1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

RC Circuit with two resistors

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Given a circuit with two resistors, R and r, and a capacitor of C, and EMF of V0 as shown in the diagram, find the voltage across the capacitor during charging. Prove that this voltage, V is given by V = V0 (r/(R+r)) (1-e-((R+r)t)/(RrC))

    2. Relevant equations

    N.A.

    3. The attempt at a solution

    This is what I have

    Loop A (with C): V0 + RI + q/c = 0 ==> I = (V0 - VC)/R
    Loop B (with r): V0 + RI + rIr = 0
    I = Ir + Ic ==> I = V/r + C(dv/dt)

    From first and third,
    (V0 - VC)/R = V/r + C(dv/dt)

    Simplify to get,
    V0 - RC(dv/dt) = V(1+R/r)
    V = (r/[R+r])(V0 - RC(dV/dt))

    The shape of the equation is getting there (I hope), but what do I do next? To get the given equation, RC(dV/dt) must be V0e-((R+r)t)/(RrC).

    RC (dV/dt) = V, solving this differential equation to get ln (V) = -t/(RC) + k, hence, RC (dV/dt) = e-t/(RC) + k. And I am totally stuck.

    Did I do something wrong somewhere? I can't think of anyway to get the RrC term in e-((R+r)t)/(RrC), not to mention the V0 and the (R+r) terms, unless k is like Rt/r. But that still does not give me a V0?
     

    Attached Files:

    • RC.png
      RC.png
      File size:
      1.6 KB
      Views:
      368
  2. jcsd
  3. Feb 13, 2012 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The differential equation is correct. You need not solve it to prove that the given V is solution, just plug in V and dV/dt. What is the derivative of the given V(t) function?


    ehild
     
  4. Feb 13, 2012 #3
    Hi ehild,

    Thanks for your reply.

    For the differential equation, V = RC (dV/dt), solved for V = e-t/(RC) + k,

    e-t/(RC) + k = RC (dV/dt)

    dV/dt = e-t/(RC) + k/RC

    Is this what you meant?
     
    Last edited: Feb 13, 2012
  5. Feb 13, 2012 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    NO.

    Your differential equation is not V = RC (dV/dt), but V = (r/[R+r])(V0 - RC(dV/dt)).


    ehild
     
  6. Feb 13, 2012 #5
    Sorry for the mistake,

    That will get me

    (V - (r/[R+r])V0)-1dV = [-(r/[R+r])RC)]-1dt

    Which gets me ln [V - (r/[R+r])V0)] = t/(RrC/[R+r])
    V - (r/[R+r])V0) = e(t/(RrC/[R+r])]
    And a little rearranging gets me V = e(t/(RrC/[R+r])] + (r/[R+r])V0)

    Hmm, did i do something wrong? Even if I subsitute V = RC(dV/dt) into the equation, i will get

    V = (r/[R+r])(V0 - e(t/(RrC/[R+r])] - (r/[R+r])V0))

    which is not equal to the answer..
     
    Last edited: Feb 13, 2012
  7. Feb 14, 2012 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You missed to add the arbitrary constant. Your result will be is the same as the given function with the condition that the initial voltage on the capacitor is zero: V(0)=0.

    Why do you substitute V=RC(dV/dt)? It is valid for a simple RC circuit only.

    ehild
     
  8. Feb 14, 2012 #7
    Hi again ehild, I added in the arbitrary constant and got the equation! Thank you so much!

    Regarding the V = RC (dV/dt), I have misunderstood the derivation, and assumed that it is a general formula for all RC circuits. Thanks for pointing that out, I'll read up on it tonight.

    Once again, thanks for your help!
     
  9. Feb 14, 2012 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You are welcome:smile:

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: RC Circuit with two resistors
  1. Two Loop RC Circuit (Replies: 7)

Loading...