// A Texbook on POWER SYSTEM ENGINEERING // A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar // DHANPAT RAI & Co. // SECOND EDITION // PART II : TRANSMISSION AND DISTRIBUTION // CHAPTER 18: POWER DISTRIBUTION SYSTEMS // EXAMPLE : 18.2 : // Page number 437-438 clear ; clc ; close ; // Clear the work space and console // Given data I = 600.0 // Constant current drawn(A) D = 8.0 // Distance b/w two sub-stations(km) V_A = 575.0 // Potential at point A(V) V_B = 590.0 // Potential at point B(V) R = 0.04 // Track resistance(ohm/km) // Calculations x = poly(0,'x') // x(km) I_A = ((-V_B+R*I*D+V_A)-(R*I)*x)/(D*R) // Simplifying V_P = V_A-I_A*R*x // Potential at P in terms of x(V) dVP_dx = derivat(V_P) // dV_P/dx x_sol = roots(dVP_dx) // Value of x(km) I_A_1 = ((-V_B+R*I*D+V_A)-(R*I)*x_sol)/(D*R) // Current drawn from end A(A) I_B = I-I_A_1 // Current drawn from end B(A) // Results disp("PART II - EXAMPLE : 18.2 : SOLUTION :-") printf("\nPoint of minimum potential along the track, x = %.2f km", x_sol) printf("\nCurrent supplied by station A, I_A = %.f A", I_A_1) printf("\nCurrent supplied by station B, I_B = %.f A \n", I_B) printf("\nNOTE: ERROR: Calculation mistake in the textbook solution")