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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 II : TRANSMISSION AND DISTRIBUTION
// CHAPTER 18: POWER DISTRIBUTION SYSTEMS
// EXAMPLE : 18.4 :
// Page number 439
clear ; clc ; close ; // Clear the work space and console
// Given data
l = 450.0 // Length of wire(m)
V_A = 250.0 // Voltage at end A(V)
V_B = 250.0 // Voltage at end A(V)
r = 0.05 // Conductor resistance(ohm/km)
i = 1.5 // Load(A/m)
I_C = 20.0 // Current at C(A)
l_C = 60.0 // Distance to C from A(m)
I_D = 40.0 // Current at D(A)
l_D = 100.0 // Distance to D from A(m)
l_E = 200.0 // Distance to E from A(m)
// Calculations
x = poly(0,"x") // Current to point D from end A(A)
AD = (I_C+x)*r*l_C+x*r*(l_D-l_C) // Drop in length AD
BD = (i*r*V_A**2/2)+(I_D-x)*r*(450-l_D) // Drop in length BD
x_sol = roots(AD-BD) // Current(A)
I_F = x_sol-I_D // Current supplied to load from end A(A)
l_F = l_E+(I_F/i) // Point of minimum potential at F from A(m)
V_F = V_B-(375.0-I_F)*(250-(l_F-200))*r/1000 // Potential at F from end B(V)
// Results
disp("PART II - EXAMPLE : 18.4 : SOLUTION :-")
printf("\nPoint of minimum potential occurs at F from A = %.2f metres", l_F)
printf("\nPotential at point F = %.2f V", V_F)
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