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First order circuits. Solving for v(t)

  1. Oct 31, 2016 #1
    1. The problem statement, all variables and given/known data
    So our lecture introduced first order circuits to us.
    We are trying to solve for v(t) in the following circuit. However in the notes he gets it into a form in which we can solve for v(t). However I can't seem to get it into the same form he has. This is more of a maths question rather than a circuits question.

    The purpose of writing it in the form that he has is to solve for v(t). Any ideas how I can get it into that form?

    054f0907b2.jpg


    2. Relevant equations

    i(t) = C dv(t)/dt


    3. The attempt at a solution

    In Pic above
     
  2. jcsd
  3. Oct 31, 2016 #2

    NascentOxygen

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    Staff: Mentor

    Presumably E is a constant here?

    The derivative of a function is unaffected if you first add (or subtract) a constant value to the function,
    so it follows that d(v(t))/dt is identical to d(v(t)–E)/dt

    It looks like your lecturer might be finding it convenient to use this fact in his derivation of v(t)?

    the derivative of a sum is equal to the sum of the individual derivatives
     
  4. Oct 31, 2016 #3
    Ah. That makes much more sense. I've been trying to figure this out for a while now. Thank you very much,
     
  5. Dec 14, 2016 #4
    Hi,

    I've just come back to revise this after properly I think understanding dynamic circuits. I am told that in the DC steady state(i.e when the source is providing a steady voltage), capacitors behave as open circuits and inductors behave as short circuits. So why doesn't that happen in this case(where E is constant).? I mean the circuit is providing a constant V and so the time derivative in i = C dv/dt should evaluate to 0 thus giving us i=0 which is an open circuit. Why don't we just replace the capacitor with a open circuit here?
     
  6. Dec 14, 2016 #5

    NascentOxygen

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    Staff: Mentor

    You can. If you do that it will tell you that the steady-state current is zero, which is true.
     
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