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Node Voltage Method: To find V1 V1 – 0.56 V1 V1 Z1 Z2 = 0 Z3 + Z4 + Z5 V1 – 0.56 -1.457462849j V1 V1 1078.194599j + 1990 + 1.77181532j V1 V1 V1 0.56 -1.457462849j 1078.194599j 1990 – 1.77181532j 1.457462849j V1 = V1= 0.55924352134 – 0.04190633479_Volts

Node Voltage Method: To find V1 V1 – 0.56 V1 V1 Z1 Z2 = 0 Z3 + Z4 + Z5 V1 – 0.56 -1.457462849j V1 V1 1078.194599j + 1990 + 1.77181532j V1 V1 V1 0.56 -1.457462849j 1078.194599j 1990 – 1.77181532j 1.457462849j V1 = V1= 0.55924352134 – 0.04190633479_Volts

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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i dont understand how we get to the last step

Node Voltage Method: To find V1 V1 – 0.56 V1 V1 Z1 Z2 Z3 + Z4 + Z5 V1 – 0.56 -1.457462849j V1 1078.194599j V1 = 0 + 1990 + 1.77181532j V1 -1.457462849j V1 1078.194599j V. 0.56 + 1990 – 1.77181532j 1.457462849j V1 = V1 = 0.55924352134 – 0.04190633479_Volts
Transcribed Image Text:Node Voltage Method: To find V1 V1 – 0.56 V1 V1 Z1 Z2 Z3 + Z4 + Z5 V1 – 0.56 -1.457462849j V1 1078.194599j V1 = 0 + 1990 + 1.77181532j V1 -1.457462849j V1 1078.194599j V. 0.56 + 1990 – 1.77181532j 1.457462849j V1 = V1 = 0.55924352134 – 0.04190633479_Volts
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