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Using mesh analysis answer the questions based on the circuit below 10 cos 10³t V 1₁=0.05-j0.75 2 ΚΩ www 1₂ = -0.035+j0.005 12 = -0.35+j0.5 ele 1 μF Phasor form of the inductor is j*400 ohms Phasor form of the inductor is j*100 ohms Phasor form of the capacitor is -j*1000 ohms Phasor form of the capacitor is -j*10 ohms Phasor form of 20*sin(1000t) is -j20 (202-90) Phasor form of 20*sin(1000t) is -j10 (10<-90) ₁=0.0025-j0.0075 16 0.4 H Iz 20 sin 10³t V

Using mesh analysis answer the questions based on the circuit below 10 cos 10³t V 1₁=0.05-j0.75 2 ΚΩ www 1₂ = -0.035+j0.005 12 = -0.35+j0.5 ele 1 μF Phasor form of the inductor is j*400 ohms Phasor form of the inductor is j*100 ohms Phasor form of the capacitor is -j*1000 ohms Phasor form of the capacitor is -j*10 ohms Phasor form of 20*sin(1000t) is -j20 (202-90) Phasor form of 20*sin(1000t) is -j10 (10<-90) ₁=0.0025-j0.0075 16 0.4 H Iz 20 sin 10³t V

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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Using mesh analysis answer the questions based on the circuit below 10 cos 10³t V ☐ 1₁=0.05-j0.75 2 ΚΩ www 1₂ = -0.035+j0.005 12 = -0.35+j0.5 H ell 1 μF HH lo 0.4 H H Phasor form of the inductor is j*400 ohms Phasor form of the inductor is j*100 ohms Phasor form of the capacitor is -j*1000 ohms Phasor form of the capacitor is -j*10 ohms Phasor form of 20* sin(1000t) is -j20 (20<-90) Phasor form of 20*sin(1000t) is-j10 (10<-90) ₁=0.0025-j0.0075 Iz 20 sin 10³t V
Transcribed Image Text:Using mesh analysis answer the questions based on the circuit below 10 cos 10³t V ☐ 1₁=0.05-j0.75 2 ΚΩ www 1₂ = -0.035+j0.005 12 = -0.35+j0.5 H ell 1 μF HH lo 0.4 H H Phasor form of the inductor is j*400 ohms Phasor form of the inductor is j*100 ohms Phasor form of the capacitor is -j*1000 ohms Phasor form of the capacitor is -j*10 ohms Phasor form of 20* sin(1000t) is -j20 (20<-90) Phasor form of 20*sin(1000t) is-j10 (10<-90) ₁=0.0025-j0.0075 Iz 20 sin 10³t V
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