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Rms voltage for capacitor charge/discharge

  • Thread starter dimpledur
  • Start date
  • #1
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Homework Statement


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)

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:

Answers and Replies

  • #2
gneill
Mentor
20,792
2,770
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]
 
  • #3
194
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Interesting. That seems a lot easier. Thanks!
 
  • #4
194
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When I do as you stated, I get an rms value of 2V. Steady state voltage in this particular instant is 1V.
 
  • #5
gneill
Mentor
20,792
2,770
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.
 
  • #6
194
0
I probably should have caught that. Thanks!
 

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