Basic Transient Question

Discussion in 'Homework Help' started by Spoon, May 18, 2010.

  1. Spoon

    Thread Starter Member

    May 18, 2009
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    Okay this is a 'basic' transient question but I'm not sure what to do for 8.3 and 8.4

    QUESTION 8 [4 MARKS]
    A 1F capacitor is connected in series with a 1Ω resistor across a 10V DC supply. A switch is also connected in series with another 1Ω resistor, which is connected in parallel with the capacitor.

    Initially, the switch is open, however at t=0 the switch is closed.
    8.1. Find the initial steady state voltage across the capacitor before the switch is closed. [1]

    I got this to be 10V ???

    8.2. Find the final steady state voltage across the capacitor once the switch is closed. [1]

    I got 5V ???

    8.3. How much time would have to pass once the switch is closed, for the voltage to reach 136.8% of it’s
    final value. [1]

    How would I go about doing this and 8.4?


    8.4. How much energy is retained in the capacitor at this time? [1]
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    63
    You are right about 8.1 and 8.2.

    For 8.3, fins the time it needs to drop from the initial 10V to 6.84V (136.8% of 5V) after the switch closes.

    For 8.4, use E=0.5*C*V^2
     
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  3. Spoon

    Thread Starter Member

    May 18, 2009
    12
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    8.4 i guess is just (1/2)CV^2

    I thought of doing 8.3 as 5(1.368) = 10(1 - e^(-T/RC)
    RC =1
    T = 1.15 seconds but this doesn't seem to be correct somehow?
     
  4. Spoon

    Thread Starter Member

    May 18, 2009
    12
    0
    never refreshed to see your reply before i posted, is my way of doing 8.3 correct?...
     
    Last edited: May 18, 2010
  5. mik3

    Senior Member

    Feb 4, 2008
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    Because you have an initial charge on the capacitor you have to use this formula:

    Vc=Vfinal(1-exp(-t/RC)+Vinitial*exp(-t/RC)
     
  6. Spoon

    Thread Starter Member

    May 18, 2009
    12
    0
    I have just looked at the provided solution and it says 1.15 seconds, the way you provided seems to get me 0.999.... Am not sure which one is correct... Or maybe I have misunderstood your formula
     
  7. mik3

    Senior Member

    Feb 4, 2008
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    When the switch is closed the time constant is not 1 but 0.5 because the resistance seen by the capacitor is the parallel combination of the two 1R resistors.
     
  8. Spoon

    Thread Starter Member

    May 18, 2009
    12
    0
    according to a computer simulation, the time taken is 0,5 seconds. I reworked RC to be 0.5, and I don't get the 0.5 seconds as the answer though??
     
  9. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    I get 0.5 sec.
     
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Mik3 is quite correct.

    Sometimes it helps to draw pictures and write things down together ....
     
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