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Voltage on capacitor

  1. Oct 28, 2012 #1
    A 22 nF capacitor is initially charged to 10V. It is then discharged by connecting a 100k resistor across it. Approximately how long does it take for the voltage on the capacitor to fall by 10%?



    v = v₀e^(–t/τ)
    RC = τ = time constant



    Im not sure if this the right equation what do you think guys?
     
  2. jcsd
  3. Oct 28, 2012 #2

    gneill

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    Staff: Mentor

    The equation looks fine. How you make use of it is what's important :smile:
     
  4. Oct 28, 2012 #3
    τ = 22nF x 100k = 2200µs = 2.2 ms

    then
    v/v₀ = 0.9 = e^(–t/2.2m)
    –t/2.2m = ln 0.9 = –0.1054
    t = 0.0002318 = 232 µs <-- answer!

    is that alright?
     
    Last edited: Oct 28, 2012
  5. Oct 28, 2012 #4

    gneill

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    Staff: Mentor

    Yes, that's fine.
     
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