- #1
ElijahRockers
Gold Member
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- 10
Homework Statement
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.
Homework Equations
LQ'' + RQ' +(1/C)Q = E
The Attempt at a Solution
(3/5)Q'' + 10Q' +30Q = 300
For the roots of characteristic equation I got:
[itex]m = -\frac{25}{3} \pm \frac{5\sqrt{7}}{3}[/itex]
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
[itex] 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[/itex]
Differentiating to find the current gives me:
[itex] 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}[/itex]
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.