Question

For our physics course, my group is presenting RLC circuits and we built one as a model. I am attempting to calculate the math on RLC circuits; however, I am uncertain what the initial current conditions are. Here is how it is laid out:

We have a battery hooked up with a certain voltage at t=0. Once t>0, the switch is turned off. We are observing the current and its behavior at time t.

I have attached some of my work with some more information.

Answer 1

@Michele_Laino @IrishBoy123 @johnweldon1993 Would any of you guys be able to assist me with this one? X)

Also, for us,
R = 0.54 ohms
L = 0.00035 H
C = 0.0001 F

Answer 2

If my basic knowledge of circuits serves me well, then I believe that the initial current at t=0 should be 0, so: \(i(0)=0\)

but then what would \(i'(0)\) be?

Answer 3

I think that your circuit can be modeled by this equation:
\[\begin{gathered}
L\dot I + RI - \frac{Q}{C} = {E_0}\cos \left( {\omega t} \right), \hfill \\
\hfill \\
I = - \dot Q \hfill \\
\end{gathered} \]
so, after a first derivative with respect tot time, we get:
\[\ddot I + \frac{R}{L}\dot I + \frac{I}{{LC}} = - \frac{{\omega {E_0}}}{L}\sin \left( {\omega t} \right)\]
and we can assume the subsequent initial conditions:
\[I\left( 0 \right) = 0,\quad \dot I\left( 0 \right) = \frac{{{E_0}}}{L}\]

Answer 4

Our professor mentioned to us that our second order differential equation will equal 0 given that we are using a small battery and not observing it's driving force. Will the second initial condition still hold for
\(\large i'(0)=\frac{V_0}{L}\)?

Answer 5

if we have no external power sources, then I have to consider a circuit wherein the cpacitor is charged, and its electrical charge, is:
\[Q = C{V_0}\]
where \(V_0\) is the initial voltage across the capacitor (initial condition for voltage), so we can write these equations:
\[I = - \dot Q,\quad Q = CV,\quad V = L\dot I + RI\]
where, I repeat, \(V\) is the voltage across the capacitor. Using those equations, we can write this ODE:
\[\frac{{{d^2}V}}{{d{t^2}}} + \left( {\frac{R}{L}} \right)\frac{{dV}}{{dt}} + \left( {\frac{1}{{LC}}} \right)V = 0\]
so the initial conditions are:
\[V\left( 0 \right) = {V_0},\quad \frac{{dV}}{{dt}}\left( 0 \right) = 0\]

Answer 6

capacitor*

Answer 7

@Michele_Laino Hmm, I found that it can also be modeled as
\[\huge i''(t)+\frac{R}{L}i'(t)+\frac{1}{LC}i=0\] Our damping factor (\(\zeta\)) was less than one, so our equation turned out to be
\[\huge i(t)=B_1e^{-\alpha t} \cos(\omega t)+B_2e^{-\alpha t} \sin(\omega t)\]
What are the initial conditions in this case?

http://www.ee.nthu.edu.tw/~sdyang/Courses/Circuits/Ch08_Std.pdf Refer to slide 39

Answer 8

I think that one condition is:
\[i\left( 0 \right) = 0\]
since, at \(t=0\) we have a charged capacitor and no current inside the circuit