Matthew's Asgn: Difference between revisions

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<math> B = \dfrac{-R}{R^2+w^2L^2} \,\!</math>
<math> B = \dfrac{-R}{R^2+w^2L^2} \,\!</math>

<math> I(s)= \dfrac{a+jb}{s+jw} + \dfrac{a-jb}{s-jw} + \dfrac{B}{s+R/L} \,\!</math>

<math> I(s)= \dfrac{(a+jb)(s-jw)+(a-jb)(s+jw)}{s^2+w^2}+\dfrac{B}{s^2+w^2} \,\!</math>

<math> I(s)= \dfrac{2as}{s^2+w^2} + \dfrac{2bw}{s^2+w^2} + \dfrac{B}{s+R/L} \,\!</math>

<math> a = \dfrac{1/2R}{R^2+w^2L^2} \,\!</math>

<math> b = \dfrac{1/2wL}{R^2+w^2L^2} \,\!</math>

<math> i(t) = \mathcal{L}^{-1} \{I(s)\} \,\!</math>

<math> i(t) = 2a \cos wt + 2b\sin wt + BE^{-Rt/L} -ie^{-Rt/L} \,\!</math>

<math> i(t) = 2 \dfrac{R}{R^2+w^2L^2} \cos wt + 2 \dfrac{wL}{R^2+w^2L^2} \sin wt + \dfrac{-R}{R^2+w^2L^2} e^{-Rt/L} -ie^{-Rt/L} \,\!</math>

This Solves the Series RL circuit for the current throughout the circuit, given initial conditions or no initial conditions

Latest revision as of 18:40, 1 November 2010

I decided that I would attempt to perform a simple analysis of a series RL circuit, which could then be used to do a more complex analysis on a basic transformer. I have always had interest in electronics, and transformers are key to basic electronics.

I decided that i would do the analysis of a RL circuit with the variables instead of given values.


Given:

V(t)=

V(s)=

I(0)=i

RLcircuit.jpg

The Laplace transform for an inductor:

=

The Laplace transform for a resistor:

=

Therefore the Resulting Equation for the system after applying the Laplace Transform:

A series of algebraic manipulations follows to come up with I(s):



We can then use partial fraction manipulation to expand the expression:


After a lot of messy math and work:

This Solves the Series RL circuit for the current throughout the circuit, given initial conditions or no initial conditions