RC circuit natural response, two capacitors

Discussion in 'Homework Help' started by stinz, Oct 19, 2008.

  1. stinz

    Thread Starter New Member

    Sep 25, 2008
    Hello, I could use some help to analyze this circuit.
    The problems are as follows:

    1. Find the current i0(t) for t>0
    2. Find the voltage v0(t) for t>=0
    3. Find the energy,w(t), stored in the circuit when t->infinity

    I've managed the first problem by finding the equivalent capacitance when t>0 which I found to be 100nF.
    Problem 2, I'm not sure where to start. Problem 3 cannot be solved without solving problem 2.

    Help would be much appreciated :)
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  2. Ratch

    New Member

    Mar 20, 2007

    You first have to determine the initial voltage on the capacitors at t=0. Then by the classical differential equation method or the Laplace method solve what the current is when the two capacitors discharge through the resistor. It will be some exponential function.

  3. hitmen

    Active Member

    Sep 21, 2008
    I am not sure if I got the correct answer cos I only deal with one capacitor question.

    for t<0 , V8k = 200V
    V150nf = 200V
    V300nf = 180V

    for t>0 io(t) = 10mA
    Vo(t) = 180V

    Total energy = E8K + E2k + E150nF + E300nf
    = 25(8) + 10(2) + 0.5(150*10^-9)(200)(200) + 0.5(300power-9)(180)(180)

    = 220W.

    My answers could be wrong. Take care.

    But I am confident that answers for t<0 is correct. Take care and get back to us cos I also wish to know the answr
  4. Ratch

    New Member

    Mar 20, 2007

    Your answer is wrong and I did take care.

    Since the capacitors can be assumed to be energizing for a long time, they can be treated as open insofar as obtaining the voltage at t=0. The current through the 2k resistor can easily be found to be 10 ma. That makes the initial voltages on C150 and C300 120 volts and 100 volts respectively. Now we can also easily find the current as C150 energizes C300 to a higher voltages and reduces it own voltage. Setting up a single loop equation, we find the current to be (10E-3)*exp(-5000t). Integrating the current with respect to time from 0 to infinity we get 2E-6 coulombs of charge. Since charge is conserved in this case, we know that what imbalance was reduced on C150 was increased on C300. Since E = Q/C, the voltage for C150 decreased 13.33 volts and increased 6.67 volts for C300. Therefore the two capacitors level off at 106.67volts. The energy can be determined by calculating the energy of the capacitors (CE^2)/2 at 106.67 volts.

    Last edited: Oct 22, 2008