=i_1+i_2[/itex]. Plugging the first equation and this one into the last equation, you get the DGL for [itex]i_1[/itex]

    [tex]r (i_1+C R \dot{i}_1)+R i_1 =U.[/tex]

    Because the capacitor is uncharged at [itex]t=0[/itex] the initial condition is [itex]i_1(0)=U_C(0)/R=0.[/itex]

    The solution of the DGL thus reads

    [tex]i_1(t)=\frac{U}{R+r} \left [1-\exp \left (-\frac{(R+r) t}{Cr R} \right ) \right ].[/tex]

    From this you can calculate all the quantities in your question. You'll see that at [itex]t=0[/itex] the capacitor acts as a short circuit and for [itex]t \rightarrow \infty[/itex] as an infinite resistance.
     
  4. Oct 25, 2011 #3
    Okay, I'm still confused. Here was my reasoning but it doesn't seem to match up:

    If the charge across the capacitor at any time is
    q(t) = (EC/2) * (1- e^(-2t/CR))
    and taking the derivative we find:
    dq/dt = E/R * (e^(-2t/CR))
    then at t=0, dq/dt is equal to E/R.

    If I_R = Q/CR then:
    I_R = Q/CR = (EC/2)*(1-e^(-2t/CR))*(1/C) = (E/2R)*(1-e^(-2t/CR))
    so at t=0, I_R = 0

    If the voltage across a capacitor is defined as V=Q/C then:
    V=Q/C=(EC/2)*(1-e^(-2t/CR))*(1/C) = (E/2)*(1-e^(-2t/CR))
    so at t=0, V=0

    But I'm afraid that reasoning is not correct?
     
  5. Oct 25, 2011 #4

    gneill

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    Staff: Mentor

    Here's a diagram of the circuit:

    attachment.php?attachmentid=40320&stc=1&d=1319565555.gif

    With these sorts of problems it's often handy to determine a few of the boundary conditions right away. These are predicated on the facts that an uncharged capacitor has zero voltage across it and looks like a short circuit initially, and when it reaches a steady state after a long time it has a constant voltage (determined by the circuit surrounding it) and zero current flowing, so it looks like an open circuit.

    So at t=0 the capacitor looks like a short circuit and the initial current into it will be E/r. Since C is a short and bypasses R, the initial current through R is zero.

    As t → ∞ the capacitor current goes to zero and the voltage on the capacitor will be determined by the voltage divider comprised of r and R. The current passing through resistor R will be determined by the series combination r + R (since the capacitor looks like an open circuit at this point).

    For all the things that change in the circuit there is a single time constant, [itex]\tau[/itex]. Once you have that and the boundary conditions, you can directly write the expressions for each of the changing values. How will you calculate the time constant?
     

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  6. Oct 26, 2011 #5

    vanhees71

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    Science Advisor
    2016 Award

    I've given the solution to the problem in my previous posting. Now, with the nice drawing by gneill it's easier to explain, how the equations came about.

    You have two loop and one independ knot equation. I take one loop (loop 1) as that containing the battery, the internal resistance and the external resistor and the other loop (loop 2) as consisting of R and C. The currents are all flowing in the direction from the + to the - pole (for convenience). The current through r is i, and the currents through R and C are [itex]i_1[/itex] and [itex]i_2[/itex] respecitvely. The plus charge, Q, is on the upper plate of the capacitor.

    Then from loop 1 you get

    [tex]E-i(r+R)=0,[/tex]

    from loop 2

    [tex]-i_1 R+Q/C=0,[/tex]

    and from each of the knots

    [itex]i=i_1+i_2.[/itex]

    To close the equations we note that

    [itex]i_2=\frac{\mathrm{d} Q}{\mathrm{d} t}[/itex] and thus from taking the time derivative of the loop-2 equation

    [tex]i_2=\dot{Q}=R C \dot{i}_1.[/tex]

    Using the knot equation in the loop-1-equation you finally arrive at the linear DGL of first order with constant coefficients, which you solve by the usual rules (full solution of the homogeneous equation + special solution of the inhomogeneous equation) as shown in my previous posting. I hope, I haven't made a mistake.

    Of course, the limiting cases are much simpler to get as indicated by gneill.
     
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