# Charge on Capacitor in RLC Circuit

Staff Emeritus

## Homework Statement

At t = 0, the charge stored on the capacitor plates is maximum in an oscillating series RLC circuit. At what time will the maximum possible energy that can be stored in the capacitor fall to one-eighth of its initial value if R = 7.20 Ω and L = 21.0 H?

The differential equation for an RLC circuit is Ld2q/dt2 + Rdq/dt + q/C = 0 and the solution to this equation is q = qmaxe−Rt/2L cos ωdt.
Assume that the damping is very weak (that is, assume the resistance R << sqrt(4L/C), so that the amplitude of the charge does not change by much during one oscillation).

Given Above

## The Attempt at a Solution

I've been trying to follow a "tutorial" on my online homework program, and after about an hour I finally managed to get most of the way through it. However, I'm now stuck on one particular part.

The tutorial wants me to determine a bunch of expressions in terms of different variables without putting any numbers in. The only I'm stuck on is finding an expression for the time.

Starting from the equations for potential energy, I eventually end up determining that e-Rt/2L = 1/√8. Solving for t by taking the natural log of both sides and then isolating t is supposed to give me ln(8)L/R. At least that's what the program tells me. But, where did the 2 in the denominator of the exponent go?! That's pretty much it. Once I figure that out I think I'll be able to solve the problem completely.

Thanks.

phyzguy
Remember that log(sqrt(x)) = log(x^(1/2)) = 1/2 log(x). So when you take the log of both sides, you get:
$$\frac{-Rt}{2L} = \frac{-log(8)}{2}$$
$$t = \frac{log(8) L}{R}$$

Staff Emeritus
I'm sorry, you've lost me. Where did the 2 on the right side, under the -log(8) come from?

Staff Emeritus
Hold on. ln(1/81/2) = ln(1)-ln(81/2) = 0-ln(8)/2
That look right?

phyzguy