1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple capacitors

  1. Mar 24, 2009 #1
    1. The problem statement, all variables and given/known data

    See attached PDF for circuit.The arangement of capacitors is attached to a voltage source. Each of the 4.0uF compactors stores a charge of 6.0uC. What charge is stored on the 6.0uF capacitor? What is the voltage of the source.

    2. Relevant equations
    Q=C(V)



    3. The attempt at a solution

    [tex]\frac{Q}{C}[/tex]=V=[tex]\frac{6.0uC}{4.0uF}[/tex]=1.5V
    This is the voltage drop across the two capacitors in series.

    Q=CV=1.5V(6.0uF)=9uC
    But this is not correct the charge on the larger capacitor should be 12uC correct?
     

    Attached Files:

  2. jcsd
  3. Mar 24, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    That's the voltage across the two 4μF capacitors, which are in parallel. It's not the voltage across the 6μF capacitor.

    Don't assume a voltage and use it to try to calculate the charge. Instead, you should be able to deduce the charge from the given information. Then you'll be able to calculate the voltage, in order to answer the second part of the question.
    That's true. The way to see that is to ask yourself what's the total charge on the two capacitors in parallel. They, in turn, are in series with the other capacitor. And the total charge on capacitor plates in series is always what?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Simple capacitors
  1. Simple Capacitors (Replies: 6)

Loading...