{"id":3087,"date":"2014-04-22T22:37:10","date_gmt":"2014-04-23T02:37:10","guid":{"rendered":"http:\/\/pages.vassar.edu\/magnes\/?p=3087"},"modified":"2014-04-22T22:53:09","modified_gmt":"2014-04-23T02:53:09","slug":"preliminary-results-rlc-circuits","status":"publish","type":"post","link":"https:\/\/pages.vassar.edu\/magnes\/2014\/04\/22\/preliminary-results-rlc-circuits\/","title":{"rendered":"Preliminary Results- RLC Circuits"},"content":{"rendered":"

Overview<\/strong><\/p>\n

In the following blog post, I discuss my progress with my project so far. I start out with a simple RLC circuit, go on to discuss the components, and derive a solution (using Mathematica) for a circuit with parameters that I have come up with. I then plot current as a function of time and discuss long-term structure in the output of the circuit, its relevance to my project, and my forward plan.<\/p>\n

Progress<\/strong><\/p>\n

In my plan I discussed a simple RLC circuit that consists of a capacitor, voltage source, resistor, and an inductor.\u00a0To easily visualize this, I have constructed a basic circuit diagram (Figure 1).<\/p>\n

\"circuitwithnumbers1\"<\/a>(Figure 1)<\/span><\/p>\n

The differential equation that governs this RLC circuit is given by<\/p>\n

\"LI''(t)+RI'(t)+\frac{1}{C}I(t)= E_{0} \omega cos(\omega t)\"\u00a0<\/span><\/span>[1]<\/span><\/p>\n

where L is the inductance, R is the resistance, C is the capacitance, I is the current, \"\omega\"\u00a0is the\u00a0resonant\u00a0angular frequency, \" E_{0}\" is\u00a0an initial value dependent on the voltage source,\u00a0and t is the time. The inductance\u00a0(L), resistance\u00a0(R),\u00a0and\u00a0capacitance\u00a0(C)are all determined by their respective circuit components,\u00a0which are easily changed out\u00a0for different ones.\u00a0\"\omega\" depends\u00a0on the inductance and capacitance, and its value is given by<\/p>\n

\"\omega =\frac{1}{\sqrt{LC}}\" [2]<\/span><\/p>\n

What we have left in terms of variables are current (I) and time (t). For our purposes, time is the independent variable, which leaves the current in the circuit at any time to be the dependent variable. Now that we have established this, we can solve our differential equation [1] for current as a function of time (\"I(t)\").<\/p>\n

In order to solve [1], I will use the equation solving powers of Mathematica. In particular, the DSolve function is used.\u00a0\u00a0When plugged into Mathematica, the input line looks like:<\/span><\/p>\n

\"\text{DSolve}\left[\frac{\text{Current}(t)}{\text{Capacitance}}+\text{Inductance} \text{Current}''(t)+\text{Resistance} \text{Current}'(t)=E_{0} \omega \cos (\omega t),\text{Current}(t),t\right]\" [3]<\/p>\n

We must set values for resistance, inductance, capacitance,\u00a0and\u00a0\" E_{0}\". For this run through of the solution I chose the following arbitrary values:<\/p>\n