Simple RC circuit question

Discussion in 'Homework Help' started by secondhandloser, Oct 26, 2008.

  1. secondhandloser

    Thread Starter Member

    Sep 30, 2008
    7
    0
    Question attached in PDF form.

    As far as what I've done, I wrote up the condition required while the voltage across the SCR is less than 5v
    x(t)=6V+(0-6)e^-t/τ

    after the switchover, I'm having difficulty writing the equation, as I don't understand what the voltage difference would be across the SCR at t=∞.
    At t=0, I know it will be 5v, because that's what it is before the changeover.

    Any help on even how to set up the equation will be appreciated- I know it's simple, and I'm just blanking on it.
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Do you have the values for R and C?
     
  3. secondhandloser

    Thread Starter Member

    Sep 30, 2008
    7
    0

    That's what I'm trying to solve for.
     
  4. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    secondhandloser,

    I will give it a try, but the hour is late and I might make a mistake.

    The voltage across C when first powered up is (6/s)(1/Cs)/(R+1/Cs) . Cleaning it up and finding the inverse Laplace we get 6(1-exp(-t/RC)) . Equating the expression to 5 volts and substituting t=1 sec we get a RC constant of 1/ln6 . Now we write a node equation (V-6)/R +VCs-5C+50E-6 = 0 . Cleaning it up and separating out V we get V = (5RC - R*50E-6 +6)/(RCs+1) . Finding the inverse and substituting t=0.1, and RC = 1/ln6, and equating the voltage to 0.1, I get R = 174 kohms. Then C = 3.2E-6 farads. I hope I did everything correctly.

    Ratch
     
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