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RC circuit, expression for voltage

  1. Oct 6, 2014 #1
    1. The problem statement, all variables and given/known data
    Assume you have a fully charged capacitor with initial condition [itex]V(0) = V_0[/itex] connected in series to two resistors [itex]R_1[/itex] and [itex]R_2[/itex]. Derive an expression for the voltage over the capacitor with respect to time.

    2. Relevant equations
    1. Kirchhoff's voltage law [itex]\sum_n V_n = 0[/itex]
    2. Ohm's law: [itex]V=RI[/itex]
    3. Capacitance of capacitor: [itex]C = \frac{Q}{V}[/itex]
    4. Current [itex]I = \frac{dQ}{dt}[/itex]

    3. The attempt at a solution
    If I "go" in the direction of the current, then the potential difference over both resistors are negative whereas it is positive over the capacitor, hence
    [tex]V(t) - V_{R_{1}} - V_{R_{2}} = 0 = V(t) - R_1 I - R_2 I[/tex].
    Which I can rewrite using the time derivative of Q (3 and 4 from above)
    [tex]= V(t) - (R_1 + R_2)\frac{dQ}{dt} = V(t) - (R_1 + R_2)C\frac{dV(t)}{dt} = 0 \implies \frac{1}{(R_1 + R_2)C}V(t) - \frac{dV(t)}{dt} = 0[/tex].
    By solving this differential equation, I get
    [tex]V(t) = V_0e^\frac{t}{(R_1+R_2)C}[/tex]
    which is wrong, because the voltage should decrease over time, so a more logical solution would be
    [tex]V(t) = V_0e^{-\frac{t}{(R_1+R_2)C}}[/tex].
    Is there something I have overlooked? Is Kirchhoff's law setup incorrectly (sign conventions)?

    Any help is appreciated.
  2. jcsd
  3. Oct 6, 2014 #2
    VC(t) + VR1(t) + VR2(t) = 0 is the equation
  4. Oct 6, 2014 #3
    Yes, I realize that has to be true for it to be correct, but I am just confused about the signs. My book (Sears and Zemansky's) mentions some sign conventions in relation to Kirchhoff's voltage law (see picture). w0QUQ.png
    So why is it + rather than - ?
  5. Oct 6, 2014 #4
    You have 3 passive elements in the circuit and no driving emf. Don't know what confuses you.
  6. Oct 6, 2014 #5


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    Staff: Mentor

    nossren, have you considered what sign should be associated with dV/dt on the capacitor? Will V be increasing or decreasing given the direction of the current flow?
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