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Сараcitor Inductor Resistor dvc = C dt dir v, = L dt VR = iRR 1/5F

Сараcitor Inductor Resistor dvc = C dt dir v, = L dt VR = iRR 1/5F

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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(Feedback and Control System) Instructions: In order to solve. The (1st picture) is the guide to solve the (2nd Picture) you need to follow the first picture enable to solve. 

Step 1: Current Loops

Step 2: Remove Integrals

Step 3: Simplify

Capacitor dvc ic = C= Inductor Resistor Vr = İRR di, v = L dt %3D dt 1/5F VE
Transcribed Image Text:Capacitor dvc ic = C= Inductor Resistor Vr = İRR di, v = L dt %3D dt 1/5F VE
Step! 工, eSid Step3 dt Step 3 1, UCt) = 301 + 3i - 3 V4) =(30+3)i, - 3ia 30-3D, + 50ig - -3 Di, + (30+50+1), =-3Di, +(80+rlia
Transcribed Image Text:Step! 工, eSid Step3 dt Step 3 1, UCt) = 301 + 3i - 3 V4) =(30+3)i, - 3ia 30-3D, + 50ig - -3 Di, + (30+50+1), =-3Di, +(80+rlia
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