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Two Capacitors connected in Series and to a 10V battery.

  1. Apr 25, 2017 #1
    1. The problem statement, all variables and given/known data
    I am going over problems for exam study - here is the question with my submitted solution. Anything helps, just trying to correct mistakes so I can study the problems.

    Two capacitors C=3mF, C=2mF are initially discharged. They are connected in series and then the two ends are connected to a 10V battery.

    a) How much charge does the battery deliver?
    b) If we connect a single capacitor to the same battery, what would be the capacitance of c so that it draws the same charge from the batteries?
    c) Find the charges on each capacitor.


    2. Relevant equations
    I know I partially have the answer correct and my TA made a few notes - just want to correct my issues.

    3. The attempt at a solution
    See work above
  2. jcsd
  3. Apr 25, 2017 #2


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    Staff: Mentor

    Your problem statement says that the two capacitors are 3 mF and 2 mF, but your work shows that they are 3 μF and 2 μF. Note that mF is millifarads (10-3 F) and μF is microfarads (10-6 F). You can access the Greek alphabet for your posts via the ##\Sigma## button in the edit window top bar menu.

    In part (a) you added the two capacitors as though they were connected in parallel rather than in series. The formula for combining capacitors in series is different from the one used to combine them in parallel (just as the formula for combining resistors in parallel is different from the formula for combining them in series).

    Note that for capacitors in series there is only a single path that current can take and it flows through both capacitors. Thus at all times the same amount of current, and hence charge, must flow into or out of both capacitors.
  4. Apr 25, 2017 #3


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    Capacitors in series, are calculated like (1/Ctot) = (1/C1) + (1/C2) + ... Think of a parallel plate capacitor. Now think of a parallel plate capacitor, with a plate inbetween them (the distance has effectively decreased). Another thing, between the capacitors, charge has nowhere to go except from one capacitor to the other. Hook a battery up: electrons (negative charge) flow from the negative terminal to plate of one cap. This repels an equal number of electrons from the opposite plate to go through the wire to the other capacitor. Then electrons leave the opposite plate of that cap and go to the positive battery terminal. Both caps have equal charge, and it is equal the how much the battery delivered. For parallel capacitors, think of just increasing the area of the parallel plates increasing: Ctot = C1 + C2 + .... An amount of charge delivered by a battery to parallel must split between the caps.
  5. Apr 26, 2017 #4
    Ok! Thanks guys!
    As for the charges on each capacitor, did I use the wrong equation since the charge for capacitor ln series should be the same?
  6. Apr 26, 2017 #5


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    Staff: Mentor

    You can use Q = C*V for the individual capacitors, but you must use the voltage that is across the individual capacitor, not the total voltage. You should find that while the two voltages differ, the charges are equal.

    That said, you should think about what you are asked to solve for in part (b). If you find the equivalent net capacitance then its charge will be Q = Ctot*V, where this time V is the source voltage since it appears across the whole capacitance. Since the source "sees" the same load whether it's the original two capacitors or the single capacitor equivalent, it must deliver the same charge in either case. So put that together with the "capacitors in series have the same charge" idea.
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