By Kirchhoffs first law we have i 1 i 2 i 3 By Kirchhoffs second law, on each loop we have E(t) Li 1 R 1 i 2 and E(t) Li 1 R 2 i 3 q C so that q CR 1 i 2 CR 2 i 3 Then i 3 q CR 1 i 2 CR 2 i 3 so that the system is Li 2 Li 3 R 1 i 2 E(t) R 1 i 2 R 2 i 3 1 C i 3 0 ...

By Kirchhoffs first law we have i 1 = i 2 + i 3 . By Kirchhoffs

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By Kirchhoff’s first law we have i₁ = i₂ + i₃. By Kirchhoff’s second law, on each loop we have E(t) = Li′₁ + R₁i₂ and E(t) = Li′₁ + R₂i₃ + q/C so that q = CR₁i₂ − CR₂i₃. Then

i₃ = q′ = CR₁i′₂ − CR₂i₃ so that the system is

Li′₂ + Li′₃ + R₁i₂ = E(t)

−R₁i′₂ + R₂i′₃ + 1/C i₃ = 0.

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A First Course in Differential Equations with Modeling Applications

ISBN: 978-1305965720

11th edition

Authors: Dennis G. Zill

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Question Posted: Jul 14, 2018 12:38 PM

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By Kirchhoffs first law we have i 1 = i 2 + i 3 . By Kirchhoffs

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A First Course in Differential Equations with Modeling Applications

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