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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 7: UNDERGROUND CABLES
+
+// EXAMPLE : 7.16 :
+// Page number 222-223
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 33.0*10**3    // Line Voltage(V)
+f = 50.0          // Frequency(Hz)
+l = 4.0           // Length(km)
+d = 2.5           // Diameter of conductor(cm)
+t = 0.5           // Radial thickness of insulation(cm)
+e_r = 3.0         // Relative permittivity of the dielectric
+PF = 0.02         // Power factor of unloaded cable
+
+// Calculations
+// Case(a)
+r = d/2.0                                         // Radius of conductor(cm)
+R = r+t                                           // External radius(cm)
+e_0 = 8.85*10**-12                                // Permittivity
+C = 2.0*%pi*e_0*e_r/log(R/r)*l*1000               // Capacitance of cable/phase(F)
+// Case(b)
+V_ph = V/3**0.5                                   // Phase voltage(V)
+I_c = V_ph*2.0*%pi*f*C                            // Charging current/phase(A)
+// Case(c)
+kVAR = 3.0*V_ph*I_c                               // Total charging kVAR
+// Case(d)
+phi = acosd(PF)                                   // Φ(°)
+delta = 90.0-phi                                  // δ(°)
+P_c = V_ph*I_c*sind(delta)/1000                   // Dielectric loss/phase(kW)
+// Case(e)
+E_max = V_ph/(r*log(R/r)*1000)                    // RMS value of Maximum stress in cable(kV/cm)
+
+// Results
+disp("PART II - EXAMPLE : 7.16 : SOLUTION :-")
+printf("\nCase(a): Capacitance of the cable, C = %.3e F/phase", C)
+printf("\nCase(b): Charging current = %.2f A/phase", I_c)
+printf("\nCase(c): Total charging kVAR = %.4e kVAR", kVAR)
+printf("\nCase(d): Dielectric loss/phase, P_c = %.2f kW", P_c)
+printf("\nCase(e): Maximum stress in the cable, E_max = %.1f kV/cm (rms)", E_max)