Can you please explain the steps to get the current values i1 , i2 , i3 , i4.I already got the 4 equations needed but cannot get the correct values of the currents.Answers are on the left bottom corner. 3AAP6.₁3Answers1₁ = -7.5 A1₂ = -2.5Ar1iz = 3.92 A14 = 2.14 Aio = 1+ = -2.14 A15Ai ²LApply KVL around super meshApply KVL on mesh 422Apply KCL at node PApply KCL at node QFSuper mesh11L1-√310-40wwwQSfis6₁₂ +21₁ + 4i₂ + 8iz - 8 i = 02₁₁ + 6₁₂ + 1² ₁₂ - 8i = 0។18₁₁ -8 ₂ + 2 1₂ + ¹0 = 0iz31₁ + 1₁ = 10iz3420wwTio13822 fir 10V10 ₁₁ -8 ₁ = -1014 131₂ = 5 + 1₁ (3)231₁ + 3i₂ +6 i ₂ - 4 ₁ ₂ = 0 - (-- 4 ₁₂ + 5₁₁₂ = −5 − (2)RE25.

The equations given are i1+3i2+6i3−4i4=0 .............................(1)−4i3+5i4=−5…

Apply KVL around P 6.₁ Answers i,= -7.5A 12 -2.5A 13 = 3.92 A 14 = 2.14 A io = -14 = -2.14 A r 1 5A 1 i ² L - Super Apply KVL on mesh 4 22 Apply KCL at node P Apply KCL at node Q Super mesh mesh 1 1 - 45 www Q - 1 √7310 (is 1380 (in 25 www 6i₂ + 21, +4₂ +8i, -8 i = 0 21, +6₂ +12, -8i =0 8₁₂-81₂ +21 + 10 = 0 4 i₁ +3₁₂ +6i₂ - 4₁₂ = 0 -(1) 10 ₁₁ -8 ₁ = -10 is i ₂ 3 i₂=5+1 (3) 3₁₁ + 1₂ = 1₂ 3 1. - 3₁₁ + 1₂ = 1₂ Tio - HOV -4₁, +5=-5-(2) 1₂ = 1₂ +3₁₂ — (4)

Apply KVL around P 6.₁ Answers i,= -7.5A 12 -2.5A 13 = 3.92 A 14 = 2.14 A io = -14 = -2.14 A r 1 5A 1 i ² L - Super Apply KVL on mesh 4 22 Apply KCL at node P Apply KCL at node Q Super mesh mesh 1 1 - 45 www Q - 1 √7310 (is 1380 (in 25 www 6i₂ + 21, +4₂ +8i, -8 i = 0 21, +6₂ +12, -8i =0 8₁₂-81₂ +21 + 10 = 0 4 i₁ +3₁₂ +6i₂ - 4₁₂ = 0 -(1) 10 ₁₁ -8 ₁ = -10 is i ₂ 3 i₂=5+1 (3) 3₁₁ + 1₂ = 1₂ 3 1. - 3₁₁ + 1₂ = 1₂ Tio - HOV -4₁, +5=-5-(2) 1₂ = 1₂ +3₁₂ — (4)

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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