Exponential decay graph

Discussion in 'Homework Help' started by Asad ahmed1, Mar 8, 2016.

  1. Asad ahmed1

    Thread Starter Member

    Feb 10, 2016
    68
    0
    how to draw waveform when the switch is closed for lone time the capacitor is behaving like an open source so 10 volts at the node and when the switch moves to 2nd position v(0)=10e^-t/tau t=Rc = 10k x 100uf
    V(0)=10e^-t
    so i plotted the graph..is this alright because voltage and current decrease exponentially.
    And energy stored is w=1/2Cv^2(1-e^(-2t/tao))
    So putting t=0

    w=1/2(100uf)(10)^2(1-1)
    w=0
     
  2. RBR1317

    Active Member

    Nov 13, 2010
    232
    48
    Whenever there is a switching transient in a resistive circuit with a single capacitor or inductor, there will be a first-order transient response as shown in the attached diagram (voltage is shown but it would be the same for current). Any first-order transient response is characterized by the following:

    1) Is the transient an exponential rise or an exponential decay?

    2) The magnitude of the transient, V_T.

    3) The capacitor or inductor time constant.

    4) The zero level, i.e. some circuits may have a constant offset from 0 and the transient response must be offset by the same amount.

    If you can answer these four questions correctly, then you are guaranteed to have the correct first-order transient response.
     
  3. Asad ahmed1

    Thread Starter Member

    Feb 10, 2016
    68
    0
    When its at switch 2 its an exponential decay but if its connected to another source then my formula would be for exponential rise
    v(t)(1-e^(-t/tau)) ?
     
  4. anhnha

    Active Member

    Apr 19, 2012
    774
    48
    No, your initial formula is correct Vo =10e^-t.
    This is not correct. What you calculated here how much energy is dissipated not what stored.
    Energy stored in a capacitor:
    upload_2016-3-9_9-49-7.png
    Where V is the voltage across the capacitor.
     
    Last edited: Mar 8, 2016
  5. RBR1317

    Active Member

    Nov 13, 2010
    232
    48
    Correct. Switch position 2 causes an exponential decay transient. If switched back to position 1 the transient would be an exponential rise in the voltage across the capacitor. It gets more interesting if it is switched back to position 1 before the capacitor has been fully discharged.
     
  6. Asad ahmed1

    Thread Starter Member

    Feb 10, 2016
    68
    0
    Thanks interesting concepts,
     
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