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-rw-r--r--3472/CH29/EX29.1/Example29_1.sce28
-rw-r--r--3472/CH29/EX29.10/Example29_10.sce35
-rw-r--r--3472/CH29/EX29.11/Example29_11.sce39
-rw-r--r--3472/CH29/EX29.4/Example29_4.sce48
-rw-r--r--3472/CH29/EX29.5/Example29_5.sce45
-rw-r--r--3472/CH29/EX29.6/Example29_6.sce40
-rw-r--r--3472/CH29/EX29.7/Example29_7.sce45
-rw-r--r--3472/CH29/EX29.8/Example29_8.sce43
-rw-r--r--3472/CH29/EX29.9/Example29_9.sce46
9 files changed, 369 insertions, 0 deletions
diff --git a/3472/CH29/EX29.1/Example29_1.sce b/3472/CH29/EX29.1/Example29_1.sce
new file mode 100644
index 000000000..6a82ffc57
--- /dev/null
+++ b/3472/CH29/EX29.1/Example29_1.sce
@@ -0,0 +1,28 @@
+// 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 3: SYMMETRICAL COMPONENTS' ANALYSIS
+
+// EXAMPLE : 3.1 :
+// Page number 487-488
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+I_R = complex(12.0,24.0) // Line current(A)
+I_Y = complex(16.0,-2.0) // Line current(A)
+I_B = complex(-4.0,-6.0) // Line current(A)
+
+// Calculations
+alpha = exp(%i*120.0*%pi/180) // Operator
+I_R0 = 1.0/3*(I_R+I_Y+I_B) // Zero sequence component(A)
+I_R1 = 1.0/3*(I_R+alpha*I_Y+alpha**2*I_B) // Positive sequence component(A)
+I_R2 = 1.0/3*(I_R+alpha**2*I_Y+alpha*I_B) // Negative sequence component(A)
+
+// Results
+disp("PART III - EXAMPLE : 3.1 : SOLUTION :-")
+printf("\nPositive sequence current, I_R1 = (%.3f + %.1fj) A", real(I_R1),imag(I_R1))
+printf("\nNegative sequence current, I_R2 = (%.3f + %.2fj) A", real(I_R2),imag(I_R2))
+printf("\nZero sequence current, I_R0 = (%.1f + %.2fj) A", real(I_R0),imag(I_R0))
diff --git a/3472/CH29/EX29.10/Example29_10.sce b/3472/CH29/EX29.10/Example29_10.sce
new file mode 100644
index 000000000..5c0a51af6
--- /dev/null
+++ b/3472/CH29/EX29.10/Example29_10.sce
@@ -0,0 +1,35 @@
+// 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 3: SYMMETRICAL COMPONENTS' ANALYSIS
+
+// EXAMPLE : 3.10 :
+// Page number 494
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+R = 20000.0 // Resistance of voltmeter(ohm)
+E_R = 100.0 // Line-to-neutral voltage(A)
+E_Y = 200.0*exp(%i*270.0*%pi/180) // Line-to-neutral voltage(A)
+E_B = 100.0*exp(%i*120.0*%pi/180) // Line-to-neutral voltage(A)
+
+// Calculations
+a = exp(%i*120.0*%pi/180) // Operator
+V_R0 = 1.0/3*(E_R+E_Y+E_B) // Zero sequence voltage(V)
+V_R1 = 1.0/3*(E_R+a*E_Y+a**2*E_B) // Positive sequence voltage(V)
+V_R2 = 1.0/3*(E_R+a**2*E_Y+a*E_B) // Negative sequence voltage(V)
+I_R1 = V_R1/R // Positive sequence current(A)
+I_R2 = V_R2/R // Negative sequence current(A)
+V_Y1 = a**2*V_R1 // Positive sequence voltage of line Y(V)
+V_Y2 = a*V_R2 // Negative sequence voltage of line Y(V)
+V_Y = V_Y1+V_Y2 // Voltmeter reading connected to the yellow line(V)
+I_Y = abs(V_Y)/R*1000 // Current through voltmeter(mA)
+
+// Results
+disp("PART III - EXAMPLE : 3.10 : SOLUTION :-")
+printf("\nVoltmeter reading connected to the yellow line, |V_Y| = %.1f V", abs(V_Y))
+printf("\nCurrent through voltmeter, I_Y = %.3f mA \n", I_Y)
+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here")
diff --git a/3472/CH29/EX29.11/Example29_11.sce b/3472/CH29/EX29.11/Example29_11.sce
new file mode 100644
index 000000000..f1930f940
--- /dev/null
+++ b/3472/CH29/EX29.11/Example29_11.sce
@@ -0,0 +1,39 @@
+// 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 3: SYMMETRICAL COMPONENTS' ANALYSIS
+
+// EXAMPLE : 3.11 :
+// Page number 495
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 400.0 // Voltage(V)
+Z_ab = 20.0 // Resistor load(ohm)
+Z_bc = -%i*40.0 // Capacitor load(ohm)
+Z_ca = 5.0+%i*10.0 // Inductor and resistance load(ohm)
+
+// Calculations
+V_ab = V // Line voltage(V)
+V_bc = V*exp(%i*-120.0*%pi/180) // Line voltage(V)
+V_ca = V*exp(%i*120.0*%pi/180) // Line voltage(V)
+I_ab = V_ab/Z_ab // Current(A)
+I_bc = V_bc/Z_bc // Current(A)
+I_ca = V_ca/Z_ca // Current(A)
+I_a = I_ab-I_ca // Line current(A)
+I_b = I_bc-I_ab // Line current(A)
+I_c = I_ca-I_bc // Line current(A)
+phi = -120.0-phasemag(I_a) // φ(°)
+P = abs(I_a*V_bc)*cosd(phi)/1000 // Wattmeter reading(kW)
+
+// Results
+disp("PART III - EXAMPLE : 3.11 : SOLUTION :-")
+printf("\nLine currents are:")
+printf("\n I_a = %.1f∠%.1f° A", abs(I_a),phasemag(I_a))
+printf("\n I_b = %.1f∠%.2f° A", abs(I_b),phasemag(I_b))
+printf("\n I_c = %.2f∠%.f° A", abs(I_c),phasemag(I_c))
+printf("\nWattmeter reading, P = %.2f kW \n", P)
+printf("\nNOTE: ERROR: Calculation mistakes in the textbook solution")
diff --git a/3472/CH29/EX29.4/Example29_4.sce b/3472/CH29/EX29.4/Example29_4.sce
new file mode 100644
index 000000000..efa29406d
--- /dev/null
+++ b/3472/CH29/EX29.4/Example29_4.sce
@@ -0,0 +1,48 @@
+// 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 3: SYMMETRICAL COMPONENTS' ANALYSIS
+
+// EXAMPLE : 3.4 :
+// Page number 489-490
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+R_bc = 5.0 // Resistance of resistor connected b/w b & c(ohm)
+R_ca = 10.0 // Resistance of resistor connected b/w c & a(ohm)
+R_ab = 20.0 // Resistance of resistor connected b/w a & b(ohm)
+V = 100.0 // Voltage of balanced system(V)
+
+// Calculations
+E_A = -V // Voltage across resistor connected b/w b & c(V)
+angle = 60.0 // Angle in delta system(°)
+E_B = V*exp(%i*60.0*%pi/180) // Voltage across resistor connected b/w c & a(V)
+E_C = V*exp(%i*-60.0*%pi/180) // Voltage across resistor connected b/w a & b(V)
+I_A = E_A/R_bc // Current flowing across resistor connected b/w b & c(A)
+I_B = E_B/R_ca // Current flowing across resistor connected b/w c & a(A)
+I_C = E_C/R_ab // Current flowing across resistor connected b/w a & b(A)
+alpha = exp(%i*120.0*%pi/180) // Operator
+I_A0 = 1.0/3*(I_A+I_B+I_C) // Zero sequence delta current(A)
+I_A1 = 1.0/3*(I_A+alpha*I_B+alpha**2*I_C) // Positive sequence delta current(A)
+I_A2 = 1.0/3*(I_A+alpha**2*I_B+alpha*I_C) // Negative sequence delta current(A)
+I_a0 = 0.0 // Zero sequence star current(A)
+I_a1 = (alpha-alpha**2)*I_A1 // Positive sequence star current(A)
+I_a2 = (alpha**2-alpha)*I_A2 // Negative sequence star current(A)
+
+// Results
+disp("PART III - EXAMPLE : 3.4 : SOLUTION :-")
+printf("\nCurrent in the resistors are:")
+printf("\n I_A = (%.f+%.fj) A", real(I_A),imag(I_A))
+printf("\n I_B = (%.f+%.2fj) A", real(I_B),imag(I_B))
+printf("\n I_C = (%.1f%.2fj) A", real(I_C),imag(I_C))
+printf("\nSequence components of currents in the resistors:")
+printf("\n Zero-sequence current, I_A0 = (%.3f+%.2fj) A", real(I_A0),imag(I_A0))
+printf("\n Positive-sequence current, I_A1 = (%.2f+%.fj) A", real(I_A1),imag(I_A1))
+printf("\n Negative-sequence current, I_A2 = (%.2f%.2fj) A", real(I_A2),imag(I_A2))
+printf("\nSequence components of currents in the supply lines:")
+printf("\n Zero-sequence current, I_a0 = %.f A", I_a0)
+printf("\n Positive-sequence current, I_a1 = %.1fj A", imag(I_a1))
+printf("\n Negative-sequence current, I_a2 = (%.1f+%.2fj) A", real(I_a2),imag(I_a2))
diff --git a/3472/CH29/EX29.5/Example29_5.sce b/3472/CH29/EX29.5/Example29_5.sce
new file mode 100644
index 000000000..bcd081621
--- /dev/null
+++ b/3472/CH29/EX29.5/Example29_5.sce
@@ -0,0 +1,45 @@
+// 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 3: SYMMETRICAL COMPONENTS' ANALYSIS
+
+// EXAMPLE : 3.5 :
+// Page number 490-491
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+E_a = 100.0 // Line to line voltage(V)
+E_b = 150.0 // Line to line voltage(V)
+E_c = 200.0 // Line to line voltage(V)
+
+// Calculations
+e_A = 1.0 // 100 V = 1 unit
+e_B = 1.5 // 150 V = 1 unit
+e_C = 2.0 // 200 V = 1 unit
+cos_alpha = (e_C**2-e_A-e_B**2)/(2*e_B)
+alpha = acosd(cos_alpha) // angle(°)
+cos_beta = (e_A+e_B*cos_alpha)/e_C
+beta = acosd(cos_beta) // angle(°)
+E_A = E_a*exp(%i*180.0*%pi/180) // Voltage(V)
+E_B = E_b*exp(%i*(180.0-alpha)*%pi/180) // Voltage(V)
+E_C = E_c*exp(%i*-beta*%pi/180) // Voltage(V)
+a = exp(%i*120.0*%pi/180) // Operator
+E_A0 = 1.0/3*(E_A+E_B+E_C) // Zero sequence voltage(V)
+E_A1 = 1.0/3*(E_A+a*E_B+a**2*E_C) // Positive sequence delta voltage(V)
+E_A1_mag = abs(E_A1) // Magnitude of positive sequence delta voltage(V)
+E_a1 = -%i/3**0.5*E_A1 // Positive sequence star voltage(V)
+E_a1_mag = abs(E_a1) // Magnitude of positive sequence star voltage(V)
+E_A2 = 1.0/3*(E_A+a**2*E_B+a*E_C) // Negative sequence delta voltage(V)
+E_A2_mag = abs(E_A2) // Magnitude of negative sequence delta voltage(V)
+E_a2 = %i/3**0.5*E_A2 // Negative sequence star voltage(V)
+E_a2_mag = abs(E_a2) // Magnitude of negative sequence star voltage(V)
+
+// Results
+disp("PART III - EXAMPLE : 3.5 : SOLUTION :-")
+printf("\nMagnitude of positive sequence delta voltage, |E_A1| = %.f V", E_A1_mag)
+printf("\nMagnitude of positive sequence star voltage, |E_a1| = %.1f V", E_a1_mag)
+printf("\nMagnitude of negative sequence delta voltage, |E_A2| = %.f V", E_A2_mag)
+printf("\nMagnitude of negative sequence star voltage, |E_a2| = %.f V", E_a2_mag)
diff --git a/3472/CH29/EX29.6/Example29_6.sce b/3472/CH29/EX29.6/Example29_6.sce
new file mode 100644
index 000000000..84f4ebd88
--- /dev/null
+++ b/3472/CH29/EX29.6/Example29_6.sce
@@ -0,0 +1,40 @@
+// 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 3: SYMMETRICAL COMPONENTS' ANALYSIS
+
+// EXAMPLE : 3.6 :
+// Page number 491-492
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 2300.0 // Rated voltage(V)
+kVA = 500.0 // kVA rating
+E_A = 2760.0*exp(%i*0*%pi/180) // Line voltage(V)
+E_B = 2300.0*exp(%i*-138.6*%pi/180) // Line voltage(V)
+E_C = 1840.0*exp(%i*124.2*%pi/180) // Line voltage(V)
+
+// Calculations
+a = exp(%i*120.0*%pi/180) // Operator
+E_A1 = 1.0/3*(E_A+a*E_B+a**2*E_C) // Positive sequence voltage(V)
+E_A2 = 1.0/3*(E_A+a**2*E_B+a*E_C) // Negative sequence voltage(V)
+E_a1 = -%i/3**0.5*E_A1 // Positive sequence star voltage(V)
+E_a2 = %i/3**0.5*E_A2 // Negative sequence star voltage(V)
+E_a0 = 0.0 // Zero sequence voltage(V)
+E_a = E_a1+E_a2+E_a0 // Symmetrical voltage component(V)
+R = V**2/(kVA*1000) // Resistance(ohm)
+I_a = abs(E_a)/R // Current in line a(A)
+E_b = a**2*E_a1+a*E_a2+E_a0 // Symmetrical voltage component(V)
+I_b = abs(E_b)/R // Current in line b(A)
+E_c = a*E_a1+a**2*E_a2+E_a0 // Symmetrical voltage component(V)
+I_c = abs(E_c)/R // Current in line c(A)
+
+// Results
+disp("PART III - EXAMPLE : 3.6 : SOLUTION :-")
+printf("\nCurrent in line a, |I_a| = %.1f A", I_a)
+printf("\nCurrent in line b, |I_b| = %.f A", I_b)
+printf("\nCurrent in line c, |I_c| = %.1f A \n", I_c)
+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here")
diff --git a/3472/CH29/EX29.7/Example29_7.sce b/3472/CH29/EX29.7/Example29_7.sce
new file mode 100644
index 000000000..9d5c41dc6
--- /dev/null
+++ b/3472/CH29/EX29.7/Example29_7.sce
@@ -0,0 +1,45 @@
+// 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 3: SYMMETRICAL COMPONENTS' ANALYSIS
+
+// EXAMPLE : 3.7 :
+// Page number 492-493
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 2300.0 // Rated voltage(V)
+kVA = 500.0 // kVA rating
+I_1 = 100.0 // Line current(A)
+I_2 = 100.0*exp(%i*180*%pi/180) // Line current(A)
+I_3 = 0 // Line current(A)
+
+// Calculations
+a = exp(%i*120.0*%pi/180) // Operator
+I_10 = 1.0/3*(I_1+I_2+I_3) // Symmetrical component of line current for phase 1(A)
+I_11 = 1.0/3*(I_1+a*I_2+a**2*I_3) // Symmetrical component of line current for phase 1(A)
+I_12 = 1.0/3*(I_1+a**2*I_2+a*I_3) // Symmetrical component of line current for phase 1(A)
+I_20 = I_10 // Symmetrical component of line current for phase 2(A)
+I_21 = a**2*I_11 // Symmetrical component of line current for phase 2(A)
+I_22 = a*I_12 // Symmetrical component of line current for phase 2(A)
+I_30 = I_10 // Symmetrical component of line current for phase 3(A)
+I_31 = a*I_11 // Symmetrical component of line current for phase 3(A)
+I_32 = a**2*I_12 // Symmetrical component of line current for phase 3(A)
+
+// Results
+disp("PART III - EXAMPLE : 3.7 : SOLUTION :-")
+printf("\nSymmetrical component of line current for phase 1:")
+printf("\n I_10 = %.1f A", abs(I_10))
+printf("\n I_11 = %.2f∠%.f° A", abs(I_11),phasemag(I_11))
+printf("\n I_12 = %.2f∠%.f° A", abs(I_12),phasemag(I_12))
+printf("\nSymmetrical component of line current for phase 2:")
+printf("\n I_20 = %.1f A", abs(I_20))
+printf("\n I_21 = %.2f∠%.f° A", abs(I_21),phasemag(I_21))
+printf("\n I_22 = %.2f∠%.f° A", abs(I_22),phasemag(I_22))
+printf("\nSymmetrical component of line current for phase 3:")
+printf("\n I_30 = %.1f A", abs(I_30))
+printf("\n I_31 = %.2f∠%.f° A", abs(I_31),phasemag(I_31))
+printf("\n I_32 = %.2f∠%.f° A", abs(I_32),phasemag(I_32))
diff --git a/3472/CH29/EX29.8/Example29_8.sce b/3472/CH29/EX29.8/Example29_8.sce
new file mode 100644
index 000000000..d5379b825
--- /dev/null
+++ b/3472/CH29/EX29.8/Example29_8.sce
@@ -0,0 +1,43 @@
+// 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 3: SYMMETRICAL COMPONENTS' ANALYSIS
+
+// EXAMPLE : 3.8 :
+// Page number 493
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+I_a = 1000.0 // Current to earth(A)
+I_b = 0 // Current(A)
+I_c = 0 // Current(A)
+
+// Calculations
+a = exp(%i*120.0*%pi/180) // Operator
+I_a0 = 1.0/3*(I_a+I_b+I_c) // Zero sequence component of current(A)
+I_b0 = I_a0 // Zero sequence component of current(A)
+I_c0 = I_a0 // Zero sequence component of current(A)
+I_a1 = 1.0/3*(I_a+a*I_b+a**2*I_c) // Positive sequence component of current(A)
+I_b1 = a**2*I_a1 // Positive sequence component of current(A)
+I_c1 = a*I_a1 // Positive sequence component of current(A)
+I_a2 = 1.0/3*(I_a+a**2*I_b+a*I_c) // Negative sequence component of current(A)
+I_b2 = a*I_a2 // Negative sequence component of current(A)
+I_c2 = a**2*I_a2 // Negative sequence component of current(A)
+
+// Results
+disp("PART III - EXAMPLE : 3.8 : SOLUTION :-")
+printf("\nZero sequence component of current for all phases are")
+printf("\n I_a0 = %.1f∠%.f° A", abs(I_a0),phasemag(I_a0))
+printf("\n I_b0 = %.1f∠%.f° A", abs(I_b0),phasemag(I_b0))
+printf("\n I_c0 = %.1f∠%.f° A", abs(I_c0),phasemag(I_c0))
+printf("\nPositive sequence component of current for all phases are")
+printf("\n I_a1 = %.1f∠%.f° A", abs(I_a1),phasemag(I_a1))
+printf("\n I_b1 = %.1f∠%.f° A", abs(I_b1),360+phasemag(I_b1))
+printf("\n I_c1 = %.1f∠%.f° A", abs(I_c1),phasemag(I_c1))
+printf("\nNegative sequence component of current for all phases are")
+printf("\n I_a2 = %.1f∠%.f° A", abs(I_a2),phasemag(I_a2))
+printf("\n I_b2 = %.1f∠%.f° A", abs(I_b2),phasemag(I_b2))
+printf("\n I_c2 = %.1f∠%.f° A", abs(I_c2),360+phasemag(I_c2))
diff --git a/3472/CH29/EX29.9/Example29_9.sce b/3472/CH29/EX29.9/Example29_9.sce
new file mode 100644
index 000000000..57fc867a4
--- /dev/null
+++ b/3472/CH29/EX29.9/Example29_9.sce
@@ -0,0 +1,46 @@
+// 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 3: SYMMETRICAL COMPONENTS' ANALYSIS
+
+// EXAMPLE : 3.9 :
+// Page number 493-494
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+I_A = 1000.0 // Current through line A(A)
+I_C = 0 // Current through line C(A)
+
+// Calculations
+I_B = 1000.0*exp(%i*180.0*%pi/180) // Current through line B(A)
+a = exp(%i*120.0*%pi/180) // Operator
+I_a0 = 1.0/3*(I_A+I_B+I_C) // Zero sequence component of current(A)
+I_b0 = I_a0 // Zero sequence component of current(A)
+I_c0 = I_a0 // Zero sequence component of current(A)
+I_a1 = 1.0/3*(I_A+a*I_B+a**2*I_C) // Positive sequence component of current(A)
+I_b1 = a**2*I_a1 // Positive sequence component of current(A)
+I_c1 = a*I_a1 // Positive sequence component of current(A)
+I_a2 = 1.0/3*(I_A+a**2*I_B+a*I_C) // Negative sequence component of current(A)
+I_b2 = a*I_a2 // Negative sequence component of current(A)
+I_c2 = a**2*I_a2 // Negative sequence component of current(A)
+
+// Results
+disp("PART III - EXAMPLE : 3.9 : SOLUTION :-")
+printf("\nCurrent in line A, I_A = %.f∠%.f° A", abs(I_A),phasemag(I_A))
+printf("\nCurrent in line B, I_B = %.f∠%.f° A", abs(I_B),phasemag(I_B))
+printf("\nCurrent in line C, I_C = %.f A", I_C)
+printf("\nSymmetrical current components of line A are:")
+printf("\n I_a0 = %.f A", abs(I_a0))
+printf("\n I_a1 = %.1f∠%.f° A", abs(I_a1),phasemag(I_a1))
+printf("\n I_a2 = %.1f∠%.f° A", abs(I_a2),phasemag(I_a2))
+printf("\nSymmetrical current components of line B are:")
+printf("\n I_b0 = %.f A", abs(I_b0))
+printf("\n I_b1 = %.1f∠%.f° A", abs(I_b1),phasemag(I_b1))
+printf("\n I_b2 = %.1f∠%.f° A", abs(I_b2),phasemag(I_b2))
+printf("\nSymmetrical current components of line C are:")
+printf("\n I_c0 = %.f A", abs(I_c0))
+printf("\n I_c1 = %.1f∠%.f° A", abs(I_c1),phasemag(I_c1))
+printf("\n I_c2 = %.1f∠%.f° A", abs(I_c2),phasemag(I_c2))
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