15 0wwia24 010 Ωww19 N12 V25 010 VisO (10)(ic – ia) – (12)(ic – is) + (25)(ic – ia) = 0%3DO (10)(ic – ia) + (25)(i. – ia) – 12 = 0O (10)(ic) + (25)(ic – ia) + 12 = 0O (10)(ic – ia) – (12)(ic – is) + 25 = 0-

15 Ω 24 N ia 10 Ω ww 19N 12 V 25 N ia 10 V ic O (10)(i. – ia) – (12)(ic – is) + (25)(ic – ia) = 0 O (10)(ic – ia) + (25)(ic – ia) – 12 = 0 O (10)(ie) + (25)(ic – ia) + 12 = 0 O (10)(ic – ia) – (12)(ic – is) + 25 = 0 +. +, ww

15 Ω 24 N ia 10 Ω ww 19N 12 V 25 N ia 10 V ic O (10)(i. – ia) – (12)(ic – is) + (25)(ic – ia) = 0 O (10)(ic – ia) + (25)(ic – ia) – 12 = 0 O (10)(ie) + (25)(ic – ia) + 12 = 0 O (10)(ic – ia) – (12)(ic – is) + 25 = 0 +. +, ww

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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Question

15 0
ww
ia
24 0
10 Ω
ww
19 N
12 V
25 0
10 V
is
O (10)(ic – ia) – (12)(ic – is) + (25)(ic – ia) = 0
%3D
O (10)(ic – ia) + (25)(i. – ia) – 12 = 0
O (10)(ic) + (25)(ic – ia) + 12 = 0
O (10)(ic – ia) – (12)(ic – is) + 25 = 0
-

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subject : electronics engineering