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First Order DE problem (electric circuit)

  1. May 11, 2008 #1
    (dq/dt) + (5/(20+t))q = 20/(20+t) ... t=0 , q=1

    (a) Find the charge on the capacitor at any time.

    This is linear so i found the integrating factor which is (20+t)^5

    and solved q(20+t)^5 = 20 integral ((20+t)^4) dt

    and got q(20+t)^5 = 4((20+t)^5) + C

    my C value i got was -9600000

    solving for q gave me q = 4 - ((9600000)/((20+t)^5))

    (b) Transient and steady state solution

    Transient = - ((9600000)/((20+t)^5))
    Steady state = 4

    (c) Find the current in the circuit at any time, hence the maximum current in the circuit.

    current = i = dq/dt

    I differentiated and got dq/dt = 48000000((20+t)^-6)

    **My problems

    Firstly does everything follow, is my logic correct?
    Secondly for the maximum current that occurs at t = 0 (presuming my equations are correct) is there anyway of showing working for that, other then stating its at t = 0? (t = 0, i = .75)
     
  2. jcsd
  3. May 11, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, everything you've done appears to be correct. As for showing that the maximum current occurs at t= 0, it is sufficient to observe that 48000000(20+t)-6 is a decreasing function. Simply observing that as t increases, the denominator increases, therefore the function decreases is enough- or if you could show that the derivative of i is never 0- therefore has a maximum at an endpoint, t= 0.
     
  4. May 11, 2008 #3
    OK so just observation is the way to really go, too easy thanks alot!
     
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