Laplace transforms: Simple Electrical Network: Difference between revisions
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<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 | <math>\Rightarrow\frac{1}{2}sI_1(s)+60I_2(s)=\frac{50}{s}</math> | ||
<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
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
Applying the Laplace transform to each equation gives
Solving for gives