# Homework Help: RLC Circuit time for Energy to drop to 20% of initial value

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1. Nov 19, 2015

### lulzury

1. The problem statement, all variables and given/known data
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

2. Relevant equations
$U(t) = \frac{Q^2}{2C}e^{\frac{-Rt}{L}}\cos^2(\omega't+\phi)$
$U_0 = \frac{Q^2}{2C}$
$\omega' = \sqrt(\frac{1}{LC} - \frac{R}{2L}^2)$
3. 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.

2. Nov 19, 2015

### ehild

Read the text carefully
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. Nov 20, 2015

### lulzury

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. Nov 20, 2015

### ehild

Then you have the equation $0.2 = e^{\frac{-Rt}{L}}$ . Solve for t in terms of R and L.

Last edited: Nov 20, 2015
5. Nov 20, 2015

### lulzury

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. Nov 20, 2015

### ehild

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. Nov 20, 2015

### lulzury

Thank you for this explanation and for your time in helping me understand this problem.

8. Nov 20, 2015

### ehild

You are welcome