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Finding the charge on a capacitor

  1. Feb 10, 2017 #1

    diredragon

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    Gold Member

    1. The problem statement, all variables and given/known data
    IMG_2502.JPG

    ##E_1=6V##
    ##E_3=5V##
    ##R_1=150Ω##
    ##R_2=R_3=100Ω##
    ##R_4=50Ω##
    ##R_5=300Ω##
    ##C_1=1.5 mF##
    ##C_2=0.5mF##
    Text:
    In the first condition when the switches 1 and 2 are open the capacitor 2 has charge on it. First the switch 1 closes and the charge flow of ##q_1=1 mC## is established to flow through the branch as shown in the picture. Then the switch 2 closes and the flow is now ##q_2=-1.4mC##. Find the initial charge on capacitor 2 (##Q_{20}##).
    2. Relevant equations
    3. The attempt at a solution

    My only problem here is with the charge division. Correct me if i'm wrong here:
    When the switch 1 closes the voltage across 1 and two must be the same so we have:
    ##q_{11}+q_{12}=q_1##
    ##\frac{q_{11}}{C_1}=\frac{Q_{20}+q_{12}}{C_2}##
    I can't find the voltage they're equal to because i don't have ##E_2## but i do have the change when the switch 2 closes. So:
    ##\frac{q_{11}+q_{21}}{C_1}=\frac{Q_{20}+q_{12}+q_{22}}{C_2}##
    The change is then just:
    ##ΔU_{c1}=\frac{q_{21}}{C_1}##
    ##ΔU_{c2}=\frac{q_{22}}{C_2}##
    And ##q_{12}+q_{21}=q_2## and from there i can find the ##q_{12}## and ##q_{22}## and have the change in the voltage while my Δ-circuit looks like this:
    IMG_2503.JPG
    Where finding the current is trivial. Is this the right track?
    Thanks
     
  2. jcsd
  3. Feb 11, 2017 #2

    diredragon

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    Gold Member

    I continued along this road and i got a wrong answer. I will now post my full work so you can see where i'm wrong.
    Strarting from the delta circuit where im trying to find the current ##ΔI_g## i have this:
    1) ##ΔU=\frac{q21}{C1}=\frac{q22}{C2}##
    2) ##q21+q22=q2=-1.4##
    ##q21=\frac{C1}{C2}q22##
    and from this i get ##q21=-0.35##,##q22=-1.05##
    so ##ΔU=-0.7 V## and i can continue finding the current ##ΔI_g##
    Using the current divider i see that:
    ##ΔU=ΔI_gR4+R2ΔI_d=ΔI_gR4+R2\frac{R3}{R2+R3+R5}ΔI_g##
    and from here ##ΔI_g=-0.01A##
    Now goind back to the original circuit, where i have the current through the branch when the switch closes i have enough information to find ##E_2## which later i can use to find the voltages across the capacitors. From mesh current analysis get that the loops that make ##ΔIg## are some ##IB and IA##
    So by arbitrary choice:
    ##ΔI_g=I_b-I_a## and by means of mesh i get ##I_b## because i only need that one. I could go the other way but i need E2 then.
    Here is the picture of the work:
    IMG_2506.JPG
    I get ##Q=2mC## while the answer is ##Q=-0.1mC##
    Where and what is wrong?
     
  4. Feb 11, 2017 #3
    I'm not sure why E2 is not given.

    Nonetheless, with Switch 1 open, C2 is just dangling. There is no voltage across it so it cannot have a charge on it. On the other hand, C1 will have a charge on it.

    This changes most of what you did initially and will affect everything beyond that point.
     
  5. Feb 11, 2017 #4
    It shouldn't change anything initially since the change is considered when the switch 2 closes so ##ΔU## remains as it is regardless of the initial charge of Q1 and it makes appearance only at the end, but i supposed that it is uncharged since its stated that C2 has charge on it.
     
  6. Feb 11, 2017 #5
    OK, I see the point about having a trapped charge on C2. Sorry about that.
     
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