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Capacitance and Voltage Confusion

  1. Oct 25, 2007 #1
    1. The problem statement, all variables and given/known data
    This is for a lab I'm doing tomorrow. We will have a 10V circuit with 4 100µF capacitors connected in series. I need to calculate the voltage drop across each capacitor. I also need to create another circuit, but I'll worry about that after I'm sure I'm doing this one properly. :)

    2. Relevant equations
    V = q/C, V = V1 + V2 + V3 + V4

    3. The attempt at a solution
    I feel like I'm doing something wrong here.

    at C1, V1 = q/(100*10^-6)
    at C2, V2 = 2q/(100*10^-6)
    at C3, V3 = 3q/(100*10^-6)
    at C4, V4 = 4q/(100*10^-6)

    Total V = 9q/(100*10^-6)

    I used that to solve for q:
    q = 1.111*10^-4C

    Then I plugged q back into each equation.

    V1 = 1.111V
    V2 = 2.222V
    V3 = 3.333V
    V4 = 4.444V

    Am I on the right track here?
  2. jcsd
  3. Oct 25, 2007 #2


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    Staff: Mentor

    From where is one getting the 1q, 2q, 3q, 4q?

    The capacitors have the same nominal capacitance.
  4. Oct 25, 2007 #3
    I have that because when capacitors are connected in series, the voltage is V = q(1/C1 + 1/C2 + 1/C3 + 1/C4), according to my text.

    So at the first capacitor, V = q/C, then at the second V = q/C + q/C = 2q/C, and so on.. or am I understanding it incorrectly?
  5. Oct 25, 2007 #4
    I apologize for bumping, but I still am not sure how to do this correctly.. any ideas?
  6. Oct 25, 2007 #5
    capacitors in series have the same charge. So they all have charge q.
  7. Oct 25, 2007 #6
    Find the fallacy on ur own! Going by ur reasoning.. at third it is.. 3q/C, and at fourth 4q/C. So, total voltage drop = q/c + 2q/C + 3q/C + 4q/C = 10q/C!
    Going by your textbook, V = q(1/C1 + 1/C2 + 1/C3 + 1/C4). Substituting, C1 = C2 = C3 = C4 = C, u will get.. V = 4q/C!!

    I think you have misinterpreted. Charge on each capacitor would remain same, because all of them were initially uncharged & they are in series. Also, individually they will have a voltage drop, Vi = qi / Ci.
  8. Oct 26, 2007 #7
    maybe it would help to simplify the circuit using equivalent capacitance and then work backwards, it might take longer, but that method usually works for me.
  9. Oct 26, 2007 #8
    Ok, so.. I used V = 4q/C, and got q = 2.5*10^-4C.

    I then recalculated the voltage drops, and got these..
    drop 1 = 2.5*10^-6V
    drop 2 = 5.0*10^-6V
    drop 3 = 7.5*10^-6V
    drop 4 = 1.0*10^-5V

    These still don't look correct (they are much smaller than anything I would be able to measure in the lab). Am I supposed to be finding q, then plugging it back into each voltage equation?

    Unfortunately, I don't understand how I would simplify it using equivalent capacitance. We've spent maybe 15-20 minutes learning about these circuits so far - I'm having a very hard time trying to understand how it works without changing stuff around.
  10. Oct 26, 2007 #9
    Since they are in series.. each one of the capacitors will have same 'q' appearing on them. +q on the positive plate and -q on the negative plate. Thus, voltage drop across each of the capacitors would be Vi = q / Ci. Here C1 = C2 = C3 = C4 = C. Thus, V1 = V2 = V3 = V4 = q/C.
    Now, V = V1 + V2 + V3 + V4. Thus, each Vi = V/4 = 2.5 volts.
  11. Oct 26, 2007 #10


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    Staff: Mentor

    Try this site - http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capac.html#c2

    As saket mentioned, the capacitors have the same charge as a consequence of having the same capacitance.
  12. Oct 26, 2007 #11
    I think I get it now.. thanks for the help and link :)
  13. Oct 26, 2007 #12
    The voltage drop would be due only to electrical series resistance (ESR) of the capacitors. Other than that, in the situation you described, there is no reason for a voltage drop.
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