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!

Electric Circuit Capacitor Charge question

  1. Jun 2, 2012 #1
    1. The problem statement, all variables and given/known data

    An RC circuit is given with a 1μF capacitor, a 25kΩ resistor and a 4μF capacitor, all in series. Initially, the 1μF capacitor has a voltage of 40V, while the 4μF capacitor has a voltage of 0V. The circuit, starting from the left and going clockwise is, 1μF (reverse polarity), the resistor and then the 4μF capacitor (regular polarity).

    The goal is to find the functions governing current and the voltage of each capacitor (total of three functions).

    i(t)=?, v1(t)=?, v4(t)=?


    2. Relevant equations

    [itex]\sum[/itex]V=0
    V=IR
    Q=CV
    I=dQ/dt

    3. The attempt at a solution

    Applying KVL I get:

    V1=Vr+V4

    Q1/C1=IR+Q2/C2

    Q1/C1=Q'R+Q2/C2

    This equation becomes very confusing to me. Firstly there are two charge functions of time. I can eliminate this problem by noting that since the circuit is a series, the current is constant throughout all elements meaning that dQ1/dt=dQ2/dt, so I can differentiate this DE and get:

    Q1'/C1=Q1''R+Q1'/C2

    Q1''+[(C1-C2)/(C1C2R)]Q1'=0

    Which leads me to conclude that:

    Q1(t)= k1+k2exp([(C2-C1)/(C1C2R)]t

    however I can go further, because since I=dQ1/dt=dQ2/dt, then Q2=∫(dQ1/dt)dt, so I get the following:

    Q1(t)= k1+k2exp([(C2-C1)/(C1C2R)]t
    Q2(t)= k3+k2exp([(C2-C1)/(C1C2R)]t
    I(t)= ([(C2-C1)/(C1C2R)]k2exp([(C2-C1)/(C1C2R)]t

    However, this leads to a problem. The initial conditions are:

    Q1(0)=C1V1i=(1μF)(40V)=40μC
    Q2(0)=C2V2i=(4μF)(0V)=0μC
    I(0)=0 A

    Which gives me:

    k1=40 μC
    k2=0 μC
    k3=0 μC

    and:

    Q1(t)= 40 μC
    Q2(t)= 0 μC
    I(t)= 0 A

    Can someone point out the error in my reasoning? I've been over it countless times and it all makes sense to me, but there is obviously a problem.
     
  2. jcsd
  3. Jun 3, 2012 #2
    Am at work right now so will try this problem later on. But on first glance....when the circuit is made, then given that the initial voltage on the 1uf cap is 40V, it should act as a voltage source of V=40V at t=0, and then exponentially discharge, no?
     
  4. Jun 4, 2012 #3

    rude man

    User Avatar
    Homework Helper
    Gold Member

    What do you mean by "reverse polarity" and "regular polarity"? A sketch showing polarity definition would help.
     
  5. Jun 5, 2012 #4
    cmmcnamara,

    I did not even bother to look at your work, because I can tell right away that you are making too big a deal out of it. It is a simple single loop equation. Set up the differential loop equations for the two caps and the resistor and solve. Forget about charge and just solve for the loop current. That's all there is to it. So set up the equations and I will let you know if you did it correctly. Then you can continue on to solve the equation for current.

    Ratch
     
  6. Jun 8, 2012 #5
    That is how I derived the differential equation for charge, I used KVL around the loop and applied the information for the caps and resistor. If I find the function for charge the voltage function for each capacitor will be easy to determine using their capacitance and the derivative of the charge function will be the current through the loop because all elements are in series. I think the problem may lie in my choice of initial conditions but I don't see it.
     
  7. Jun 8, 2012 #6
    cmmcnamara,

    Here is the loop equation. Remember V=(1/C)∫i(t)dt

    (1/C1)∫i(t)dt+(1/C2)∫i(t)dt+i(t)*R+40=0

    Can you solve this differential equation?

    Ratch
     
  8. Jun 10, 2012 #7
    I'm not I understand your loop equation. There is no voltage source in the loop, so I'm not understanding where the 40V term is coming from.
     
  9. Jun 10, 2012 #8
    cmmcnamara,


    Sure there is. One of the caps is energized to 40 volts, isn't it?

    Ratch
     
  10. Jun 10, 2012 #9
    Yes initially it is so wouldn't that be part of the initial conditions?
     
  11. Jun 10, 2012 #10
    cmmcnamara,

    Yes, it is. Mathematically, a constant voltage source in series with a unenergized cap is the same as a cap energized at the same voltage with no voltage source in series.

    Ratch
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Electric Circuit Capacitor Charge question
Loading...