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RC circuits and potential difference

  1. May 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Circuit is at this link: http://www.colorado.edu/physics/phys1120/phys1120_sp08/hws/capa8_figs/gian1946.gif

    1. What is the potential at point a with S open (let V=0 at the negative terminal of the source, and assume it's a long time after the circuit was connected to the potential difference source).

    2. What is the potential at point b with the switch open? Again assume it's a long time after the circuit was connected.

    3. After the switch is closed for a long time, what is the final potential of point b?

    I know what the answers are (some I solved through a happy coincidence, and some I was told the answer to-- this problem has been discussed on PF and yahoo answers already) but I don't understand why the answers are what they are. I think I'm confused about the foundation of this problem. See questions below.



    2. Relevant equations

    V = IR
    C = Q/C
    Capacitors in series have same charge
    RC circuits are time-dependent
    Resistors in series are additive



    3. The attempt at a solution

    1. I got this one right myself, but when I looked back at it I confused myself. The answer is, as you can find elsewhere on the internet:

    Va = ItotR2 = VtotR2/(R1+R2) from V = IR, Itot = Vtot/Rtot, and the knowledge that resistors in series are additive and that the same current goes through both of them.

    However, why is the voltage across a the voltage across R2, and not R1? Also, am I right in thinking that all current goes through the resistors, since it's been a long time since the circuit was connected and the capacitors are fully charged?

    2. Using Vtot = V1 + V2 and V = Q/C and Q1 =Q2 when in series, Vb = V2 = Q/C2 and so forth... you can get Q in terms of the total voltage and total capacitance.

    My confusion is related to above: why do we use C2, and not C1 in the equation Vb = Q/C2?

    3. Same as #1, because the capacitors become fully charged and all I again goes through the resistors. This one I understand.

    So, I think I'm missing something really elementary but it would be helpful to me if someone lets me know my dumb mistake.
     
  2. jcsd
  3. May 7, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    The potential of interest is the one across R2 because you are told that the zero reference is the negative terminal of V, which happens to be connected to the bottom of R2. So the potential at a is thus the potential across R2. You could also have used the potential across R1 if you used KVL up through the supply V and down through R1 to reach a. Bit of a roundabout way to go about it though.
    Its the nature of capacitors in series that the potential divides across them in inverse relation to their capacitances. You should prove this for yourself by taking two capacitors in series and placing some charge Q on each of them (since the capacitors are in series they must both experience the same current, so their charges must always be the equal). Note the ratios of the individual potentials to the total potential across the branch (sum of the two potentials). Compare to the ratios of the individual capacitances to the sum of the capacitances.
     
  4. May 7, 2012 #3
    Got it, thanks! I missed the importance of defining where zero voltage was. And I did use Kirchhoff's; it was, as you say, a roundabout way to do it that got me the right answer but left me confused.

    Thanks again!
     
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