These two nonlinear equations are the power-flow equations for the simple system shown below. fi (x1, x2, u) = 4ux2 sin x₁ +0.6 = 0 f2 (x1, x2, u) = 4x2 - 4ux2 cos x₁ +0.3 = 0 Setting u=1.0 and using the initial estimates of x0) = 0 rad and x₁ one iteration. O 0.3 rad PB1 O -0.150 rad Qg1 O O rad O -0.425 rad 1 IV₁/0° rele ooo j0.25 j6.0 (0) (1) x2 = 1.0, calculate x₁ with just 2 1 |V₂\/82 Pd2 0.6 ૨2 0.3 To load

These two nonlinear equations are the power-flow equations for the simple system shown below. fi (x1, x2, u) = 4ux2 sin x₁ +0.6 = 0 f2 (x1, x2, u) = 4x2 - 4ux2 cos x₁ +0.3 = 0 Setting u=1.0 and using the initial estimates of x0) = 0 rad and x₁ one iteration. O 0.3 rad PB1 O -0.150 rad Qg1 O O rad O -0.425 rad 1 IV₁/0° rele ooo j0.25 j6.0 (0) (1) x2 = 1.0, calculate x₁ with just 2 1 |V₂\/82 Pd2 0.6 ૨2 0.3 To load

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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Characteristic equation of a system is given below. -5s6-2s5- 10s1-9s3-2s2-4s – 10=0 .Which of the following statement is correct? Oall the coefficients are negative. Hence we can conclude that system is stable. Necessary condition of RH criteria is satisfied. Sufficient condition has to be checked with RH table. Necessary condition of RH criteria is not satisfied. Hence system is unstable O all the coefficients are negative. Hence we can conclude that system is unstable.
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