Source Switching RC circuit

Discussion in 'Homework Help' started by Jess_88, Apr 30, 2011.

  1. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    Hi

    I need to fined i(t) for te following circuit.
    [​IMG]

    I think I know what I'm doing, I'm just a little stuck on using
    I(t) = V/R + C (dv/dt)

    so this is what I have so far-
    V(0-) = 30V(3Ω/(6Ω + 3 Ω) = V(0) = V(0+)
    = 10V

    Removing the capacitor
    Rth = 6Ω + 3 Ω

    \tau = Rth x C = 2 x 9

    At t>0
    V(∞) = 12V(3Ω/9Ω)
    V(t) = V(∞) + [V(0) - V(∞)]e^(t/\tau)
    = 4 + [10 - 4]e^(t/18)
    = 4 + 6e^(t/18)

    this is were i started getting confused.

    so I(t) = V(0)/Rth + C (dv/dt) right??

    = (10/9) + 2(d/dt (4 + 6e^(t/18)))
    =10/9 + 2((6)(1/18)e^(t/18))
    =10/9 + (2/3)e^(t/18)

    dose that look right?

    looking at an example, it looks like its on the right path... but I dont understand the rule that was used in the differentiation in the example.

    isn't
    d/dx ke^x = k((d/dx)x)e^x ?

    thanks a bunch guys :)
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Rth would not be 9Ω.

    From the point of view of the capacitor it "sees" an equivalent resistance of 3Ω in parallel with 6Ω. Consider the DC source as a short for this purpose.

    The other way to view this is to reduce the 12V source and the two resistors to a Thevenin equivalent at the connections to the capacitor.
     
    Jess_88 likes this.
  3. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    ah ok, I see. Thanks :)

    so, with Rth = 2ohms used in my calculation would my procedure to answering the question be ok?

    I feel like I could be completely miss interpreting the correct method in applying the formulas.
     
  4. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    I just noticed that I didn't include two minor things for the original problem statement.
    1) the circuit is at position "a" for a long time before switching to "b"
    2)calculate i(t) for ALL t>0

    this shouldn't change much... but just incase.

    So I have redone the question.
    hope its right this time everything is ok.

    V(0-) = 30(3/(6+3)) = V(0) = V(0+)
    = 10V

    removing the dc source to determine the Req for the time constant ( time const i'm denoting by s, because i cant fined the math symbols in quick post).

    Req = 3//6 = 2

    s = RC = 4

    V(inf) = 12(3/9) = 4

    V(t) = V(inf) + [V(0) - V(inf)]e^(-t/s)
    = 4 + [10 - 4]e^(-t/4)

    i(t) = v(inf)/R +C(dv/dt)
    =4/3 + 2(6 x (-1/4)e^(-t/4))
    =4/3 +-3e^(-t/4)

    which looks good to me, but in a similar example there appears to be another differential equation...

    did I miss something out?
    did I differentiate correctly?

    thanks guys :)
     
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