Capacitor charging another capactior

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

  1. Vincent9009

    Thread Starter New Member

    Mar 2, 2015
    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
    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
    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

    Well-Known Member

    Jun 17, 2014

    Looks good.
  5. darrough


    Jan 18, 2015
    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


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