Transients

Discussion in 'Homework Help' started by Steve1992, Nov 29, 2006.

  1. Steve1992

    Thread Starter Senior Member

    Apr 7, 2006
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    I saw this question on transients:

    0.1uF cap. in series with 10k resistor.
    Combination connected to 100V dc.

    Q. 'The initial rate of rise of the pd across the capacitor?'
    A. '100 kV/s'


    I assumed the initial voltage across cap. would be zero?
     
  2. Papabravo

    Expert

    Feb 24, 2006
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    It is. What do you think "rate of rise" refers to?
     
  3. Dave

    Retired Moderator

    Nov 17, 2003
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    Papabravo is correct. The initial steady-state voltage across the capacitor is 0V, however the voltage across the capacitor will increase at a rate of 100kV per second (no longer in steady-state conditions) so every second the voltage across the capacitor would in theory increase by 100kV. It would in fact reach steady state conditions of 100V across the capacitor very quickly; 1ms if I have done my maths right.

    Dave
     
  4. Papabravo

    Expert

    Feb 24, 2006
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    Also note that, as in all cases where the solution to a differential equation is an exponential function, the rate of rise decreases monotonically as the voltage approaches it's steady state value asymptotically.
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    The time constant is 1mS. The voltage on the cap will theoretically never reach 100V, because, as Papabravo says, it will approach that value asymptotically. In five time constants, it will charge to 99.326V, which is generally accepted as fully charged.
     
  6. Dave

    Retired Moderator

    Nov 17, 2003
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    You are correct about the RC time constant and the fact that the voltage across the capacitor approaches the 100V source voltage asymptotically. The general point I was making was that the rate of rise of the voltage across the capacitor is the intermediate transition from the initial steady-state 0V to a steady-state ~100V. This is a simplification of the reality but demonstrates the point raised in the OP.

    Thanks for the clarification.

    Dave
     
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