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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 q= SO -R₁ L L E(t) = Li,' + di₂ dt E di₂ dt diz dt Ai₁ +L + L Then 13 = q = CR₁1₂' - CR₂13' di -R₂ + 00000 L Using Kirchhoff's first law we have i₁ = 12 + 13. By Kirchhoff's second law, on each loop we have E(t) = Li₁' + dt di₂ diz -R₁ + R₂ + dt dt di 3 dt di 3 dt + R₁/₂ = E(t) so that in terms of C, R₁, R₂, 1₂ and i3 we have the following. 11/13 + 1/1/3= = 0. iz R₂ 12 = 0. R₁ + R₁₂ = E(t) + C C

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 q= SO -R₁ L L E(t) = Li,' + di₂ dt E di₂ dt diz dt Ai₁ +L + L Then 13 = q = CR₁1₂' - CR₂13' di -R₂ + 00000 L Using Kirchhoff's first law we have i₁ = 12 + 13. By Kirchhoff's second law, on each loop we have E(t) = Li₁' + dt di₂ diz -R₁ + R₂ + dt dt di 3 dt di 3 dt + R₁/₂ = E(t) so that in terms of C, R₁, R₂, 1₂ and i3 we have the following. 11/13 + 1/1/3= = 0. iz R₂ 12 = 0. R₁ + R₁₂ = E(t) + C C

Introductory Circuit Analysis (13th Edition)
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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Show that a system of differential equations that describes the currents i₂(t) and i3(t) in the electrical network shown in the figure below is q= SO L L -R₁ E(t) = Li₁' + -R₁ dt E di₂ dt Ai, di ₂ dt + R₂ di₂ dt Then 13 = q' = CR₁₂ - CR₂/3' di₂ di 3 +L ooooo L Using Kirchhoff's first law we have i₁ = 12 +13. By Kirchhoff's second law, on each loop we have E(t) = Li₁' + +2 di 3 + L + R₁1₂ + R₂ so that in terms of C, R₁, R₂, 12 and i3 we have the following. dt di 3 dt + dt dig +13 dt + R₁/2 1 1₂ 13 R₂ R₁ = 0. = E(t) = E(t) = 0. + C
Transcribed Image Text:Show that a system of differential equations that describes the currents i₂(t) and i3(t) in the electrical network shown in the figure below is q= SO L L -R₁ E(t) = Li₁' + -R₁ dt E di₂ dt Ai, di ₂ dt + R₂ di₂ dt Then 13 = q' = CR₁₂ - CR₂/3' di₂ di 3 +L ooooo L Using Kirchhoff's first law we have i₁ = 12 +13. By Kirchhoff's second law, on each loop we have E(t) = Li₁' + +2 di 3 + L + R₁1₂ + R₂ so that in terms of C, R₁, R₂, 12 and i3 we have the following. dt di 3 dt + dt dig +13 dt + R₁/2 1 1₂ 13 R₂ R₁ = 0. = E(t) = E(t) = 0. + C
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