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RC Circuit - parallel plate

  1. Sep 30, 2015 #1
    If, by some means, the distance between the parallel plates of the capacitor is reduced to half of what it was before, would the time constant change? If yes, then what would be the new time constant?
  2. jcsd
  3. Sep 30, 2015 #2
    IS the time constant .00021? and 3.16?

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  4. Sep 30, 2015 #3
    I am not sure a capacitor alone has a time constant... It is in combination with a resistor that then you have a time content tau=RC... Should the resistance remain constant and the distant between plates reduced by half, then the time constant doubles... See the definition for capacitance C
  5. Sep 30, 2015 #4
    .00021? More like .00012 looks to me...and it is the same for both charge and discharge
  6. Sep 30, 2015 #5
    How did you get .00012? though?
  7. Oct 1, 2015 #6
    Look up the definition for time constant... It is the time it takes the system response to go from "zero" to 63.2% of its final, asymptotic value.
    So, you go to the beginning of the signal in the picture, then you follow it for a Delta of .316 V (0.5*63.2/100) on the y-axis and then you look up what the corresponding Delta of time was on the x-axis
  8. Oct 1, 2015 #7


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    Look up the equation for the capacitance of a parallel plate capacitor. Then the equation for an RC time constant.
  9. Oct 1, 2015 #8
    I'm not sure what I'm doing wrong though, because with that I'm getting a completely different answer. I realize its 63% until tmax. and the settings would be .5V. But how are you getting the other parts?
  10. Oct 1, 2015 #9
    Focusing on the charging (left) half of the signal in the attached picture...

    You can see that from the point where it starts to the point that it achieves an asymptotic value is about a delta of 0.5V, right? You know that it is a delta of 0.5V because the beginning of the signal is on one grid line and the "end" (top) is 5 gridlines higher up...and each gridline is equivalent to 0.1 V as per the legend.

    Now, since a time constant is the time that passes from "zero" to the time the signal achieves 63.2% of its asymptotic value, we take the entire zero-to-max delta of 0.5V and multiply that by 63.2%....yielding 0.316V.

    Then, we go back to the beginning of the signal - put your left index there - then, with your right index, you follow along the signal until the voltage delta between your left and right index is 0.316 V (vertical distance)...Stop!..Now, using the vertical gridlines, determina what is the horizontal distance between your figners and how much time it is as per the legend ( 0.001 seconds per gridline).
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