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I 1 Z₂ = 3+ jwC j50 x 10 x 10-3 Z₁ = 8 + jwL = 8 + j50 x 0.2 = (8+j10) f Thus, The input impedance is = 3+ = -j10 + ZinZ₁ + Z₂ Z3 = -10 + (44+j14)(11-18) 11² + 8² (3-j2)(8 +j10) 11 +j8 = -j10+ 3.22 - j1.07 2 = (3-j2) 2 Zin= 3.22-j11.07 2

I 1 Z₂ = 3+ jwC j50 x 10 x 10-3 Z₁ = 8 + jwL = 8 + j50 x 0.2 = (8+j10) f Thus, The input impedance is = 3+ = -j10 + ZinZ₁ + Z₂ Z3 = -10 + (44+j14)(11-18) 11² + 8² (3-j2)(8 +j10) 11 +j8 = -j10+ 3.22 - j1.07 2 = (3-j2) 2 Zin= 3.22-j11.07 2

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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Could someone explain these last steps in detail, where the 11 +j8 comes from and the step after
j50 X X 2 x 10 1 1 jwC j50 x 10 x 107 Z₂ = 8 + jwL = 8 + j50 × 0.2 = (8 +j10) f Thus, Z₂ = 3 + The input impedance is JWC = -j10 + 3+ ZinZ₁ + Z₂ Z3 = -10 + (44+j14)(11-18) 11² +8² = = (3-j2) f (3-j2)(8 +j10) 11 +j8 Zin 3.22j11.07 2 = -j10 +3.22-j1.07 2
Transcribed Image Text:j50 X X 2 x 10 1 1 jwC j50 x 10 x 107 Z₂ = 8 + jwL = 8 + j50 × 0.2 = (8 +j10) f Thus, Z₂ = 3 + The input impedance is JWC = -j10 + 3+ ZinZ₁ + Z₂ Z3 = -10 + (44+j14)(11-18) 11² +8² = = (3-j2) f (3-j2)(8 +j10) 11 +j8 Zin 3.22j11.07 2 = -j10 +3.22-j1.07 2
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