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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 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS
+
+// EXAMPLE : 4.10 :
+// Page number 519-520
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+MVA_A = 30.0        // Alternator rating(MVA)
+kV_A = 11.0         // Alternator rating(kV)
+X_1 = 2.5           // Reactance to positive sequence current(ohm)
+X_2 = 0.8*X_1       // Reactance to negative sequence current(ohm)
+X_0 = 0.3*X_1       // Reactance to zero sequence current(ohm)
+
+// Calculations
+// Case(a)
+a = exp(%i*120.0*%pi/180)                 // Operator
+Z_1 = %i*X_1                              // Positive sequence impedance(ohm)
+Z_2 = %i*X_2                              // Negative sequence impedance(ohm)
+Z_0 = %i*X_0                              // Zero sequence impedance(ohm)
+Z_02 = Z_0*Z_2/(Z_0+Z_2)                  // Impedance(ohm)
+E_a = kV_A*1000/3**0.5                    // Phase voltage(V)
+I_a1 = E_a/(Z_1+Z_02)                     // Positive sequence current(A)
+I_a2 = -Z_0/(Z_0+Z_2)*I_a1                // Negative sequence current(A)
+I_a0 = -Z_2/(Z_0+Z_2)*I_a1                // Zero sequence current(A)
+I_0 = I_a0                                // Zero sequence current(A)
+I_a = I_a0+I_a1+I_a2                      // Line current(A)
+I_b = I_0+a**2*I_a1+a*I_a2                // Line current(A)
+I_c = I_0+a*I_a1+a**2*I_a2                // Line current(A)
+// Case(b)
+I_n = 3*abs(I_0)                          // Current through ground(A)
+// Case(c)
+V_a2 = Z_02*I_a1                          // Negative sequence voltage(V)
+V_a = 3*abs(V_a2)                         // Voltage of healthy phase to neutral(V)
+
+// Results
+disp("PART III - EXAMPLE : 4.10 : SOLUTION :-")
+printf("\nCase(a): Currents in the faulted phase are")
+printf("\n         I_a = %.f A", abs(I_a))
+printf("\n         I_b = %.f∠%.1f° A", abs(I_b),phasemag(I_b))
+printf("\n         I_c = %.f∠%.1f° A", abs(I_c),phasemag(I_c))
+printf("\nCase(b): Current through ground, I_n = %.f A", I_n)
+printf("\nCase(c): Voltage of healthy phase to neutral, V_a = %.f V\n", V_a)
+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here")