# Uncharging RLC circuit, critical & overdamped case

1. Jun 5, 2006

### wimvd

When a non-driven RLC circuit discharges you have several possibilities: underdamped, critical & overdamped. The general solution of the differential equation is Q(t) = A*exp(L1*t)+B*exp(L2*t) with L1 and L2 roots of the characteristic polynom being (-R+sqrt(R^2-4*L/C))/(2*L) and (-R-sqrt(R^2-4*L/C))/(2*L)
In the critical case R^2-4*L/C is zero, in the overdamped case it is larger than zero.

Now in class we were told that in the critical case the charge always approaches faster to zero than in any overdamped case. I tried to figure out why but I don't seem to find the answer. I thought it had to do with the fact that in the critical case the characteristic polynom only has one root and in the overdamped it has two but I can't seem to find any mathematical proof of the charge going faster to zero.

If anyone has an idea as to how to prove this, it would be most welcome. :)

Thanks.

2. Jun 5, 2006

### vsage

You're sort of right, but I'm not sure you understand why you're right. Consider the critically damped case. There is a double characteristic root of -R/(2*L), and so the circuit discharges toward zero at some rate.

Now consider the overdamped case (you can simply adjust the capacitor to a larger value to obtain this while keeping R and L constant)... you then have two roots: one of which is more negative than -R/(2*L) and one that is less negative than -R/(2*L). The charge over time will then be the sum of two decaying exponentials, one that decays more fast than e^(-R/(2*L)) and one that decays less fast. The one that decays more fast will die off quicker, but after that point the charge in the circuit will be governed by the exponential which decayed less quickly, therefore the overdamped case will always decay slower than the critically damped case (assuming you are just adjusting capacitances).

Edit: Minor correction. You don't have the restriction of leaving R and L constant to prove my point, you just need to have -R/2L as a constant to maintain the ratio.

Last edited by a moderator: Jun 5, 2006
3. Jun 5, 2006

### Staff: Mentor

And keep in mind that in the underdamped case, the charge on the cap reaches zero faster than in the other two cases. But then it keeps on going negative, turns around, crosses zero again, etc. as it damps out to a final DC zero coulombs. The critically damped case gets the charge on the cap (and hence the voltage) to zero the quickest for the case where it does not swing negative.