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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 : 3.11 :
// Page number 495
clear ; clc ; close ; // Clear the work space and console
// Given data
V = 400.0 // Voltage(V)
Z_ab = 20.0 // Resistor load(ohm)
Z_bc = -%i*40.0 // Capacitor load(ohm)
Z_ca = 5.0+%i*10.0 // Inductor and resistance load(ohm)
// Calculations
V_ab = V // Line voltage(V)
V_bc = V*exp(%i*-120.0*%pi/180) // Line voltage(V)
V_ca = V*exp(%i*120.0*%pi/180) // Line voltage(V)
I_ab = V_ab/Z_ab // Current(A)
I_bc = V_bc/Z_bc // Current(A)
I_ca = V_ca/Z_ca // Current(A)
I_a = I_ab-I_ca // Line current(A)
I_b = I_bc-I_ab // Line current(A)
I_c = I_ca-I_bc // Line current(A)
phi = -120.0-phasemag(I_a) // φ(°)
P = abs(I_a*V_bc)*cosd(phi)/1000 // Wattmeter reading(kW)
// Results
disp("PART III - EXAMPLE : 3.11 : SOLUTION :-")
printf("\nLine currents are:")
printf("\n I_a = %.1f∠%.1f° A", abs(I_a),phasemag(I_a))
printf("\n I_b = %.1f∠%.2f° A", abs(I_b),phasemag(I_b))
printf("\n I_c = %.2f∠%.f° A", abs(I_c),phasemag(I_c))
printf("\nWattmeter reading, P = %.2f kW \n", P)
printf("\nNOTE: ERROR: Calculation mistakes in the textbook solution")
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