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

AI Thread Summary
In an oscillating series RLC circuit, the time required for the maximum energy in the capacitor to drop to 20% of its initial value can be derived from the energy equation U(t) = (Q²/2C)e^(-Rt/L)cos²(ω't). The key step involves recognizing that at maximum energy, cos(ω't) equals 1, simplifying the equation to 0.2 = e^(-Rt/L). By solving this equation for time t in terms of resistance R and inductance L, one can find the desired time. The discussion emphasizes that the exponential decay factor approximates the energy drop, even if it doesn't occur at an exact local maximum. Understanding this relationship is crucial for solving the problem effectively.
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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.
 
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lulzury said:

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.
 
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} ##
 
lulzury said:
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.
 
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ehild said:
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?
 
lulzury said:
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.
 
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ehild said:
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.
 
lulzury said:
Thank you for this explanation and for your time in helping me understand this problem.
You are welcome :oldsmile:
 

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