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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 : 7.14 :
+// Page number 219-220
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+c_s = 0.90        // Capacitance b/w all conductors(µF)
+C_0 = 0.4         // Capacitance b/w two conductor(µF)
+V = 11.0*10**3    // Line Voltage(V)
+f = 50.0          // Frequency(Hz)
+
+// Calculations
+C_s = c_s/3.0                              // Capacitance measured(µF)
+C_c = (C_0-C_s)/2.0                        // Capacitance(µF)
+C_a = 3.0/2*(C_c+(1/3.0)*C_s)              // Capacitance b/w any two conductors(µF)
+C_b = 2.0*C_c+(2.0/3)*C_s                  // Capacitance b/w any two bounded conductors and the third conductor(µF)
+C_o = 3.0*C_c+C_s                          // Capacitance to neutral(µF)
+I_c = 2.0*%pi*f*C_o*V/3**0.5*10**-6        // Charging current(A)
+
+// Results
+disp("PART II - EXAMPLE : 7.14 : SOLUTION :-")
+printf("\nCase(a): Capacitance between any two conductors = %.3f µF", C_a)
+printf("\nCase(b): Capacitance between any two bounded conductors and the third conductor = %.1f µF", C_b)
+printf("\nCase(c): Capacitance to neutral, C_0 = %.2f µF", C_o)
+printf("\n         Charging current taken by cable, I_c = %.3f A \n", I_c)
+printf("\nNOTE: ERROR: Calculation mistakes in textbook answer")