# Solving a Second Order Circuit for Capacitor Voltage

1. Oct 10, 2014

### Vishera

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

2. Relevant equations

Here is the technique I am using:

3. The attempt at a solution

I understand how to solve the problem using the technique provided by the solution but I was wondering where I messed up in the technique that I used. I prefer the second technique because there is less memorization.

2. Oct 10, 2014

### Staff: Mentor

Did you have a question? The provided solution looks pretty thorough except that they didn't explicitly state the steady-state solution.

3. Oct 10, 2014

### Vishera

The problem is that the provided solution doesn't derive the differential equations from scratch. It uses equations that we are supposed to memorize. I would prefer to just solve the problem from scratch instead of plugging in and chugging. I am trying to derive the differential equations from scratch and I did so but my final solution is incorrect.

4. Oct 10, 2014

### Staff: Mentor

When I do nodal analysis I see:
$$\frac{v}{R} + \frac{1}{L} \int v\;dt + C \frac{dv}{dt} = 0$$
Clear the integral by differentiating the whole thing:
$$\frac{1}{R} \frac{dv}{dt} + \frac{1}{L} v + C \frac{d^2 v}{dt^2} = 0$$
$$\frac{d^2 v}{dt^2} + \frac{1}{R C} \frac{dv}{dt} + \frac{1}{L C} v = 0$$
Plugging in component values and changing notation:
RC = 1
LC = 1/4

thus:

v'' + v' + 4v = 0

So you have complex conjugate roots, not a double real root. This makes sense since the circuit is underdamped.

5. Oct 10, 2014

### Vishera

I just did it this way and it seems to be correct. Is there anything I did wrong with my method?

Actually, I think I figured out what I did wrong. I substituted C=0.25 instead of C=1 and I substituted L=1 instead of L=0.25. Opps.

Last edited: Oct 10, 2014