Answered: in the solution he continues to say.…

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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in the solution he continues to say. kvl in L1 2+1-vr2=0 vr2=3v r2=v/1=3A and by kcl a2=4A please explain the solution and how it was done as i didnt' understand the process. please do not re-solve just explain the given solution. expand on it for it to be more detailed and explain each process step by step as to help me understand what happened and how i should deal with these type of questions in the future. please do not leave any details unexplained no matter how small.Thank you!

(2)
E₁ (E
PE1=-2W
in
R₁ R₂
J2
2.
PRI= V. I,
PRI= R. J. ²
2
E, (!)
zm
I=+1
=
70
J₁ = A
A 2
R2
↑
P= I.V
P= I₁.E
Az
I₁ = P
E,
V=R.. I₁
A 2
↑ A₂
here I₁=-1A
=₁=
IX
-2.
-2
€₁
-2
El
€₁=-20
E1= 20
Since in given
8170
but we
Convert
the direction of In

Given the circuit below with R₁ = R₂ = 10, compute the values for E₁ and
A₂ (with A₂ > E₁ > 0), by assuming that:
E₁
مند
the power delivered by the voltage source is PE₁ = -2 W.
the power delivered by the current source is PA2 = 12 W.
the power absorbed by the resistor R₁ is PR₁ = 1 W

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