..andNowZ₁ = x₁ + 38 = 1·2+38 = 5522₂ = x₂+4·1=09+4·1=552}Z₁ = (5) 2; Z₂ = (5) 2; Z= (50+ j40) 2210⁰ = 5774 210° volts =E₂Referring to Fig. 12.87, we have,1₁ Z₁ + (1₁+1₂)Z = E₁and1₂ *Z₁ + (1₁ + 1₂) ZE₂Subtracting eq. (ii) from eq. (i), we have,1₁ - 1₂..1₁-1₂ =Adding eqs. (i) and (), we have,=1₁ + 1₂-10,000√√312000=E₁-E₂Z₁-1242 + j1003j5(200+j250) A20⁰ = (6928+j0) volts1001₁ + 1₂On solvin eqs. (iii) and (iv), we have,=E₁ + E₂2 Z+Z₁12614+j103(5686+j1003) - (6928+ j0)j5= (200+j250) A(5686+j1003) + (6928+ j0)2 (50+ j40) + j512654 24.5°785131-2240-4°96-4 2-35-9° A = (78-1-j56-6) A(78-1-j56-6) A(5686+j1003) volts169 A ; I, = 164 A597...(1).... (iii)...(iv)Hello expert, I want you to explain howhe got the law inside the circle. Can youexplain step by step and a clear line, please?

Answered: Image /qna-images/answer/2b2d3e73-441e-4c68-881d-a82dc29fed32.jpg

Referring to Fig. 12.87, we have, 1₁ Z₁ + (1₁ +1₂)Z = E₁ 1₂ *Z₁ + (1₁ +1₂)Z = E₂ and Subtracting eq. (ii) from eq. (i), we have, 1₁-1₂ :. 1₁-1₂ Adding eqs. (i) and (), we have, = 1₁ + 1₂ E₁-E₂ Z₁ -1242 + j1003 j5 (200+j250) A = = (5686+j1003) - (6928+ j 0) j5 = (200+j250) A E₁ + E₂ 2 Z+Z₁ 12614+j103 (5686+j1003) + (6928+ j0) 2 (50+ j40) + j5 12654 24.5⁰ 85 131-2240-4⁰ = 96-42-35-9° A = (78-1-i56-6) A ...(1) ... (iii)

Referring to Fig. 12.87, we have, 1₁ Z₁ + (1₁ +1₂)Z = E₁ 1₂ *Z₁ + (1₁ +1₂)Z = E₂ and Subtracting eq. (ii) from eq. (i), we have, 1₁-1₂ :. 1₁-1₂ Adding eqs. (i) and (), we have, = 1₁ + 1₂ E₁-E₂ Z₁ -1242 + j1003 j5 (200+j250) A = = (5686+j1003) - (6928+ j 0) j5 = (200+j250) A E₁ + E₂ 2 Z+Z₁ 12614+j103 (5686+j1003) + (6928+ j0) 2 (50+ j40) + j5 12654 24.5⁰ 85 131-2240-4⁰ = 96-42-35-9° A = (78-1-i56-6) A ...(1) ... (iii)

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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