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B Time constant and capacitors

  1. Jun 6, 2016 #1
    So the rate at which a capacitor charges and discharges is dependent on resistance in a circuit and the magnitude of capacitance of the capacitor? So the time constant is equal to RC. So using this equation where Q=Qoe-t/RC ,time constant is the time taken (when the capacitor is discharging) for charge on a capacitor (Q) to decrease to 37% of Qo ( i.e. charge on the capacitor when it is is fully charged). But I've been reading around and what I don't get is how the time constant is also equal to the time taken for the charge (Q) on a charging capacitor to increase by 63% of Qo . If this makes any sense, would be good if someone could maybe explain it mathematically aswell i.e. how 'e' is involved?
     
  2. jcsd
  3. Jun 6, 2016 #2

    cnh1995

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    Have you studied calculus?
     
  4. Jun 6, 2016 #3
    Um haven't studied it in much detail, but I could give it a go
     
  5. Jun 6, 2016 #4

    cnh1995

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    Well, KVL equation for an RC circuit is a differential equation. Solving that differential equation, you get exponential expressions for charge, current and voltage.
     
  6. Jun 6, 2016 #5
    Ok so... how does this relate to the 63% of Qo?
     
  7. Jun 6, 2016 #6

    cnh1995

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    Or in other words, time taken to lose 63% of Q0.
    You can see that the time taken to gain a charge of 0.63Qo (while charging) is equal to the time taken to lose the same charge of 0.63Q0(while discharging). It's obvious, isn't it?
     
  8. Jun 6, 2016 #7

    cnh1995

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    If you are asking where 63% comes from, it comes from the constant 'e-1' in the solution to the differential equation of the RC circuit.
     
  9. Jun 6, 2016 #8
    the time constant is the time taken to get to within 37% of the final value.
    In decay the final value is 0%!
    In growth the final value is 100%
     
  10. Jun 6, 2016 #9
    Ok that was pretty obvious now I look at it, thank you for explaining it otherwise I probably wouldn't have got there lol
     
  11. Jun 6, 2016 #10
    Yep this makes sense, thanks!
     
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