Homework Help: Non-homogenous Diff EQ, LRC circuit

1. Jun 18, 2012

ElijahRockers

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

Consider an LRC circuit. L = 3/5, R = 10, C = 1/30, E(t) = 300, Q(0)=0, I(0)=0
i) Find charge and current.
ii) Find maximum charge on the capacitor.

2. Relevant equations

LQ'' + RQ' +(1/C)Q = E

3. The attempt at a solution

(3/5)Q'' + 10Q' +30Q = 300

For the roots of characteristic equation I got:

$m = -\frac{25}{3} \pm \frac{5\sqrt{7}}{3}$

This leads to the solution to the homogenous part, and the particular solution is simply Q=10.

So, after applying initial conditions to find the constants, the full general solution came out to

$Q(t) = (5- \frac{25}{\sqrt{7}})e^{(-\frac{25}{3} + \frac{5\sqrt{7}}{3})t} + (\frac{25}{\sqrt{7}}-15)e^{(-\frac{25}{3}-\frac{5\sqrt{7}}{3})t} +10$

Differentiating to find the current gives me:

$I(t) = (-\frac{25}{3} + \frac{5\sqrt{7}}{3})(5- \frac{25}{\sqrt{7}})e^{(-\frac{25}{3} + \frac{5\sqrt{7}}{3})t} + (-\frac{25}{3}-\frac{5\sqrt{7}}{3})(\frac{25}{\sqrt{7}}-15)e^{(-\frac{25}{3}-\frac{5\sqrt{7}}{3})t}$

So that's part i) finished.

But I'm confused on part ii)....

It's asking for the maximum charge on the capacitor, and I assume charge is given by Q(t).
So to find the maximum charge, do I set I=0 and solve for t? Since it says I(0)=0, then wouldn't t=0, which means Q=0?

I'm pretty sure there is more to it then that, but I'm confused.

2. Jun 18, 2012

vela

Staff Emeritus
You should try plotting Q(t) to see what's going on.

The extrema of Q(t) may occur at points where I(t)=0. You don't know if they're maxima or minima without looking closer, though. The extrema can also occur at other places as well.

3. Jun 18, 2012

LCKurtz

I didn't check your numbers. But the capacitor is going to charge up and everything will come to a halt, won't it? You need to look what happens as $t\rightarrow\infty$.

4. Jun 18, 2012

ElijahRockers

According to my trusty graphing calculator, The graph increases sharply until it levels off at Q=10, at approximately t=3.

OH. Ok I think I get it... so the lim as t-> inf of Q = 10. duh. For some reason my mind just didn't want to register that extra +10 on the end of the equation, that has been giving me trouble the whole time.

Thank you.

So since I(0) = 0, then Q(t) has a minimum at t=0.

What about the maximum? It doesn't really have a formal max because it is an open interval in the positive direction, but it converges to Q=10... is that enough to say the max=10?

5. Jun 18, 2012

klondike

From the distinct real roots, you know that the 2nd order system is overdamped, so there is no over shoot in the step response. Hence, the max Q occurs at steady state while t->infy.

On the other hand, if you have a complex conjugate roots (underdamp), the max Q is the peak overshoot of the step response.

6. Jun 18, 2012

vela

Staff Emeritus
A mathematician would say it doesn't have a maximum. Everyone else would say Qmax=10.