Capacitor charging another capactior

Discussion in 'Homework Help' started by Vincent9009, Mar 2, 2015.

  1. Vincent9009

    Thread Starter New Member

    Mar 2, 2015
    3
    0
    I have been doing some calculation on capacitors in different circuits lately and everything has been going fine, but this one got me really stuck. As a result I have been scratching my head on this problem for a couple of days now and need some pointers on how to solve it.

    I want to create a function which shows voltage over time for ex. C1.

    The initial condition (t=0) are:
    V(C1) = 4 V => Q(C1) = 2,72 mC
    V(C2) = 0 V
    C1 = C2 = 680 microF
    R1 = 4,7 Kohm

    I have been thinking:
    V(C1) - V(R1) - V(C2) = 0
    V(C1) - R*dq/dt - dq/C = 0

    And this is where I get stuck, Im not really sure how to proceed or even if I'm going in the right direction.
    I can solve this if it had been only a capacitor decharging itself or a power supply charging a capacitor.
    But when it is two capacitors, the voltage over both of them will alter over time and I cant see a solution for this.

    Added a quick drawing for the layout of the circuit.
     
  2. Vincent9009

    Thread Starter New Member

    Mar 2, 2015
    3
    0
    I finally figured it out.

    The function for Vc1(t):
    Vc1(t) = 4V - 2V(e^(-t/1,598s))
    t = time in seconds

    And if someone is wondering what the function for the other components are, I'll post them too:

    Vc2(t) = 2V(1-e^(-t/1,598s))

    Vr1(t) = 4V(e^(-t/1,598s))

    I had messed up the total storage for the capacitors, I considered them connected in parallel, but they act like they are connected in a series.
     
    Last edited: Mar 3, 2015
  3. Vincent9009

    Thread Starter New Member

    Mar 2, 2015
    3
    0
    ops my bad, typed in the wrong function for Vc1(t)
    The correct one are:

    Vc1(t) = 2V + 2V(1-e^(-t/1,598s)) = 2V(1+e^(-t/1,598s))
     
    Last edited: Mar 3, 2015
  4. MrAl

    Distinguished Member

    Jun 17, 2014
    3,618
    760
    Hello,

    Looks good.
     
  5. darrough

    Member

    Jan 18, 2015
    86
    19
    The two capacitors could be combined into one capacitor. The formula is 1 / ( 1 / C1 + 1 / C2 ). The time constant for a capacitor is R * C. The current falls off exponentially and in 5 time constants it is less than 1% of its original value.
     
  6. WBahn

    Moderator

    Mar 31, 2012
    20,064
    5,665
    This description applies to circuits subjected to a step input, which is not the case here.
     
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