Show that a system of differential equations that describes the currents i(t) and i3(t) in the electrical network shown in the figure below isdi 3dt9 =soE(t) = Li₁' +LdizLdtdi 2-R₁ dtdizdtdiz-R1 dtAi₁E+ LThen13 = 9' = CR₁12 - CR₂13'+ R₂00000LUsing Kirchhoff's first law we have i₁ = 12 + 13. By Kirchhoff's second law, on each loop we haveE(t) = Li₁' ++L+ R₂so that in terms of C, R₁, R₂, 12 and 13 we have the following.di3dtdi 3dt+di 3dt+ R₁₂ = E(t)iz R₂1/13R₁+ R₁/2 = E(t)+= 0.= 0.C

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Show that a system of differential equations that describes the currents i(t) and i3(t) in the electrical network shown in the figure below is di 3 dt q= so diz L dt di 2 -R₁ dt E(t) = Li₁' + L Ai₁ E diz -R1 dt +L Then 13 = q' = CR₁12 - CR₂13' + R₂ 00000 L diz di3 dt dt Using Kirchhoff's first law we have i₁ = 12 + 13. By Kirchhoff's second law, on each loop we have E(t) = Li₁' + +L + R₂ so that in terms of C, R₁, R₂, 12 and 13 we have the following. di 3 dt + di3 dt + R₁₂ = E(t) iz R₂ 2/3 R₁ + R₁/2 = E(t) + = 0. = 0. IC

Show that a system of differential equations that describes the currents i(t) and i3(t) in the electrical network shown in the figure below is di 3 dt q= so diz L dt di 2 -R₁ dt E(t) = Li₁' + L Ai₁ E diz -R1 dt +L Then 13 = q' = CR₁12 - CR₂13' + R₂ 00000 L diz di3 dt dt Using Kirchhoff's first law we have i₁ = 12 + 13. By Kirchhoff's second law, on each loop we have E(t) = Li₁' + +L + R₂ so that in terms of C, R₁, R₂, 12 and 13 we have the following. di 3 dt + di3 dt + R₁₂ = E(t) iz R₂ 2/3 R₁ + R₁/2 = E(t) + = 0. = 0. IC

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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