Laplace transforms: Simple Electrical Network: Difference between revisions

From Class Wiki
Jump to navigation Jump to search
Line 20: Line 20:
<math>0.006(s\mathcal{L}{i_2}-i_2(0))+\mathcal{L}\left\{i_2\right\}-\mathcal{L}\left\{i_1\right\}=0</math>
<math>0.006(s\mathcal{L}{i_2}-i_2(0))+\mathcal{L}\left\{i_2\right\}-\mathcal{L}\left\{i_1\right\}=0</math>


<math>\Rightarrow\frac{1}{2}s\mathcal{L}\left\{i_1\right\}+60\mathcal{L}\left\{i_2\right\}=\frac{50}{s}</math>
<math>\Rightarrow\frac{1}{2}sI_1(s)+60I_2(s)=\frac{50}{s}</math>


<math>-200\mathcal{L}\left\{i_1\right\}+(s+200)\mathcal{L}\left\{i_2\right\}=0</math>
<math>-200I_1(s)+[s+200]I_2(s)=0</math>


Solving for <math>\mathcal{L}\left\{i_1\right\}</math> gives
Solving for <math>\mathcal{L}\left\{i_1\right\}</math> gives

Revision as of 23:00, 30 November 2009

Problem Statement

Using the formulas

E(t)=LdiRdt+RiC
RCdiRdt+iRiC=0

Solve the system when V0 = 50 V, L = 0.5 h, R = 60 Ω, C = 10-4 f, and the currents are initially zero.

Solution

Solve the system when V0 = 50 V, L = 0.5 h, R = 60 Ω, C = 10^{-4} f, and the currents are initially zero. Substituting numbers into the equations, we have

0.5di1dt+60i2=50

60(104)di2dt+i2i1=0

Applying the Laplace transform to each equation gives

12(s{i1}i1(0))+60{i2}=50

0.006(si2i2(0))+{i2}{i1}=0

12sI1(s)+60I2(s)=50s

200I1(s)+[s+200]I2(s)=0

Solving for {i1} gives