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Rms voltage for capacitor charge/discharge

  1. Nov 4, 2011 #1
    1. The problem statement, all variables and given/known data
    Hello,
    I was wondering if anyone could confirm my work for the following graph:
    Untitled-6.jpg
    I'm supposed to find the rms for 1 complete cycle (0, 10ms)

    The time constant is 1ms.
    Charging phase, v=1-e^(-t/1ms)=1-e^(-1000t)
    discharging phase, v=e^(-1000t)

    3. The attempt at a solution

    First, I find the rms contribution for the charging stage.
    [tex]\sqrt{\frac{1}{10^{-3}}\int ^{5(10)^{-3}}_{0}{(1-e^{-1000t})^2dt}}=\sqrt{\frac{1}{10^{-3}}(t-\frac{e^{-2000t}}{2000}+\frac{e^{-1000t}}{500}|^{5(10)^{-3}}_{0})}=0.59274 V[/tex]
    Next, I found the contribution of the discharge phase:
    [tex]\sqrt{\frac{1}{10^{-3}}\int ^{5(10)^{-3}}_{0}{(e^{-1000t})^2dt}}=\sqrt{\frac{1}{10^{-3}}(-\frac{e^{-2000t}}{2000}|^{5(10)^{-3}}_{0})}=0.707091 V[/tex]
    and the effective rms for the entire cycle is
    [tex]v_{rms}=\sqrt{(0.59274 V)^2+(0.707091 V)^2}=0.923 V[/tex]

    Or do I have to multiply each of the functions underneatht he the sqrt by a factor of 5(10)^-3 ?
     
    Last edited: Nov 4, 2011
  2. jcsd
  3. Nov 4, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    Add the means (over the cycle length) of the voltage-squared contributions of each section of the curve prior to taking the square root. So:
    [tex] A = \frac{1}{10^{-3}}\int ^{5(10)^{-3}}_{0}{(1-e^{-1000t})^2dt} [/tex]
    [tex] B = \frac{1}{10^{-3}}\int ^{5(10)^{-3}}_{0}{(e^{-1000t})^2dt} [/tex]
    [tex] v_{rms} = \sqrt{A + B} [/tex]
     
  4. Nov 4, 2011 #3
    Interesting. That seems a lot easier. Thanks!
     
  5. Nov 4, 2011 #4
    When I do as you stated, I get an rms value of 2V. Steady state voltage in this particular instant is 1V.
     
  6. Nov 4, 2011 #5

    gneill

    User Avatar

    Staff: Mentor

    Oops. The period of the waveform is 10ms or 10-2s, not 10-3s. So the constant multiplying the integrals should be 1/10-2 = 100.

    I think you'll find that the RMS value should be 0.634V.
     
  7. Nov 4, 2011 #6
    I probably should have caught that. Thanks!
     
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