RC time constant calculation

Discussion in 'Homework Help' started by ihaveaquestion, May 8, 2009.

  1. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
    314
    0
    I just wanted confirmation on this one:

    http://img8.imageshack.us/img8/8641/scan0058vzn.jpg

    Supposed to just calculate the time constant RC

    I think it goes something like this:

    After a long period of time the current source is going to act like an open-circuit and thus eliminate the resistor that is in series with it.

    After a long period of time the voltage source acts like a short circuit and also effectively gets rid of the resistor in parallel with it.

    The effective circuit after a long period of time and thus my RC calculation is simply based off the capacitor in series with only the 1k ohm resistor to the upper right of the capacitor (since the other two are gone).

    RC = 1000*(1*10^-6)

    correct?
     
  2. Darren Holdstock

    Active Member

    Feb 10, 2009
    262
    11
    Curious circuit; I wonder what the context is? You're correct - assuming perfect current and voltage sources, then the resistor is series with i(t) and the resistor in parallel with v(t) are both redundant. But I'm not sure that this is the full answer required. What is the exact question pertaining to this circuit?
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Given the capacitor voltage never reaches a steady state value, the concept of time constant probably has no relevance in this example.
     
  4. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
    314
    0
    Teacher today just said 'when you're calculating time constant you get rid of the sources' haha.... I wasn't even going to bother asking her why. For the test, I'm just going to do that and figure it out later.
     
  5. steinar96

    Active Member

    Apr 18, 2009
    239
    4
    The time constant is always the effective resistance (the resistance the capacitor sees in the circuit) times it's capacitance as you propably now.

    So if you remove the sources you can calculate the resistance the capacitor sees.
     
  6. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    Assuming the sources are step functions at time T=0, it reaches steady state with a time constant of 1ms, as the OP correctly calculated.
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    As Darren implied, to find the time constant, replace the current source with an open circuit and the voltage source with a short circuit.
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Yep - obviously I had brain meltdown.
     
  9. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    That sounds worse than a brain fart, which I frequently emit.:D
     
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Nice one Ron!:D

    Is that the latest source greenhouse gas?:rolleyes:
     
  11. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    Yes you are correct. Replace the current source with an open circuit, the voltage source with a short circuit and calculate the resistance 'seen' by the capacitor.
     
Loading...