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RLC Circuit time for Energy to drop to 20% of initial value

  • #1
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Homework Statement


In an oscillating series RLC circuit, with resistance R and inductance L, find the time required for the maximum energy in the capacitor during an oscillation to fall to 1/5 its initial value. Assume q = Q at t = 0

Homework Equations


## U(t) = \frac{Q^2}{2C}e^{\frac{-Rt}{L}}\cos^2(\omega't+\phi) ##
## U_0 = \frac{Q^2}{2C} ##
## \omega' = \sqrt(\frac{1}{LC} - \frac{R}{2L}^2) ##

The Attempt at a Solution


## 0.5 \frac{Q^2}{2C} = \frac{Q^2}{2C}e^{\frac{-Rt}{L}}\cos^2(\omega't+\phi) ##
## 0.5 = e^{\frac{-Rt}{L}}\cos^2(\omega't+\phi) ##

Since q = Q @ t=0, I dropped the phase angle:
## 0.5 = e^{\frac{-Rt}{L}}\cos^2(\omega't) ##

I'm not even sure on how to begin solving this without getting that t out of the cosine squared. To do this I would either need to add some other ## \sin(t)^2 ## to get ## \cos(t)^2 + \sin(t)^2 = 1 ## on the resulting function or I would need to find some expression ## \cos(t) = \frac{adj}{hyp} ##

Any hints, tips? Thanks.
 

Answers and Replies

  • #2
ehild
Homework Helper
15,408
1,810

Homework Statement


In an oscillating series RLC circuit, with resistance R and inductance L, find the time required for the maximum energy in the capacitor during an oscillation to fall to 1/5 its initial value. Assume q = Q at t = 0

Homework Equations


## U(t) = \frac{Q^2}{2C}e^{\frac{-Rt}{L}}\cos^2(\omega't+\phi) ##
## U_0 = \frac{Q^2}{2C} ##
## \omega' = \sqrt(\frac{1}{LC} - \frac{R}{2L}^2) ##

The Attempt at a Solution


## 0.5 \frac{Q^2}{2C} = \frac{Q^2}{2C}e^{\frac{-Rt}{L}}\cos^2(\omega't+\phi) ##
## 0.5 = e^{\frac{-Rt}{L}}\cos^2(\omega't+\phi) ##



Since q = Q @ t=0, I dropped the phase angle:
## 0.5 = e^{\frac{-Rt}{L}}\cos^2(\omega't) ##

I'm not even sure on how to begin solving this without getting that t out of the cosine squared. To do this I would either need to add some other ## \sin(t)^2 ## to get ## \cos(t)^2 + \sin(t)^2 = 1 ## on the resulting function or I would need to find some expression ## \cos(t) = \frac{adj}{hyp} ##

Any hints, tips? Thanks.
Read the text carefully
find the time required for the maximum energy in the capacitor during an oscillation to fall to 1/5 its initial value
Assuming weak damping, what is cos(ω't) when the energy of the capacitor is maximum?
The energy should drop to 1/5 of its initial value. You worked with 1/2.
 
  • #3
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0
Sorry that was a mistake on my part. That should be:

## 0.2 = e^{\frac{-Rt}{L}}\cos^2(\omega't) ##
## cos(\omega t) = 1 ## when the energy of the capacitor is maximum which is why I dropped it on the equation for ## U_0 = \frac{Q^2}{2C} ##
 
  • #4
ehild
Homework Helper
15,408
1,810
Sorry that was a mistake on my part. That should be:

## 0.2 = e^{\frac{-Rt}{L}}\cos^2(\omega't) ##
## cos(\omega t) = 1 ## when the energy of the capacitor is maximum which is why I dropped it on the equation for ## U_0 = \frac{Q^2}{2C} ##
Then you have the equation ## 0.2 = e^{\frac{-Rt}{L}} ## . Solve for t in terms of R and L.
 
Last edited:
  • #5
8
0
Then you have the equation 0.2=

Then you have the equation ## 0.2 = e^{\frac{-Rt}{L}} ## . Solve for t in terms of R and L.
Thanks, I think I see. To clarify, do we throw the cosine portion away because @ ## 1/5 U_0 ## we'll be at a local peak for the decaying energy in the RLC circuit? How do we know this?
 
  • #6
ehild
Homework Helper
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Thanks, I think I see. To clarify, do we throw the cosine portion away because @ ## 1/5 U_0 ## we'll be at a local peak for the decaying energy in the RLC circuit? How do we know this?
Assuming the period of the cosine term is much shorter than the time constant L/R of the decay, the local maxima of the energy are where the cosine is maximum. You can find it if you take the derivative of the energy with respect time and make it equal to zero. You get a term with the factor R/L and the other with the factor ω'. If R/(Lω') is much less than 1, the local extrema are where sin(ω't)=0, that is cos(2(ω't)=1.
The exponential factor is envelope of the cos2 factor, so the time when the exponential is 1/5 does not need to be exactly at a local maximum, but it is within half period distance from it. For short period, it is a good approximation.
 
  • #7
8
0
Assuming the period of the cosine term is much shorter than the time constant L/R of the decay, the local maxima of the energy are where the cosine is maximum. You can find it if you take the derivative of the energy with respect time and make it equal to zero. You get a term with the factor R/L and the other with the factor ω'. If R/(Lω') is much less than 1, the local extrema are where sin(ω't)=0, that is cos(2(ω't)=1.
The exponential factor is envelope of the cos2 factor, so the time when the exponential is 1/5 does not need to be exactly at a local maximum, but it is within half period distance from it. For short period, it is a good approximation.
Thank you for this explanation and for your time in helping me understand this problem.
 
  • #8
ehild
Homework Helper
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Thank you for this explanation and for your time in helping me understand this problem.
You are welcome :oldsmile:
 

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