# Differential Equation Setup for an RLC Circuit

I'm getting confused on how to setup the following differential equation problem:

You have a series circuit with a capacitor of $0.25*10^{-6}$ F, a resistor of $5*10^{3}$ ohms, and an inductor of 1H.

A 12 volt battery is connected and is closed at $t=0$. I need to find the equation for the charge of the capacitor at time $t$.

Based on the information given in the book I am using, I would think to setup the equation as follows:

$LQ''+RQ'+\frac1cQ=E(t)$

$L$, the inductance, would be $1$

$R$ is resistance and is $5*10^3$

Finally, capitance is $C=0.25*10^{-6}$

$Q(t)$ would represent the charge of the capacitor at time $t$, which is the solution to my problem.

The issue I am having is with the 12 volt battery that is connected. Normally, I would think to set $E(t)$ to 0 because $E(t)$ is the impressed voltage, but obviously with the battery, the impressed voltage would be 12 volts and not 0.

I'm guessing that my equation for $E(t)$ is wrong (and should be a function of time), but could someone point me in the right direction to complete the setup of the problem?

• Your equation for $E(t)$ is a function of time, because $Q$ is a function of time. – Trevor Wilson Feb 25 '13 at 23:58
• If you're doubt is about the physical analysis to get to the right equation, you may get better help in physics.SE than here. If you're question is about solving that equation, the this is the place, but it looks like the first one. – MyUserIsThis Feb 25 '13 at 23:59
• An alternative to physics.SE would be electronics.SE – Dilip Sarwate Feb 26 '13 at 0:07
• I'm not really interested in the physics of it, but rather the differential equation side of things-- I'm not an electrical engineering major ;) – Andrew M Feb 26 '13 at 0:25

Using KVL, we would have ($V_S$ is your $E(t)$ = Voltage Source = constant DC $12 V$ source):

$$\tag 1 V_R + V_L + V_C = V_S \Rightarrow iR + L\frac{di}{dt} + \frac{1}{C}\int idt = V_S$$

Differentiating $(1)$ wrt $t$, we get:

$$R \frac{di}{dt} + L\frac{d^{2}i}{dt^2} + \frac{1}{C} i = 0$$

That is:

$$\frac{d^{2}i}{dt^2} + \frac{R}{L} \frac{di}{dt} + \frac{1}{LC} i = 0$$

This a 2nd-order ODE with constant coefficients.

You would of course solve the homogeneous and non-homogeneous case.

Maybe these notes will also be helpful.

Make sense?

• Great answer, and a helpful link, to boot! +1 – Namaste Apr 29 '13 at 0:42
• @Amzoti Link broken. Do you still have the notes? Or can you maybe find it again? Thanks. – the swine May 4 '17 at 23:14
• @theswine the notes are here for all future reference – cadams Sep 18 '17 at 20:31