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// A Texbook on POWER SYSTEM ENGINEERING
// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
// DHANPAT RAI & Co.
// SECOND EDITION
// PART III : SWITCHGEAR AND PROTECTION
// CHAPTER 3: SYMMETRICAL COMPONENTS' ANALYSIS
// EXAMPLE : 3.6 :
// Page number 491-492
clear ; clc ; close ; // Clear the work space and console
// Given data
V = 2300.0 // Rated voltage(V)
kVA = 500.0 // kVA rating
E_A = 2760.0*exp(%i*0*%pi/180) // Line voltage(V)
E_B = 2300.0*exp(%i*-138.6*%pi/180) // Line voltage(V)
E_C = 1840.0*exp(%i*124.2*%pi/180) // Line voltage(V)
// Calculations
a = exp(%i*120.0*%pi/180) // Operator
E_A1 = 1.0/3*(E_A+a*E_B+a**2*E_C) // Positive sequence voltage(V)
E_A2 = 1.0/3*(E_A+a**2*E_B+a*E_C) // Negative sequence voltage(V)
E_a1 = -%i/3**0.5*E_A1 // Positive sequence star voltage(V)
E_a2 = %i/3**0.5*E_A2 // Negative sequence star voltage(V)
E_a0 = 0.0 // Zero sequence voltage(V)
E_a = E_a1+E_a2+E_a0 // Symmetrical voltage component(V)
R = V**2/(kVA*1000) // Resistance(ohm)
I_a = abs(E_a)/R // Current in line a(A)
E_b = a**2*E_a1+a*E_a2+E_a0 // Symmetrical voltage component(V)
I_b = abs(E_b)/R // Current in line b(A)
E_c = a*E_a1+a**2*E_a2+E_a0 // Symmetrical voltage component(V)
I_c = abs(E_c)/R // Current in line c(A)
// Results
disp("PART III - EXAMPLE : 3.6 : SOLUTION :-")
printf("\nCurrent in line a, |I_a| = %.1f A", I_a)
printf("\nCurrent in line b, |I_b| = %.f A", I_b)
printf("\nCurrent in line c, |I_c| = %.1f A \n", I_c)
printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here")
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