Q. No. 5Construct the equivalent system of second-order ODE for the problem given in Q. No. 4, in thelight of Kirchhoff Current Law, can be given as .1.I – 31, + 1813 + 21² – 121,1 – 141 = 0with the initial conditionsП(0) — 2, I;(0) — 4%3D2.I – 21, + 1717 + 21;² – 121,1 – 1411 = 0%3Dwith the initial conditions%3D3.K – 41, + 1617 + 21{² – 121, 1 – 141 = 0%3Dwith the initial conditions1 (0) = 2, I(0) = 4

KIRCHHOFF’S LAW KCL: The algebraic sum of all currents at a node is zero. KCL is…

Construct the equivalent system of second-order ODE for the problem given in Q. No. 4, in the light of Kirchhoff Current Law, can be given as . 1. I – 31, + 1817 + 21;² – 121,1 – 141, = 0 with the initial conditions I (0) = 2, I{(0) = 4

Construct the equivalent system of second-order ODE for the problem given in Q. No. 4, in the light of Kirchhoff Current Law, can be given as . 1. I – 31, + 1817 + 21;² – 121,1 – 141, = 0 with the initial conditions I (0) = 2, I{(0) = 4

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Question

Q. No. 5
Construct the equivalent system of second-order ODE for the problem given in Q. No. 4, in the
light of Kirchhoff Current Law, can be given as .
1.
I – 31, + 1813 + 21² – 121,1 – 141 = 0
with the initial conditions
П(0) — 2, I;(0) — 4
%3D
2.
I – 21, + 1717 + 21;² – 121,1 – 1411 = 0
%3D
with the initial conditions
%3D
3.
K – 41, + 1617 + 21{² – 121, 1 – 141 = 0
%3D
with the initial conditions
1 (0) = 2, I(0) = 4

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