Laplace transforms: Simple Electrical Network: Difference between revisions
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==Solution== |
==Solution== |
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Solve the system when V0 = 50 V, L = 0.5 h, R = 60 Ω, C = 10^{-4} f, and the currents are initially zero. |
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Substituting numbers into the equations, we have |
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<math>0.5\frac{di_1}{dt}+60i_2=0</math> |
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<math>60(10^{-4})\frac{di_2}{dt}+i_2-i_1=0</math> |
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Applying the Laplace transform to each equation gives |
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<math>\frac{1}{2}(s^2\mathcal{L}{I_1}(s)-i_1(0))+60\mathcal{L}\left\{i_2\right\}</math> |
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<math></math> |
Revision as of 16:52, 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