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RC circuit

  1. May 28, 2012 #1
    Supposing we have an RC circuit (see figure)
    As we increase R, what will happen to the visualized signal of E and V(capacitor) and V(resistor)?


    And a general question about measuring the (Tao) [the one that equals to RC] BUT FROM THE OSCILLOSCOPE..

    They mentioned it is the time for which V(resistor) decreases to half its initial value (but I still don't get it)

    Thanks in advance
     
  2. jcsd
  3. May 28, 2012 #2
    Hi M.next.... I cannot see your diagram !!!!! (I can picture what you mean but....)
    One thing to note... RC is called the time constant (tau) but this is NOT the time taken for the voltage across R to 1/2.
    The time for it to 1/2 is 0.693RC (ln2 x RC) so measuring the time to become 1/2 from the oscilloscope (not difficult !!!) enables you to calculate RC
    Hope this helps..... your diagram would be useful !!!
     
  4. May 28, 2012 #3
    I donno why it is not uploading :/!!!
    Concerning the tau, please elaborate supposing you can imagine an oscilloscope signal..
     
  5. May 28, 2012 #4
    If you have a simple R in series with a C connected to a battery of voltage E....then when the switch is closed the current will be a max (= E/R) and will decrease exponentially. The voltage across R will also decrease exponentially from a value of E.
    When the voltage across R has decreased to E/2 then the time taken = 0.693RC.
    If you can display the V across R on a CRO you should be able to measure the time for V to become V/2.
    I am not certain what you mean by....'as we increase R....!!!
    Hope this helps.
    As an example: if you have 1 microfarad in series with 1 megohm then time constant = 1 second so it will take 0.693s for the voltage across R to become 1/2
     
  6. May 29, 2012 #5
    Thank you very much truesearch
     
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