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a) R = 100ΚΩ+47ΚΩ= C=Cy 1 2πfCr XC = - 1 147ΚΩ 1 1 + 0.01μFT 0.0022 F 1.8μF 1 Σπ(100Hz)(1.8μF) -=884 Ω Z= R – jXc = 147ΚΩ – 884Ω = 147ΚΩ<0.344°

a) R = 100ΚΩ+47ΚΩ= C=Cy 1 2πfCr XC = - 1 147ΚΩ 1 1 + 0.01μFT 0.0022 F 1.8μF 1 Σπ(100Hz)(1.8μF) -=884 Ω Z= R – jXc = 147ΚΩ – 884Ω = 147ΚΩ<0.344°

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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Can you please check my calculations.

a) R = 100ΚΩ+47ΚΩ= 147ΚΩ C = CT= XC = 1 1 1 0.01μF 0.0022μF 1 2nfСT + 1.8μF 1 2π(100Hz)(1.8μF) -=884 Ω Z=R-jXc = 147KN – 884N = 147kQ<0.344°
Transcribed Image Text:a) R = 100ΚΩ+47ΚΩ= 147ΚΩ C = CT= XC = 1 1 1 0.01μF 0.0022μF 1 2nfСT + 1.8μF 1 2π(100Hz)(1.8μF) -=884 Ω Z=R-jXc = 147KN – 884N = 147kQ<0.344°
50/0° V f= 100 Hz (a) C₁ C₂ R₁ www 100 kn 0.01 μF 0.022 μF R₂ 47 ΚΩ
Transcribed Image Text:50/0° V f= 100 Hz (a) C₁ C₂ R₁ www 100 kn 0.01 μF 0.022 μF R₂ 47 ΚΩ
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