1. Limited time only! Sign up for a free 30min personal 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!

Find the magnitude of charge stored in the entire network

  1. Oct 1, 2014 #1
    Problem: In the circuit shown in the figure below, the capacitors C1 and C2 have
    a capacitance of 45µF. C3 and C4 have a capacitance of 90µF. The two
    terminals A and B on the left have a potential difference of 25V

    (Sorry - had to quickly draw it, could not save/find image)

    (a) Find the magnitude of charge stored in the entire network.
    (b) Find the magnitude of charges stored in each capacitor.
    (c) Find the energy stored in each capacitor. Identify the capacitor with
    the largest energy content.
    (d) The C3 capacitor is now removed from the circuit, leaving a break
    in the wire at its position. What is the voltage drop across the three
    remaining capacitors?
    (e) Describe how the maximal amount of energy can be stored using
    these four capacitors. Draw the resulting circuit and state how much
    energy is stored in your circuit using the same potential difference of

    Ceq series=(1/c1+1/c2...)^-1
    Ceq parallel=c1+c2....
    Series capacitors have equal charges and different voltages while parallel's have different charges and equal voltages.
    U=(CV^2)/2 (Energy of capacitors)


    a) Q1234=C1234(V)

    Knowing that C1 and C3 are in series , C13=3e-5F
    C2 and C4 in parallel, thus C24 = 1.34e-4F
    C13 and C24 are in series thus C1234=2.45e-5F

    Thus Q1234=C1234*V=2.45e-5F*25V=6.1e-4C

    b)Since C1 and C3 are series, Q1=Q3=Q13=Q1234=C1234(V)=2.45e-5F*25V=6.1e-4C

    I am not sure how to then solve Q2 and Q4 assuming I did Q1, Q3 correctly.

    c) U=(CV^2)/2 , I realize that C for each is already given but am not sure what are the individual voltages.

    d) I am unsure how to approach this

    e) Since Ceq parallel is greater then the individual capacitance then it would make sense for it to be an all parallel circuit where the new Energy is equal to U=(CV^2)/2 where C=C1+C2+C3+C4.

  2. jcsd
  3. Oct 1, 2014 #2


    User Avatar
    Homework Helper
    Gold Member

    C1 and C3 are not in series.
  4. Oct 1, 2014 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    C1 and C3 are not in series. When two elements are in series, the current through one element has to go through the other. The current through C1, however, can go through C2 or C3. In contrast, C3 and C4 are in series. There's nowhere else for the current through C3 to go other than through C4.

  5. Oct 1, 2014 #4
    Thanks Vela, that made a lot of sense. I'll give it another go.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Find the magnitude of charge stored in the entire network