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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 : 4.11 :
+// Page number 520-521
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+n = 6.0       // Number of alternator
+kV_A = 6.6    // Alternator rating(kV)
+X_1 = 0.9     // Positive sequence reactance(ohm)
+X_2 = 0.72    // Negative sequence reactance(ohm)
+X_0 = 0.3     // Zero sequence reactance(ohm)
+Z_n = 0.2     // Resistance of grounding resistor(ohm)
+
+// Calculations
+E_a = kV_A*1000/3**0.5               // Phase voltage(V)
+// Case(a)
+Z_1_a = %i*X_1/n                     // Positive sequence impedance when alternators are in parallel(ohm)
+Z_2_a = %i*X_2/n                     // Negative sequence impedance when alternators are in parallel(ohm)
+Z_0_a = %i*X_0/n                     // Zero sequence impedance when alternators are in parallel(ohm)
+I_a_a = 3*E_a/(Z_1_a+Z_2_a+Z_0_a)    // Fault current assuming 'a' phase to be fault(A)
+// Case(b)
+Z_0_b = 3*Z_n+%i*X_0                 // Zero sequence impedance(ohm)
+I_a_b = 3*E_a/(Z_1_a+Z_2_a+Z_0_b)    // Fault current(A)
+// Case(c)
+Z_0_c = %i*X_0                       // Zero sequence impedance(ohm)
+I_a_c = 3*E_a/(Z_1_a+Z_2_a+Z_0_c)    // Fault current(A)
+
+// Results
+disp("PART III - EXAMPLE : 4.11 : SOLUTION :-")
+printf("\nCase(a): Fault current if all alternator neutrals are solidly grounded, I_a = %.f A", imag(I_a_a))
+printf("\nCase(b): Fault current if one alternator neutral is grounded & others isolated, I_a = %.1f∠%.1f° A", abs(I_a_b),phasemag(I_a_b))
+printf("\nCase(c): Fault current if one alternator neutral is solidly grounded & others isolated, I_a = %.2fj A\n", imag(I_a_c))
+printf("\nNOTE: ERROR: Calculation mistakes in the textbook solution")