Laplace transforms: Simple Electrical Network: Difference between revisions

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==Solution==
==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

<math>0.5\frac{di_1}{dt}+60i_2=0</math>

<math>60(10^{-4})\frac{di_2}{dt}+i_2-i_1=0</math>

Applying the Laplace transform to each equation gives

<math>\frac{1}{2}(s^2\mathcal{L}{I_1}(s)-i_1(0))+60\mathcal{L}\left\{i_2\right\}</math>

<math></math>

Revision as of 17:52, 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