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Homework Help: Non-Complex Capacitor Potential Difference question

  1. Mar 8, 2017 #1
    1. The problem statement, all variables and given/known data
    Three capacitors, with capacitances C1 = 4.0 μF, C2 = 3.0 μF, and C3 = 2.0 μF, are connected to a 24 -V voltage source, as shown in the figure. What is the charge on capacitor C2?
    http://imgur.com/x4LpU3t http://imgur.com/x4LpU3t

    a) What is the equivalent capacitance of the three capacitors ?

    b) What is the potential difference across C1 ?

    c) What is the charge on C2 ?

    d) What is the electrical energy stored on C3 ?

    2. Relevant equations
    Parallel share V
    Series Share q

    3. The attempt at a solution

    a) So I know that C2 and C3 are parallel so C23=5
    Then, C23 and C1 and in series, so Ceq=2.2μF

    b) Find Charge of C1 and use that to find Potential Difference. V=5.2E-2V

    c) Charge on C2 is 5.2E-2V / 3μF = 1.73E4C

    d) U =.5*C*V^2 so U=5.76E-4J

    Also, this isn't relevant but I dont think it justifies it's own post. How is W=VQ the same as W=EdQ? I appreciate any help
    Last edited: Mar 8, 2017
  2. jcsd
  3. Mar 8, 2017 #2


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    Staff: Mentor

    Hi Tyler R,

    Welcome to Physics Forums!

    It would have been helpful if the image you supplied contained the circuit diagram. But from your shown work it seems that the circuit looks something like this:


    Your answer for (a) looks good.

    For (b) how did you find the charge on C1? The potential difference you found for C1 does not look right.

    For (c), how did you decide that the potential difference across C2 is 5.2E-2 V? Shouldn't the sum of the potential differences across C1 and C2 be equal to the supply voltage of 24 V?

    For (d) it would appear that you used the supply voltage of 24 V in your energy expression. How would 24 V appear across C3?
  4. Mar 8, 2017 #3
    I found the charge by multiplying 2.08E-7 by 2.2E-6 because C1 is in series to the Ceq

    I found the potential by using the charge with the capacitance of C1

    I decided the V across C2 would be equal to that of C1 because V is the same in series and C23 and C1 are in series.

    Part d it wouldnt let me edit, but I changed it to U=.5*1.73E4*5.2E-2 which gave me 3.398E2J

    It seems I must be missing a core concept about Voltage across parallel and series circuits, but how would I find potential without charge?

    Edit: by the way thank you and I thought I included the picture, but yours is nearly spot on.

  5. Mar 8, 2017 #4
    For Part B

    I now found the charge by multiplying 4*2.2 resulting in 8.8E-6

    Using the charge I can calculate potential to be 2.2E6V
  6. Mar 8, 2017 #5


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    Staff: Mentor

    What is 2.08E-7 ? Where did it come from?
    That's incorrect. Series capacitors share the same charge but not the same potential difference. Parallel capacitors share the same potential difference.
    Can you explain those numbers?
    Parallel components share the same potential difference (voltage). Series components share the same current.

    If you can find the potential difference across other components in a series circuit, KVL can be used to find a "missing" potential difference, as the sum of the all the potential differences around a closed path must be zero.
    Great. By the way, it's preferable to upload critical images to the PF server rather than host them off-site. Inserted images are better than links.
  7. Mar 8, 2017 #6


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    Staff: Mentor

    You need to explain where these numbers are coming from. Also, given that the source voltage is 24 V you shouldn't find any potential differences larger than that anywhere in the circuit!
  8. Mar 8, 2017 #7
    Thanks for the reply and that is good to know! I will for sure do that next time.

    I think I see what you're saying.

    The total charge of the circuit is Q=C*V which is Q=2.2μF*24 which gives 5.28E-5C

    This means the potential at C1 is: V=Q/C ---> 5.28E-5C/4E-6 = 13.2V

    Because C1 and C23 are in series the charges are the same so

    The potential at C23 is V=Q/C ----> 5.28E-5C/5E-6 which is 10.56V

    Since C2 and C3 are parallel the V is the same which means charge of c2 is

    Q=3E-6*10.56V= 3.168E-5C

    The electrical energy stored on c3 is U=.5*C*V^2 --> U= .5 * 2E-6F * 10.56V

    which ends up being 10.056E-5J, but that doesnt seem to be correct
    Last edited: Mar 8, 2017
  9. Mar 8, 2017 #8
    The electrical energy stored on c3 we can find the charge first

    2E-6*10.56 = 2.112E-5C

    But I still get a wrong answer for U

    Thanks for all your help so far
    Last edited: Mar 8, 2017
  10. Mar 8, 2017 #9


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    Staff: Mentor

    Good so far!
    You haven't squared the voltage: ##E = \frac{1}{2} C V^2##

    A suggestion: When you work one of these multi-part questions where calculations from one part feed into the next and so on, be sure to retain extra digits in all intermediate values. These "guard digits" will prevent rounding and truncation errors from creeping into your significant figures as the calculations progress. Values should only be rounded at the end for presentation.
  11. Mar 8, 2017 #10
    Thanks :) You are awesome. Have a great night
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