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RC circuits question

  1. Nov 12, 2011 #1
    1. The problem statement, all variables and given/known data
    S1 switch - always close.
    S2 switch is closed, until the capacitor voltage reaches 2V.
    given: R,C,V values.
    V0 = 0V.
    Need to find Vout while S1+S2 are closed and until S2 opens.


    2. Relevant equations
    Vc(t) = (Vi - Vf)*(exp -t/TAU) + Vf


    3. The attempt at a solution
    I have found the Vc(t) equation (until S2 opens) and got:
    Vc(t) = 2/3V1*(1-exp(-3t/5RC))
    but now i am alittle bit confused.
    can't i tell that by KVL for the large loop (throught S2) each R has V1/3 voltage?
    or i can't do it because we are in the charging stage?
    thanks.
     

    Attached Files:

  2. jcsd
  3. Nov 12, 2011 #2

    rude man

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    If S1 is always closed, why show it? It's just a wire ...
     
  4. Nov 12, 2011 #3
    because in the next question it opens...
     
  5. Nov 12, 2011 #4

    rude man

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    Oh, OK.

    Where is V0 and what is V1?
     
  6. Nov 12, 2011 #5
    V0 is the capacitor initial voltage.
    V1 is the source voltage
     
  7. Nov 12, 2011 #6

    rude man

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    Vout = V1(1/2.5 - 1/3)exp(-t/T) + V1/3

    T = {R + R||2R}C.

    Initially you have C charged to V1 so the Thevenin equivalent is V1 connected to the rest of the circuit by R/2. So Vout(0+) = V1R/(2.5R). At t = ∞ you have C out of the picture so Vout(∞) = V1/3.

    You know the output must take the form Vout = a*exp(-t/T) + b. So you fit a and b:
    Vout()+) = a + b, V(∞) = b.

    Then T is by inspection R + R||2R since that is the total impedance C sees going to hard voltages (impedance = 0).
     
  8. Nov 12, 2011 #7

    rude man

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    If V0 is initial capacitor voltage, it is not zero. It is V1.

    I assume V1 is constant, -∞ < t < +∞
    S1 closed.
    S2 open, t < 0 and closed, t > 0.

    If this is incorrect you need to define the problem better.

    See you in about 8 hrs.
     
  9. Nov 12, 2011 #8

    NascentOxygen

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    Series resistors [any number of] can be considered to form a simple proportional voltage divider ONLY if all carry the same current. Clearly, that is true in your case ONLY if zero current is going into the capacitor branch. This condition (of zero capacitor current) becomes true only once the capacitor has reached full charge. So that potential divider sets the voltage the capacitor eventually could attain, but until then, one of the resistors carries more current than do the other two.
     
    Last edited: Nov 12, 2011
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