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Charge on capacitors each with a battery

  1. Mar 24, 2010 #1
    1. The problem statement, all variables and given/known data
    Three capacitors C1–C3, all initially uncharged, are placed in the circuit shown. The capacitances are C1=8.9 μF, C2=20 μF, C3=12.2 μF, and the battery voltages are V1=18 V, V2=10 V, V3=18 V.

    What is the magnitude q1 of the charge on capacitor C1 once equilibrium has been established? Answer (given but I can't get it on my own): 0.000121 C

    What is the magnitude q2 of the charge on capacitor C2 once equilibrium has been established? Answer (given but I can't get it on my own): 0.000287 C

    What is the magnitude q3 of the charge on capacitor C3 once equilibrium has been established? Answer (given but I can't get it on my own): 0.000166 C


    2. Relevant equations

    Potential across each pair of battery / capacitor is the same.

    3. The attempt at a solution

    Equate the potential gain / drop from each branch to one another. But only get two equations (third is redundant) and have three unknowns.

    Thanks in advance for any help.
     
    Last edited: Mar 25, 2010
  2. jcsd
  3. Mar 25, 2010 #2

    ehild

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    Use conservation of charge.

    ehild
     
  4. Mar 25, 2010 #3
    Hey,

    What I did to work through this problem was to first determine which way the currents were going from the sources and then write an equation based on Kirchhoff's circuit law for the current distribution. From there you can rewrite the capacitors in their appropriate forms (series or parallel equivalents) and start solving for variables.

    The one thing that had me caught up for quite some time was what the total voltage was... pay close attention to the circuit diagram.

    Cheers
     
  5. Mar 25, 2010 #4
    Thanks for this message ehild, can you give me a bit more detail on the steps I would need to do in order to get the final solution?

    Thanks again
     
  6. Mar 25, 2010 #5
    Thanks RTjones. I have tried this but am still not getting the answer. Did you get the final answers?

    Thanks again for any further help you can provide.
     
  7. Mar 25, 2010 #6
    Yes I did get the correct answers. What I ended up doing to get the correct answer was using the EQn V = Q/Ceq. Where Ceq is the correct form for capacitance and Q is the total charge (which is subsequently the charge on capacitor 2). The part that was tripping me up though was that V is 28, not 18 or 10...the reason being because V1 and V2 are in series.

    Using this V with the equation above you can solve for Qtotal (aka Q2) and then work your way back to the other charges via relating different paths to one another. Let me know if you're still struggling, I'd be glad to help...
     
  8. Mar 25, 2010 #7

    ehild

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    You can change the position of the third battery with capacitor 3, it will not alter the voltages. If the upper plates of all capacitors are connected, no charge can flow on them, so the sum of charges on the three plates is zero. On the plates, the individual charges can be either positive or negative. If the capacitors are charged to Q1, Q2, Q3, respectively, there is -Q1 charge on the right plate of C1, +Q2 on the upper plate of C2 and -Q3 on the upper plate of C3. See picture.

    ehild
     
    Last edited: Jun 29, 2010
  9. Mar 25, 2010 #8
    Ok, thanks again for your help.
     
  10. Mar 25, 2010 #9
    Thanks, it was that fact that the charges must add up to zero due to plates being connecged that gave me my third equation.
     
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