20 files changed, 1329 insertions, 0 deletions
diff --git a/1092/CH8/EX8.1/Example8_1.sce b/1092/CH8/EX8.1/Example8_1.sce
new file mode 100755
index 000000000..25d7c9dcf
--- /dev/null
+++ b/1092/CH8/EX8.1/Example8_1.sce
@@ -0,0 +1,72 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-1
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// 3- phase Y-connected synchronous motor
+P = 20 ; // No. of poles
+hp = 40 ; // power rating of the synchronous motor in hp
+V_L = 660 ; // Line voltage in volt
+beta = 0.5 ; // At no-load, the rotor is retarded 0.5 mechanical degree from 
+// its synchronous position.
+X_s = 10 ; // Synchronous reactance in ohm
+R_a = 1.0 ; // Effective armature resistance in ohm
+
+// Calculations
+// case a
+funcprot(0); // To avoid this message "Warning : redefining function: beta"
+alpha = P * (beta/2); // The rotor shift from the synchronous position in
+// electrical degrees.
+
+// case b
+V_p = V_L / sqrt(3); // Phase voltage in volt
+E_gp = V_p ; // Generated voltage/phase at no-load in volt (given)
+E_r = (V_p - E_gp*cosd(alpha)) + %i*(E_gp*sind(alpha));
+ // Resultant emf across the armature per phase in V/phase
+E_r_m = abs(E_r);//E_r_m=magnitude of E_r in volt
+E_r_a = atan(imag(E_r) /real(E_r))*180/%pi;//E_r_a=phase angle of E_r in degrees
+
+// case c
+Z_s = R_a + %i*X_s ; // Synchronous impedance in ohm
+Z_s_m = abs(Z_s);//Z_s_m=magnitude of Z_s in ohm
+Z_s_a = atan(imag(Z_s) /real(Z_s))*180/%pi;//Z_s_a=phase angle of Z_s in degrees
+
+I_a  = E_r / Z_s ; // Armature current/phase in A/phase
+I_a_m = abs(I_a);//I_a_m=magnitude of I_a in A
+I_a_a = atan(imag(I_a) /real(I_a))*180/%pi;//I_a_a=phase angle of I_a in degrees 
+
+// case d
+theta = I_a_a ; // Phase angle between V_p and I_a in degrees
+P_p = V_p * I_a_m * cosd(theta); // Power per phase drawn by the motor from the bus
+P_t = 3*P_p ; // Total power drawn by the motor from the bus
+
+// csae e
+P_a = 3 * (I_a_m)^2 * R_a ; // Armature power loss at no-load in W
+P_d = (P_t - P_a)/746 ; // Internal developed horsepower at no-load
+
+// Display the results
+disp("Example 8-1 Solution : ");
+printf(" \n a: alpha = %d degrees (electrical degrees)\n",alpha );
+
+printf(" \n b: E_gp = %d V also, as given ",E_gp);
+printf(" \n    E_r in V/phase = ");disp(E_r);
+printf(" \n    E_r = %.1f <%.1f V/phase \n",E_r_m, E_r_a );
+
+printf(" \n c: Z_s in ohm/phase = ");disp(Z_s);
+printf(" \n    Z_s = %.2f <%.1f ohm/phase \n",Z_s_m, Z_s_a );
+printf(" \n    I_a in A/phase = ");disp(I_a);
+printf(" \n    I_a = %.2f <%.2f A/phase \n ",I_a_m, I_a_a);
+
+printf(" \n d: P_p = %.2f W/phase ",P_p );
+printf(" \n    P_t = %.2f W ",P_t);
+printf(" \n    Note: Slight variations in power values is due to slight variations");
+printf(" \n          in V_p , I_a and theta values from those of the textbook\n");
+
+printf(" \n e: P_a = %.f W ",P_a );
+printf(" \n    P_d = %d hp ", P_d );
diff --git a/1092/CH8/EX8.10/Example8_10.sce b/1092/CH8/EX8.10/Example8_10.sce
new file mode 100755
index 000000000..dfdd17198
--- /dev/null
+++ b/1092/CH8/EX8.10/Example8_10.sce
@@ -0,0 +1,56 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-10
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA = 10000 ; // kVA rating of a system
+cos_theta = 0.65 ; // power factor of the system
+sin_theta = sqrt( 1 - (cos_theta)^2 );
+cos_theta_b = 0.85 ; // Raised PF
+sin_theta_b = sqrt( 1 - (cos_theta_b)^2 );
+cost = 60 ; // cost of the synchronous capacitor to improve the PF in dollars/kVA
+// neglect the losses in the synchronous capacitor
+
+// Calculations
+// case a : For unity PF
+// at the original load
+kW_a = kVA * cos_theta ; //
+theta = acosd(cos_theta) ; // Power factor angle of the system in degrees
+kvar = kVA * sind(theta) ; // Reactive power in kvar
+kVA_a = kvar ;
+cost_cap_a = kvar * cost ; // Cost of raising the PF to unity PF in dollars
+
+// case b
+theta_b = acosd(cos_theta_b) ; // Power factor angle of the system in degrees
+kVA_b = kW_a / cos_theta_b ; // kVA value reduction
+kvar_b = kVA_b * sind(theta_b) ; // final kvar value reduced
+kvar_add = kvar - kvar_b ; // kvar of correction added
+
+cost_cap_b = kvar_add * cost ; // Cost of raising the PF to 0.85 PF in dollars
+
+// Display the results
+
+disp("Example 8-10 Solution : ");
+printf(" \n    Note : Slight variations in the kvar and cost values are due to  ");
+printf(" \n    non-approximation of theta values while calculating in scilab.\n");
+printf(" \n a: At the original load,\n");
+printf(" \n    kW = %d kW at theta = %.1f degrees \n", kW_a , theta );
+printf(" \n    kvar = %.3f kvar\n\n    For unity PF,",kvar);
+printf(" \n    kVA of synchronous capacitor = %.3f kVA (neglecting losses)\n",kVA_a);
+printf(" \n    Cost of synchronous capacitor = $%.f \n\n",cost_cap_a );
+
+printf(" \n b: For %.2f, PF = cos(%.1f), the total power,",cos_theta_b, theta_b);
+printf(" \n    %.f kW,remains the same. Therefore,\n ",kW_a);
+printf(" \n    kVA of final system reduced to = %.f kVA \n",kVA_b);
+printf(" \n    kvar of final system reduced to = %.f kvar \n    Therefore,",kvar_b);
+
+printf(" \n    kvar of correction added = %.3f kvar\n ",kvar_add);
+printf(" \n    kVA of synchronous capacitor = %.3f kVA (neglecting losses)\n",kvar_add);
+printf(" \n    Cost of synchronous capacitor = $%.f",cost_cap_b );
+printf(" \n    or less than half the cost in part(a)");
diff --git a/1092/CH8/EX8.11/Example8_11.sce b/1092/CH8/EX8.11/Example8_11.sce
new file mode 100755
index 000000000..ed54ffbd6
--- /dev/null
+++ b/1092/CH8/EX8.11/Example8_11.sce
@@ -0,0 +1,31 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-11
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+S_conjugate = 1000 ; // Apparent complex power in kVA
+cos_theta = 0.6 ; // lagging PF
+sin_theta = sqrt( 1 - (cos_theta)^2 );
+
+// Calculations
+// case a
+P_o = S_conjugate * cos_theta ; // Active power dissipated by the load in kW
+
+// case b
+jQ_o = S_conjugate * sin_theta ; // Inductive reactive quadrature power - 
+// - drawn from and returned to the supply
+
+// Display the results
+
+disp("Example 8-11 Solution : ");
+printf(" \n a: Active power \n    P_o = %d kW \n ", P_o );
+
+printf(" \n b: Inductive reactive quadrature power \n    +jQ_o in kvar =  \n");disp(%i*jQ_o);
+
+printf(" \n c: The original power triangle is shown in Fig.8-26a.");
diff --git a/1092/CH8/EX8.12/Example8_12.sce b/1092/CH8/EX8.12/Example8_12.sce
new file mode 100755
index 000000000..6efcd28af
--- /dev/null
+++ b/1092/CH8/EX8.12/Example8_12.sce
@@ -0,0 +1,49 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-12
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+S_conjugate = 1000 ; // Apparent complex power in kVA
+cos_theta_f = 0.8 ; // lagging PF
+sin_theta_f = sqrt( 1 - (cos_theta_f)^2 );
+
+// Calculated values from Ex.8-11
+P_o = 600 ; // Active power dissipated by the load in kW
+Q_o = 800 ; // Inductive reactive quadrature power - 
+// - drawn from and returned to the supply 
+
+// Calculations :
+ 
+// case a
+P_f = S_conjugate * cos_theta_f ; // Active power dissipated by the load in kW
+
+// case b
+Q_f = S_conjugate * sin_theta_f ; // Reactive quadrature power drawn from -
+//  - and returned to the supply
+
+// case c
+P_a = P_f - P_o ; // Additional active power in kW that may be supplied to -
+// - new customers
+
+// case d
+jQ_a = %i * ( Q_f ) - %i * ( Q_o ); // Correction kvar required to raise PF -
+// -from 0.6 to o.8 lagging
+
+// case e
+S_c_conjugate = 0 - jQ_a ; // Rating of correction capacitors needed for case d
+
+// Display the results
+
+disp("Example 8-12 Solution : ");
+printf(" \n a: P_f = %d kW \n ", P_f );
+printf(" \n b: +jQ_f in kvar = ");disp(%i*Q_f);
+printf(" \n c: P_a = %d kW \n ", P_a );
+printf(" \n d: jQ_a in kvar =   ");disp(jQ_a)
+printf(" \n e: S_c_conjugate = %d kVA \n ", abs(S_c_conjugate) );
+printf(" \n f: The power tabulation grid is shown in Fig.8-26b.");
diff --git a/1092/CH8/EX8.13/Example8_13.sce b/1092/CH8/EX8.13/Example8_13.sce
new file mode 100755
index 000000000..3ddf06d9b
--- /dev/null
+++ b/1092/CH8/EX8.13/Example8_13.sce
@@ -0,0 +1,59 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-13
+
+clear; clc; close; // Clear the work space and console.
+
+// Ex.8-12 PF
+cos_theta = 0.6 ; // PF lagging
+
+// Given data
+S_conjugate = 1000 ; // Apparent complex power in kVA
+cos_theta_f = 1.0 ; // unity PF
+sin_theta_f = sqrt( 1 - (cos_theta_f)^2 );
+
+// Calculated values from Ex.8-11
+P_o = 600 ; // Active power dissipated by the load in kW
+Q_o = 800 ; // Inductive reactive quadrature power - 
+// - drawn from and returned to the supply 
+
+// Calculations :
+ 
+// case a
+P_f = S_conjugate * cos_theta_f ; // Active power dissipated by the load in kW
+
+// case b
+Q_f = S_conjugate * sin_theta_f ; // Reactive quadrature power drawn from -
+//  - and returned to the supply
+
+// case c
+P_a = P_f - P_o ; // Additional active power in kW that may be supplied to -
+// - new customers
+
+// case d
+jQ_a = %i * ( Q_f ) - %i * ( Q_o ); // Correction kvar required to raise PF -
+// -from 0.6 to o.8 lagging
+Q_a = -abs(jQ_a); // 
+
+// case e
+S_c_conjugate = 0 - jQ_a ; // Rating of correction capacitors needed for case d
+
+// Display the results
+
+disp("Example 8-13 Solution : ");
+printf(" \n a: P_f = %d kW \n ", P_f );
+printf(" \n b: +jQ_f in kvar = ");disp(%i*Q_f);
+printf(" \n c: P_a = %d kW \n ", P_a );
+printf(" \n d: jQ_a in kvar =   ");disp(jQ_a)
+printf(" \n e: S_c_conjugate = %d kVA \n ", abs(S_c_conjugate) );
+printf(" \n f: The power tabulation grid is shown below.\n");
+printf(" \n    \t\t P \t ±jQ \t S* ");
+printf(" \n    \t\t(kW) \t(kvar) \t(kVA) \t cosӨ ");
+printf(" \n    ___________________________________________");
+printf(" \n    Original : \t %d \t +j%d \t %d \t %.1f ",P_o ,Q_o ,S_conjugate,cos_theta);
+printf(" \n    Added : \t %d \t %dj \t __ \t __",P_a ,Q_a );
+printf(" \n    Final : \t %d \t +j%d \t %d \t %.1f",P_f ,Q_f ,S_conjugate,cos_theta_f);
diff --git a/1092/CH8/EX8.14/Example8_14.sce b/1092/CH8/EX8.14/Example8_14.sce
new file mode 100755
index 000000000..abb791b02
--- /dev/null
+++ b/1092/CH8/EX8.14/Example8_14.sce
@@ -0,0 +1,69 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-14
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P_o = 2000 ; // load in kW drawn by a factory
+cos_theta_o = 0.6 ; // PF lagging
+sin_theta_o = sqrt( 1-  (cos_theta_o)^2 );
+cos_theta_f = 0.85 ; // final PF lagging required
+sin_theta_f = sqrt( 1-  (cos_theta_f)^2 );
+P_a = 275 ; // Losses in the synchronous capacitor in kW
+
+// Calculations
+// case a
+S_o_conjugate = P_o / cos_theta_o ; // Original kVA drawn from the utility 
+
+// case b
+Q_o = S_o_conjugate * sin_theta_o ; // Original lagging kvar
+
+// case c
+P_f = P_o + P_a ; // Final system active power consumed from the utility in kW
+
+// case d
+S_f_conjugate = P_f / cos_theta_f ; // Final kVA drawn from the utility 
+S_f_conjugate_a = acosd(cos_theta_f); // Phase angle of S_f_conjugate in degrees 
+
+// case e
+jQ_f = S_f_conjugate * sin_theta_f ; // Final lagging kvar
+jQ_a = %i*(jQ_f) - %i*(Q_o); // Correction kvar produced by the synchronous capacitor
+Q_a = abs(jQ_a); // Magnitude of jQ_a in kvar 
+
+// case f
+P = P_a ;
+S_a_conjugate = P -%i*(abs(jQ_a)); // kVA rating of the synchronous capacitor
+S_a_conjugate_m = abs(S_a_conjugate);//S_a_conjugate_m = magnitude of S_a_conjugate in kVA
+S_a_conjugate_a = atan(imag(S_a_conjugate) /real(S_a_conjugate))*180/%pi;
+//S_a_conjugate_a=phase angle of S_a_conjugate in degrees
+PF_f = cosd(S_a_conjugate_a); // PF
+
+// Display the results
+disp("Example 8-14 Solution : ");
+printf(" \n a: S*o = %.1f kVA \n",S_o_conjugate );
+
+printf(" \n b: Q*o in kvar = " );disp(%i*Q_o);
+
+printf(" \n c: P*f = %.f kW \n",P_f );
+
+printf(" \n d: S*f = %.1f <%.1f kVA\n ",S_f_conjugate,S_f_conjugate_a );
+
+printf(" \n e: jQ_f in kvar = ");disp(%i*jQ_f);
+printf(" \n    -jQ_a in kvar = ");disp(jQ_a);
+
+printf(" \n f: S*a = %.f <%.2f kVA ", S_a_conjugate_m , S_a_conjugate_a );
+printf(" \n    (cos(%.2f) = %.3f leading)\n",S_a_conjugate_a,PF_f);
+
+printf(" \n g: Power tabulation grid : \n ");
+printf(" \n    \t\t P \t ±jQ \t S* ");
+printf(" \n    \t\t(kW) \t(kvar) \t(kVA) \t cosӨ ");
+printf(" \n    ___________________________________________");
+printf(" \n    Original : \t %d \t +j%.f  %.1f  %.1f lag",P_o ,Q_o ,S_o_conjugate,cos_theta_o);
+printf(" \n    Added    : \t %d \t -%.fj  %.f \t %.3f lead",P_a ,Q_a,S_a_conjugate_m,cosd(S_a_conjugate_a) );
+printf(" \n    Final    : \t %d \t +j%.f  %.1f  %.2f lag",P_f ,jQ_f ,S_f_conjugate,cos_theta_f);
+
diff --git a/1092/CH8/EX8.15/Example8_15.sce b/1092/CH8/EX8.15/Example8_15.sce
new file mode 100755
index 000000000..d598f602a
--- /dev/null
+++ b/1092/CH8/EX8.15/Example8_15.sce
@@ -0,0 +1,87 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-15
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P_o = 2275 ; // Original kVA 
+Q_o = 1410 ; // Original kvar 
+S_f_conjugate = 3333.3 ; // final kVA of the load
+S_o_conjugate = P_o + %i*Q_o ; // Load of the alternator in kVA
+S_o_conjugate_m = abs(S_o_conjugate);//S_o_conjugate_m = magnitude of S_o_conjugate in kVA
+S_o_conjugate_a = atan(imag(S_o_conjugate) /real(S_o_conjugate))*180/%pi;
+//S_o_conjugate_a=phase angle of S_o_conjugate in degrees
+
+disp("Example 8-15");
+printf(" \n  Power tabulation grid : \n ");
+printf(" \n  \t\t P \t\t ±jQ \t\t S* ");
+printf(" \n  \t\t(kW) \t\t(kvar) \t\t(kVA) \t\t cosӨ ");
+printf(" \n  ________________________________________________________________________");
+printf(" \n  Original : \t%d \t\t j%.f \t\t %.1f \t%.2f lag",real(S_o_conjugate) ,imag(S_o_conjugate) ,S_o_conjugate_m,cosd(S_o_conjugate_a));
+printf(" \n  Added : \t0.8x \t\t j0.6x \t\t x \t\t0.80 lag" );
+printf(" \n  Final :    (%d + 0.8x) \tj(%.f + 0.6x)   %.1f \t0.841 lag\n",real(S_o_conjugate) ,imag(S_o_conjugate),S_f_conjugate );
+
+// Calculations
+// case a
+// Assume x is the additional kVA load. Then real and quadrature powers are 0.8x and j0.6x
+// respectively,as shown. Adding each column vertically and using the Pythagorean theorem,
+// we may write (2275 + 0.8x)^2 + (1410 + 0.6x)^2 = (3333.3)^2, and solving this eqution yields
+// the quadratic x^2 + 5352x -3947163 = 0. Applying the quadratic yields the added kVA load:
+x = poly(0,'x'); // Defining a polynomial with variable 'x' with  root at 0
+p = -3947163 + 5352*x + x^2 
+a = 1 ; // coefficient of x^2
+b = 5332 ; // coefficient of x
+c = -3947163 ; // constant
+
+// Roots of p
+x1 = ( -b + sqrt (b^2 -4*a*c ) ) /(2* a);
+x2=( -b - sqrt (b^2 -4*a*c ) ) /(2* a);
+
+// case b
+P_a = 0.8*x1 ; // Added active power of the additional load in kW
+Q_a = 0.6*x1 ; // Added reactive power of the additional load in kvar
+
+// case c
+P_f = P_o + P_a ; // Final active power of the additional load in kW
+Q_f = Q_o + Q_a ; // Final reactive power of the additional load in kvar 
+
+// case d
+PF = P_f / S_f_conjugate ; // Final power factor
+// Validity check
+S_conjugate_f = P_f + %i*Q_f ; // Final kVA of the load 
+S_conjugate_f_m = abs(S_conjugate_f);//S_conjugate_f_m = magnitude of S_conjugate_f in kVA
+S_conjugate_f_a = atan(imag(S_conjugate_f) /real(S_conjugate_f))*180/%pi;
+//S_conjugate_f_a=phase angle of S_conjugate_f in degrees
+
+// Display the results
+
+disp(" Solution : ")
+
+printf(" \n a: The given data is shown in the above power tabulation grid.Assume");
+printf(" \n    x is the additional kVA load. Then real and quadrature powers are");
+printf(" \n    0.8x and j0.6x respectively,as shown.Adding each column vertically");
+printf(" \n    and using the Pythagorean theorem, we may write");
+printf(" \n    (2275 + 0.8x)^2 + (1410 + 0.6x)^2 = (3333.3)^2, and solving this");
+printf(" \n    equation yields the quadratic as follows : \n");
+printf(" \n    x^2 + 5332x -3947163 = 0. \n ")
+printf(" \n    Applying the quadratic yields the added kVA load:");
+printf(" \n    Roots of quadratic Eqn p are \n ");
+printf(" \n    x1 = %.2f \n    x2 = %.2f ", x1, x2 );
+printf(" \n    Consider +ve value of x for added kVA so");
+printf(" \n    x = S*a = %.2f kVA \n ", x1 );
+
+printf(" \n b: P_a = %.1f kW \n ", P_a );
+printf(" \n    Q_a in kvar = \n");disp(%i*Q_a);
+
+printf(" \n c: P_f = %.1f kW \n ", P_f );
+printf(" \n    Q_f in kvar = \n");disp(%i*Q_f);
+
+printf(" \n d: PF = cosθ_f = %.3f lagging \n ", PF );
+printf(" \n    Validity check\n    S*f = ");disp(S_conjugate_f);
+printf(" \n    S*f = %.1f <%.2f kVA \n",S_conjugate_f_m,S_conjugate_f_a);
+printf(" \n    PF = cos(%.1f) = %.3f lagging",S_conjugate_f_a ,cosd(S_conjugate_f_a));
diff --git a/1092/CH8/EX8.16/Example8_16.sce b/1092/CH8/EX8.16/Example8_16.sce
new file mode 100755
index 000000000..37ef37246
--- /dev/null
+++ b/1092/CH8/EX8.16/Example8_16.sce
@@ -0,0 +1,45 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-16
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// Calculated values as per Ex.8-15 are as follows
+S_o_conjugate = 2676.5*exp(%i*31.79*(%pi/180)); // Original kVA rating
+S_o_conjugate_m = abs(S_o_conjugate);//S_o_conjugate_m = magnitude of S_o_conjugate in kVA
+S_o_conjugate_a = atan(imag(S_o_conjugate) /real(S_o_conjugate))*180/%pi;
+//S_o_conjugate_a=phase angle of S_o_conjugate in degrees
+
+S_a_conjugate = 658.86*exp(%i*36.87*(%pi/180)); // Added kVA rating
+S_a_conjugate_m = abs(S_a_conjugate);//S_a_conjugate_m = magnitude of S_a_conjugate in kVA
+S_a_conjugate_a = atan(imag(S_a_conjugate) /real(S_a_conjugate))*180/%pi;
+//S_a_conjugate_a=phase angle of S_a_conjugate in degrees
+
+S_f_conjugate = -3333.3*exp(%i*32.792687*(%pi/180)); // Final kVA rating
+S_f_conjugate_m = abs(S_f_conjugate);//S_f_conjugate_m = magnitude of S_f_conjugate in kVA
+S_f_conjugate_a = atan(imag(S_f_conjugate) /real(S_f_conjugate))*180/%pi;
+//S_f_conjugate_a=phase angle of S_f_conjugate in degrees
+
+// Calculations
+kVA_total = S_o_conjugate + S_a_conjugate + S_f_conjugate ; // Tellegan's theorem
+kVA_total_m = abs(kVA_total);//kVA_total_m = magnitude of kVA_total in kVA
+kVA_total_a = atan(imag(kVA_total) /real(kVA_total))*180/%pi;
+//kVA_total_a=phase angle of kVA_total in degrees
+
+// Display the result
+disp("Example 8-16 Solution : ");
+printf(" \n From the solution to Ex.8-15, we have ");
+printf(" \n S*o = %.1f <%.2f kVA \n ", S_o_conjugate_m,S_o_conjugate_a );
+printf(" \n S*a = %.1f <%.2f kVA \n ", S_a_conjugate_m,S_a_conjugate_a );
+printf(" \n S*f = %.1f <%.2f kVA \n ", S_f_conjugate_m,S_f_conjugate_a );
+
+printf(" \n Validity check ");
+printf(" \n S*o + S*a + S*f =  ");
+disp(S_o_conjugate),printf("  +"),disp(S_a_conjugate),printf("  +"),disp(S_f_conjugate);
+printf(" \n   = %d ",kVA_total );
+printf(" \n Hence, Tellegen`s theorem is proved");
diff --git a/1092/CH8/EX8.17/Example8_17.sce b/1092/CH8/EX8.17/Example8_17.sce
new file mode 100755
index 000000000..04e50959c
--- /dev/null
+++ b/1092/CH8/EX8.17/Example8_17.sce
@@ -0,0 +1,68 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-17
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kW = 40000 ; // Load on a factory in kW
+PF = 0.8 ; // power factor lagging of the load
+cos_theta = PF; 
+sin_theta = sqrt( 1 - (cos_theta)^2 );
+hp = 7500 ; // power rating of the induction motor in hp
+PF_IM = 0.75 ; // power factor lagging of the induction motor
+cos_theta_IM = PF_IM; 
+sin_theta_IM = sqrt( 1 - (cos_theta_IM)^2 );
+eta = 91*(1/100) ; // Efficiency of IM
+PF_SM = 1 ; // power factor  of the synchronous motor
+
+// Calculations
+kVA_original = kW / PF ; // Original kVA 
+kvar_original = kVA_original * sin_theta ; // Original kvar
+
+kW_IM = ( hp * 746 ) / ( 1000 * eta ) ; // Induction motor kW
+kVA_IM = kW_IM / PF_IM ; // Induction motor kVA
+kvar_IM = kVA_IM * sin_theta_IM ; // Induction motor kvar
+
+kvar_final = kvar_original - kvar_IM ; // final kvar
+kVA_final = kW + %i*(abs(kvar_final)); // final kVA
+kVA_final_m = abs(kVA_final);//kVA_final_m = magnitude of kVA_final in kVA
+kVA_final_a = atan(imag(kVA_final) /real(kVA_final))*180/%pi;
+//kVA_final_a=phase angle of kVA_final in degrees
+
+PF_final = cosd(kVA_final_a); // Final power factor
+
+// Display the result
+disp("Example 8-17 Solution : ");
+printf(" \n The synchronous motor operates at the same efficiency as the IM");
+printf(" \n that has been replaced, and therefore the total power of the system");
+printf(" \n is unchanged. The solution involves construction of table that shows ")
+printf(" \n the original condition of the system, the change, and the final condition.\n");
+printf(" \n Original kVA = %d kVA \n ", kVA_original );
+printf(" \n Original kvar = \n" );disp(%i*kvar_original);
+
+printf(" \n Induction motor kW = %d kW \n ", kW_IM );
+printf(" \n Induction motor kVA = %.f kVA \n ", kVA_IM );
+printf(" \n Induction motor kvar = ");disp(%i*kvar_IM)
+
+printf(" \n Final kvar = ");disp(%i*kvar_final);
+printf(" \n Final kVA = " );disp(kVA_final);
+printf(" \n Final kVA = %f <%.2f kVA \n ",kVA_final_m,kVA_final_a);
+
+printf(" \n Final PF = %.3f lagging \n ", PF_final );
+
+printf(" \n __________________________________________________________________________");
+printf(" \n Power tabulation grid : \n ");
+printf(" \n \t\t P \t\t ±jQ \t\t S* ");
+printf(" \n \t\t(kW) \t\t(kvar) \t\t(kVA) \t\t cosӨ ");
+printf(" \n __________________________________________________________________________");
+printf(" \n Original : \t%d \t\tj%.f \t\t%.1d \t\t %.1f lag",kW ,kvar_original ,kVA_original,PF);
+printf(" \n Removed  : \t-%.f \t\t-(+j%.f) \t%.f \t\t %.2f lag",kW_IM,kvar_IM,kVA_IM,PF_IM);
+printf(" \n Added    : \t+%.f  \t\t   0     \t%.1f \t\t 1.0 ",kW_IM,kW_IM);
+printf(" \n Final    : \t%d \t\tj%.f \t\t%.1f \t %.3f lag",kW ,kvar_final ,kVA_final_m,PF_final);
+printf(" \n __________________________________________________________________________");
+
diff --git a/1092/CH8/EX8.18/Example8_18.sce b/1092/CH8/EX8.18/Example8_18.sce
new file mode 100755
index 000000000..f191dd590
--- /dev/null
+++ b/1092/CH8/EX8.18/Example8_18.sce
@@ -0,0 +1,83 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-18
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kW = 40000 ; // Load on a factory in kW
+PF = 0.8 ; // power factor lagging of the load
+cos_theta = PF; 
+sin_theta = sqrt( 1 - (cos_theta)^2 );
+
+PF_SM = 0.8 ; // power factor leading of the synchronous motor
+cos_theta_SM = PF_SM; 
+sin_theta_SM = sqrt( 1 - (cos_theta_SM)^2 );
+hp = 7500 ; // power rating of the induction motor in hp
+
+PF_IM = 0.75 ; // power factor lagging of the induction motor
+cos_theta_IM = PF_IM; 
+sin_theta_IM = sqrt( 1 - (cos_theta_IM)^2 );
+
+eta = 91*(1/100) ; // Efficiency of IM
+
+// Calculations
+kVA_original = kW / PF ; // Original kVA 
+kvar_original = kVA_original * sin_theta ; // Original kvar
+
+
+kW_IM = ( hp * 746 ) / ( 1000 * eta ) ; // Induction motor kW
+kVA_IM = kW_IM / PF_IM ; // Induction motor kVA
+kvar_IM = kVA_IM * sin_theta_IM ; // Induction motor kvar
+
+// case a
+kW_SM = ( hp * 746 ) / ( 1000 * eta ) ; // Synchronous motor kW
+kVA_SM = kW_SM / PF_SM ; // Synchronous motor kVA
+kvar_SM = kVA_SM * sin_theta_SM ; // Synchronous motor kvar
+
+kvar_final = kvar_original - kvar_IM - kvar_SM ; // final kvar
+kVA_final = kW + %i*(abs(kvar_final)); // final kVA
+kVA_final_m = abs(kVA_final);//kVA_final_m = magnitude of kVA_final in kVA
+kVA_final_a = atan(imag(kVA_final) /real(kVA_final))*180/%pi;
+//kVA_final_a=phase angle of kVA_final in degrees
+
+PF_final = cosd(kVA_final_a); // Final power factor
+
+// Display the result
+disp("Example 8-18 Solution : ");
+
+printf(" \n Original kVA = %d kVA \n ", kVA_original );
+printf(" \n Original kvar = \n" );disp(%i*kvar_original);
+printf(" \n a:");
+printf(" \n Synchronous motor kW = %d kW \n ", kW_SM );
+printf(" \n Synchronous motor kVA = %.f kVA \n ", kVA_SM );
+printf(" \n Synchronous motor kvar = ");disp(-%i*kvar_SM)
+
+printf(" \n Final kvar = ");disp(%i*kvar_final);
+printf(" \n Final kVA = " );disp(kVA_final);
+printf(" \n Final kVA = %f <%.2f kVA \n ",kVA_final_m,kVA_final_a);
+
+printf(" \n Final PF = %.3f lagging \n ", PF_final );
+
+printf(" \n __________________________________________________________________________");
+printf(" \n Power tabulation grid : \n ");
+printf(" \n \t\t P \t\t ±jQ \t\t S* ");
+printf(" \n \t\t(kW) \t\t(kvar) \t\t(kVA) \t\t cosӨ ");
+printf(" \n __________________________________________________________________________");
+printf(" \n Original : \t%d \t\tj%.f \t\t%.1d \t\t %.1f lag",kW ,kvar_original ,kVA_original,PF);
+printf(" \n Removed  : \t-%.f \t\t-(+j%.f) \t%.f \t\t %.2f lag",kW_IM,kvar_IM,kVA_IM,PF_IM);
+printf(" \n Added    : \t+%.f  \t\t-j%.2f     \t%.1f \t\t %.1f lead",kW_SM,abs(kvar_SM),kVA_SM,PF_SM);
+printf(" \n Final    : \t%d \t\tj%.2f \t%.1f \t %.3f lag",kW ,kvar_final ,kVA_final_m,PF_final);
+printf(" \n __________________________________________________________________________\n\n");
+
+printf(" \n b: ");
+printf(" \n In Ex.8-17, a 6148 kVA, unity PF, 7500 hp synchronous motor is needed.");
+printf(" \n In Ex.8-18, a 7685 kVA, 0.8 PF leading, 7500 hp synchronous motor is needed.\n");
+printf(" \n \t Ex.8-18b shows that a 0.8 PF leading,7500 hp synchronous motor ");
+printf(" \n must be physically larger than a unity PF,7500 hp synchronous motor ");
+printf(" \n because of its higher kVA rating.");
+
diff --git a/1092/CH8/EX8.19/Example8_19.sce b/1092/CH8/EX8.19/Example8_19.sce
new file mode 100755
index 000000000..4284f9aad
--- /dev/null
+++ b/1092/CH8/EX8.19/Example8_19.sce
@@ -0,0 +1,70 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-19
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA_load = 500 ; // Load of 500 kVA
+PF_load = 0.65 ; // Load operates at this PF lagging
+cos_theta_load = PF_load ;
+sin_theta_load = sqrt(1 - (cos_theta_load)^2);
+hp = 200 ; // power rating of the system in hp
+eta = 88*(1/100); // Efficiency of the system after adding the load
+PF_final = 0.85 ; // Final lagging PF after adding the load
+
+// Calculations
+kW_original = kVA_load * cos_theta_load ; // Original kW
+kvar_original = kVA_load * sin_theta_load ; // Original kvar
+
+kW_SM = ( hp * 746 ) / ( 1000 * eta ) ; // Synchronous motor kW
+
+// case a
+kW_final = kW_original + kW_SM ; // final kW of the system with the motor added
+kVA_final = kW_final / PF_final ; // final kVA of the system with the motor added
+PF_system = kW_final / kVA_final ; // Final PF of the system with the motor added
+cos_theta_system = PF_system ; // Final PF of the system with the motor added
+sin_theta_system  = sqrt(1 - (cos_theta_system )^2);
+
+kvar_final = kVA_final * sin_theta_system ; // final kvar of the system with the motor added
+
+// case b
+kvar_SM = %i*kvar_final - %i*kvar_original ; // kvar rating of the sychronous motor
+
+kVA_SM = kW_SM + kvar_SM ; // kVA rating of the sychronous motor
+kVA_SM_m = abs(kVA_SM);//kVA_SM_m = magnitude of kVA_SM in kVA
+kVA_SM_a = atan(imag(kVA_SM) /real(kVA_SM))*180/%pi;
+//kVA_SM_a=phase angle of kVA_SM in degrees
+
+PF_SM = cosd(kVA_SM_a); // PF of the sychronous motor
+
+// Display the result
+disp("Example 8-19 Solution : ");
+
+printf(" \n Original kW = %.f kW \n ", kW_original );
+printf(" \n Original kvar = %.f kvar\n",kvar_original );
+printf(" \n Synchronous motor kW = %.1f kW \n ", kW_SM );
+
+printf(" \n a: Final kW = %.1f kW",kW_final);
+printf(" \n    Final kVA of the system = %.f kVA",kVA_final);
+printf(" \n    System PF = %.2f lagging",PF_system);
+printf(" \n    Final kvar of the system = j%d (lagging)kvar\n\n",kvar_final);
+
+printf(" \n b: Synchronous motor kvar = -%.2fj(leading)kvar\n",abs(kvar_SM));
+printf(" \n    Synchronous motor kVA = " );disp(kVA_SM);
+printf(" \n    Synchronous motor kVA = %.f <%.1f kVA \n ", kVA_SM_m , kVA_SM_a );
+printf(" \n    Synchronous motor PF = cos(%.1f) = %.3f leading \n ",kVA_SM_a,PF_SM );
+
+printf(" \n    ________________________________________________");
+printf(" \n    Power tabulation grid : \n ");
+printf(" \n    \t\t P \t ±jQ \t S* ");
+printf(" \n    \t\t(kW) \t(kvar) \t(kVA) \t cosӨ ");
+printf(" \n    ________________________________________________");
+printf(" \n    Original : \t %d \t +j%.f   %.1d \t %.2f lag",kW_original,kvar_original,kVA_load,PF_load);
+printf(" \n    Added    : \t %.1f \t -%.1fj  %.f \t %.4f lead",kW_SM ,abs(kvar_SM),kVA_SM_m,PF_SM);
+printf(" \n    Final    : \t %.1f \t +j%.f   %.f     %.2f lag",kW_final,kvar_final,kVA_final,PF_final);
+printf(" \n    ________________________________________________");
diff --git a/1092/CH8/EX8.2/Example8_2.sce b/1092/CH8/EX8.2/Example8_2.sce
new file mode 100755
index 000000000..19fcee1bc
--- /dev/null
+++ b/1092/CH8/EX8.2/Example8_2.sce
@@ -0,0 +1,74 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-2
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// 3- phase Y-connected synchronous motor
+P = 20 ; // No. of poles
+hp = 40 ; // power rating of the synchronous motor in hp
+V_L = 660 ; // Line voltage in volt
+beta = 5 ; // At no-load, the rotor is retarded 0.5 mechanical degree from 
+// its synchronous position.
+X_s = 10 ; // Synchronous reactance in ohm
+R_a = 1.0 ; // Effective armature resistance in ohm
+
+// Calculations
+// case a
+funcprot(0); // To avoid this message "Warning : redefining function: beta"
+alpha = P * (beta/2); // The rotor shift from the synchronous position in
+// electrical degrees.
+
+// case b
+V_p = V_L / sqrt(3); // Phase voltage in volt
+E_gp = V_p ; // Generated voltage/phase at no-load in volt (given)
+E_r = (V_p - E_gp*cosd(alpha)) + %i*(E_gp*sind(alpha));
+E_r_m = abs(E_r);//E_r_m=magnitude of E_r in volt
+E_r_a = atan(imag(E_r) /real(E_r))*180/%pi;//E_r_a=phase angle of E_r in degrees
+
+// case c
+Z_s = R_a + %i*X_s ; // Synchronous impedance in ohm
+Z_s_m = abs(Z_s);//Z_s_m=magnitude of Z_s in ohm
+Z_s_a = atan(imag(Z_s) /real(Z_s))*180/%pi;//Z_s_a=phase angle of Z_s in degrees
+
+I_a  = E_r / Z_s ; // Armature current/phase in A/phase
+I_a_m = abs(I_a);//I_a_m=magnitude of I_a in A
+I_a_a = atan(imag(I_a) /real(I_a))*180/%pi;//I_a_a=phase angle of I_a in degrees 
+
+// case d
+theta = I_a_a ; // Phase angle between V_p and I_a in degrees
+P_p = V_p * I_a_m * cosd(theta); // Power per phase drawn by the motor from the bus
+P_t = 3*P_p ; // Total power drawn by the motor from the bus
+
+// csae e
+P_a = 3 * (I_a_m)^2 * R_a ; // Armature power loss at no-load in W
+P_d = (P_t - P_a)/746 ; // Internal developed horsepower at no-load
+
+// Display the results
+disp("Example 8-2 Solution : ");
+printf(" \n a: alpha = %d degrees (electrical degrees)\n",alpha );
+
+printf(" \n b: E_gp = %d V also, as given ",E_gp);
+printf(" \n    E_r in V/phase = ");disp(E_r);
+printf(" \n    E_r = %d <%.1f V/phase \n",E_r_m, E_r_a );
+
+printf(" \n c: Z_s in ohm/phase = ");disp(Z_s);
+printf(" \n    Z_s = %.2f <%.1f ohm/phase \n",Z_s_m, Z_s_a );
+printf(" \n    I_a in A/phase = ");disp(I_a);
+printf(" \n    I_a = %.2f <%.2f A/phase \n ",I_a_m, I_a_a);
+
+printf(" \n d: P_p = %.2f W/phase ",P_p );
+printf(" \n    P_t = %.2f W ",P_t);
+printf(" \n    Note: Slight variations in power values is due to slight variations");
+printf(" \n          in V_p , I_a and theta values from those of the textbook\n");
+
+
+printf(" \n e: P_a = %.f W ",P_a );
+printf(" \n    P_d = %.1f hp ", P_d );
+
+
diff --git a/1092/CH8/EX8.20/Example8_20.sce b/1092/CH8/EX8.20/Example8_20.sce
new file mode 100755
index 000000000..1d5154286
--- /dev/null
+++ b/1092/CH8/EX8.20/Example8_20.sce
@@ -0,0 +1,56 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-20
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+f_a = 400 ; // Frequency of the alternator in Hz
+f_m = 60 ; // Frequency of the motor in Hz
+
+// Calculations
+Pole_ratio = f_a / f_m ; // Ratio of no. of poles in alternator to that of motor
+// Subscript 1 below indicates 1st combination
+P_a1 = 40 ; // first combination must have 40 poles on the alternator
+P_m1 = 6 ; // first combination must have 6 poles on the synchronous motor at a speed
+S_m1 = (120*f_m) / P_m1 ; // Speed of the motor in rpm
+
+// Subscript 2 below indicates 2nd combination
+P_a2 = 80 ; // second combination must have 40 poles on the alternator
+P_m2 = 12 ; // second combination must have 12 poles on the synchronous motor at a speed
+S_m2 = (120*f_m) / P_m2 ; // Speed of the motor in rpm
+
+// Subscript 13below indicates 3rd combination
+P_a3 = 120 ; // third combination must have 40 poles on the alternator
+P_m3 = 18 ; // third combination must have 18 poles on the synchronous motor at a speed
+S_m3 = (120*f_m) / P_m3 ; // Speed of the motor in rpm
+
+// Display the result
+disp("Example 8-20 Solution : ");
+
+printf(" \n Since P_a/P_m = f_a/f_m = %d/%d, or %d/%d, the ratio of",f_a,f_m,f_a/20,f_m/20);
+printf(" \n f_a/f_m determines the combinations of poles and speed.\n");
+printf(" \n Only even multiples of the above ratio are possible,since poles ");
+printf(" \n are always in pairs, hence first three combinations are as follows \n");
+
+printf(" \n The first combination must have %d poles on the alternator and ",P_a1);
+printf(" \n %d poles on the sychronous motor at a speed = %d rpm.\n",P_m1,S_m1);
+
+printf(" \n The second combination must have %d poles on the alternator and ",P_a2);
+printf(" \n %d poles on the sychronous motor at a speed = %d rpm.\n",P_m2,S_m2);
+
+printf(" \n The third combination must have %d poles on the alternator and ",P_a3);
+printf(" \n %d poles on the sychronous motor at a speed = %d rpm.\n",P_m3,S_m3);
+
+printf(" \n ___________________________________________________________________");
+printf(" \n Combination \t  Alternator Poles \t  Motor Poles \t Speed (rpm)");
+printf(" \n             \t         P_a       \t      P_m      \t     S      ");
+printf(" \n ___________________________________________________________________");
+printf(" \n First \t\t:\t  %d\t\t       %d \t    %d",P_a1,P_m1,S_m1);
+printf(" \n Second\t\t:\t  %d\t\t       %d \t    %d",P_a2,P_m2,S_m2);
+printf(" \n Third \t\t:\t  %d\t\t       %d \t    %d",P_a3,P_m3,S_m3);
+printf(" \n ___________________________________________________________________");
diff --git a/1092/CH8/EX8.3/Example8_3.sce b/1092/CH8/EX8.3/Example8_3.sce
new file mode 100755
index 000000000..a72a2afc2
--- /dev/null
+++ b/1092/CH8/EX8.3/Example8_3.sce
@@ -0,0 +1,110 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-3
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// 3- phase Y-connected synchronous motor
+P = 6 ; // No. of poles
+hp = 50 ; // power rating of the synchronous motor in hp
+V_L = 440 ; // Line voltage in volt
+X_s = 2.4 ; // Synchronous reactance in ohm
+R_a = 0.1 ; // Effective armature resistance in ohm
+alpha = 20 ; // The rotor shift from the synchronous position in
+// electrical degrees.
+E_gp_a = 240 ; // Generated voltage/phase in volt when the motor is under-excited(case a)
+E_gp_b = 265 ; // Generated voltage/phase in volt when the motor is under-excited(case b)
+E_gp_c = 290 ; // Generated voltage/phase in volt when the motor is under-excited(case c)
+
+// Calculations
+V_p = V_L / sqrt(3); // Phase voltage in volt
+// case a
+E_ra = (V_p - E_gp_a * cosd(alpha)) + %i*(E_gp_a * sind(alpha));
+E_ra_m = abs(E_ra);//E_ra_m=magnitude of E_ra in volt
+E_ra_a = atan(imag(E_ra) /real(E_ra))*180/%pi;//E_ra_a=phase angle of E_ra in degrees
+
+Z_s = R_a + %i*X_s ; // Synchronous impedance in ohm
+
+I_ap1  = E_ra / Z_s ; // Armature current/phase in A/phase
+I_ap1_m = abs(I_ap1);//I_ap1_m=magnitude of I_ap1 in A
+I_ap1_a = atan(imag(I_ap1) /real(I_ap1))*180/%pi;//I_ap1_a=phase angle of I_ap1 in degrees 
+
+cos_theta_a = cosd(I_ap1_a); // Power factor
+Ia_m1 = abs(I_ap1_m); // Absoulte value of magnitude of I_ap1
+
+P_d1 = 3 * (E_gp_a*Ia_m1) * cosd(160 - I_ap1_a); // // Internal developed power in W
+// 160 + I_ap1_a is the angle between E_gp_a and I_ap1
+Pd1 = abs(P_d1); // Consider absolute value of power in W for calculating hp
+
+Horse_power1 = Pd1 / 746 ; // Horsepower developed by the armature in hp 
+
+// case b
+E_rb = (V_p - E_gp_b * cosd(alpha)) + %i*(E_gp_b * sind(alpha));
+E_rb_m = abs(E_rb);//E_rb_m=magnitude of E_rb in volt
+E_rb_a = atan(imag(E_rb) /real(E_rb))*180/%pi;//E_rb_a=phase angle of E_rb in degrees
+
+I_ap2  = E_rb / Z_s ; // Armature current/phase in A/phase
+I_ap2_m = abs(I_ap2);//I_ap2_m=magnitude of I_ap2 in A
+I_ap2_a = atan(imag(I_ap2) /real(I_ap2))*180/%pi;//I_ap2_a=phase angle of I_ap2 in degrees 
+
+cos_theta_b = cosd(I_ap2_a); // Power factor
+Ia_m2 = abs(I_ap2_m); // Absoulte value of magnitude of I_ap2
+
+P_d2 = 3 * (E_gp_b*Ia_m2) * cosd(160 - I_ap2_a); // // Internal developed power in W
+// 160 + I_ap2_a is the angle between E_gp_b  and I_ap2
+Pd2 = abs(P_d2); // Consider absolute value of power in W for calculating hp
+
+Horse_power2 = Pd2 / 746 ; // Horsepower developed by the armature in hp 
+
+// case c
+E_rc = (V_p - E_gp_c * cosd(alpha)) + %i*(E_gp_c * sind(alpha));
+E_rc_m = abs(E_rc);//E_rc_m=magnitude of E_rc in volt
+E_rc_a = atan(imag(E_rc) /real(E_rc))*180/%pi;//E_rc_a=phase angle of E_rc in degrees
+
+I_ap3  = E_rc / Z_s ; // Armature current/phase in A/phase
+I_ap3_m = abs(I_ap3);//I_ap3_m=magnitude of I_ap3 in A
+I_ap3_a = atan(imag(I_ap3) /real(I_ap3))*180/%pi;//I_ap3_a=phase angle of I_ap3 in degrees 
+
+cos_theta_c = cosd(I_ap3_a); // Power factor
+Ia_m3 = abs(I_ap3_m); // Absoulte value of magnitude of I_ap3
+
+P_d3 = 3 * (E_gp_c*Ia_m3) * cosd(160 - I_ap3_a); // // Internal developed power in W
+// 160 + I_ap3_a is the angle between E_gp_c  and I_ap3
+Pd3 = abs(P_d3); // Consider absolute value of power in W for calculating hp
+
+Horse_power3 = Pd3 / 746 ; // Horsepower developed by the armature in hp 
+
+// Display the results
+disp("Example 8-3 Solution : ");
+disp("Slight variations in power values are because of non-approximation of I_a & cos(E_gp,I_a) values during power calculations in scilab ")
+printf(" \n a: V_p = %.f <0 V \n ",V_p);
+printf(" \n    E_r in V = ");disp(E_ra);
+printf(" \n    E_r = %.2f <%.2f V \n ",E_ra_m,E_ra_a);
+printf(" \n    I_ap in A = ");disp(I_ap1);
+printf(" \n    I_ap = %.2f <%.2f A \n", I_ap1_m , I_ap1_a );
+printf(" \n    cos(theta) = %.4f lagging \n ", cos_theta_a );
+printf(" \n    P_d = %d W drawn from bus(motor operation)\n", P_d1 );
+printf(" \n    Horsepower = %.1f hp \n\n", Horse_power1 );
+
+printf(" \n b: E_r in V = ");disp(E_rb);
+printf(" \n    E_r = %.2f <%.2f V \n ",E_rb_m,E_rb_a);
+printf(" \n    I_ap in A = ");disp(I_ap2);
+printf(" \n    I_ap = %.2f <%.2f A \n", I_ap2_m , I_ap2_a );
+printf(" \n    cos(theta) = %.4f = %.f(unity PF) \n ", cos_theta_b, cos_theta_b );
+printf(" \n    P_d = %d W drawn from bus(motor operation)\n", P_d2 );
+printf(" \n    Horsepower = %.1f hp \n\n", Horse_power2 );
+
+printf(" \n c: E_r in V = ");disp(E_rc);
+printf(" \n    E_r = %.2f <%.2f V \n ",E_rc_m,E_rc_a);
+printf(" \n    I_ap in A = ");disp(I_ap3);
+printf(" \n    I_ap = %.2f <%.2f A \n", I_ap3_m , I_ap3_a );
+printf(" \n    cos(theta) = %.4f leading \n ", cos_theta_c );
+printf(" \n    P_d = %d W drawn from bus(motor operation)\n", P_d3 );
+printf(" \n    Horsepower = %.1f hp \n\n", Horse_power3 );
+
+
diff --git a/1092/CH8/EX8.4/Example8_4.sce b/1092/CH8/EX8.4/Example8_4.sce
new file mode 100755
index 000000000..d3af817aa
--- /dev/null
+++ b/1092/CH8/EX8.4/Example8_4.sce
@@ -0,0 +1,90 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-4
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// Y-connected synchronous dynamo
+P = 2 ; // No. of poles
+hp = 1000 ; // power rating of the synchronous motor in hp
+V_L = 6000 ; // Line voltage in volt
+f = 60 ; // Frequency in Hz
+R_a = 0.52 ; // Effective armature resistance in ohm
+X_s = 4.2 ; // Synchronous reactance in ohm
+P_t = 811 ; // Input power in kW
+PF = 0.8 ; // Power factor leading
+
+// Calculations
+V_p = V_L / sqrt(3); // Phase voltage in volt
+
+// case a
+cos_theta = PF ; // Power factor leading
+I_L = (P_t*1000) / ( sqrt(3) * V_L * cos_theta); // Line armature current in A
+I_ap = I_L ; // Phase armature current in A
+
+// case b
+Z_p = R_a + %i * X_s ; // Impedance per phase in ohm
+Z_p_m = abs(Z_p);//Z_p_m=magnitude of Z_p in ohm
+Z_p_a = atan(imag(Z_p) /real(Z_p))*180/%pi;//Z_p_a=phase angle of Z_p in degrees
+
+// case c
+Ia_Zp = I_L * Z_p_m ; 
+E_r = Ia_Zp ; 
+
+// case d
+theta = acosd(0.8); // Power factor angle in degrees
+
+// case e
+funcprot(0); // Use  to avoid this message "Warning : redefining function: beta"                    . 
+beta = Z_p_a ; // 
+deba = beta + theta // Difference angle at 0.8 leading PF in degrees
+
+// case f
+//  Generated voltage/phase in volt 
+E_gp_f = sqrt( (E_r)^2 + (V_p)^2 - 2*E_r*V_p*cosd(deba) );
+
+// case g
+//  Generated voltage/phase in volt 
+E_gp_g = ( V_p + Ia_Zp * cosd(180-deba) ) + %i * ( Ia_Zp * sind(180-deba) ); 
+E_gp_g_m = abs(E_gp_g);//E_gp_g_m=magnitude of E_gp_g in volt
+E_gp_g_a = atan(imag(E_gp_g) /real(E_gp_g))*180/%pi;//E_gp_g_a=phase angle of E_gp_g in degrees
+
+// case h
+IaZp = Ia_Zp * expm(%i * Z_p_a * (%pi/180) ); // voltage generated by alternator 1 in volt
+IaZp_m = abs(IaZp);//IaZp_m=magnitude of IaZp in A
+IaZp_a = atan(imag(IaZp) /real(IaZp))*180/%pi;//IaZp_a=phase angle of IaZp in degrees
+IaRa = IaZp_m*cosd(IaZp_a); // Real part of IaZp
+IaXs = IaZp_m*sind(IaZp_a); // Imaginery part of IaZp
+
+cos_theta = PF ; //
+sin_theta = sqrt( 1 - (cos_theta)^2 );
+//  Generated voltage/phase in volt 
+E_gp_h = ( V_p * cos_theta - IaRa ) + %i * ( V_p * sin_theta + IaXs);
+E_gp_h_m = abs(E_gp_h);//E_gp_h_m=magnitude of E_gp_h in volt
+E_gp_h_a = atan(imag(E_gp_h) /real(E_gp_h))*180/%pi;//E_gp_h_a=phase angle of E_gp_h in degrees
+
+// Display the results
+disp("Example 8-4 Solution : ");
+printf(" \n a: I_L = %.2f \n    I_ap = %.2f A \n", I_L, I_ap );
+
+printf(" \n b: Z_p in ohm = ");disp(Z_p);
+printf(" \n    Z_p = %.3f <%.2f ohm \n ", Z_p_m , Z_p_a );
+
+printf(" \n c: IaZp = %.1f V \n    E_r = %.1f V \n ",Ia_Zp , E_r );
+
+printf(" \n d: Power factor angle,\n    theta = %.2f degrees leading \n ", theta );
+
+printf(" \n e: Difference angle,\n    deba = %.2f degrees \n ", deba );
+
+printf(" \n f: E_gp = %.f V \n ", E_gp_f );
+ 
+printf(" \n g: E_gp in V = ");disp(E_gp_g );
+printf(" \n    E_gp = %d <%.2f V \n",E_gp_g_m , E_gp_g_a );
+
+printf(" \n h: E_gp in V = ");disp(E_gp_h);
+printf(" \n    E_gp = %.f <%.2f V",E_gp_h_m, E_gp_h_a );
diff --git a/1092/CH8/EX8.5/Example8_5.sce b/1092/CH8/EX8.5/Example8_5.sce
new file mode 100755
index 000000000..db33e8140
--- /dev/null
+++ b/1092/CH8/EX8.5/Example8_5.sce
@@ -0,0 +1,51 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-5
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// Y-connected synchronous dynamo
+P = 2 ; // No. of poles
+hp = 1000 ; // power rating of the synchronous motor in hp
+V_L = 6000 ; // Line voltage in volt
+f = 60 ; // Frequency in Hz
+R_a = 0.52 ; // Effective armature resistance in ohm
+X_s = 4.2 ; // Synchronous reactance in ohm
+P_t = 811 ; // Input power in kW
+PF = 0.8 ; // Power factor leading
+
+// Calculated values
+E_gp = 3687 ; // Generated voltage/phase in volt 
+V_p = V_L / sqrt(3); // Phase voltage in volt
+E_r = 412.8 ; // Resultant EMF across armature/phase in volt
+deba = 119.81 ; // Difference angle at 0.8 leading PF in degrees
+theta = 36.87 ; // Power factor angle in degrees
+IaXs = 409.7 ; // Voltage drop across synchronous reactance in volt
+IaRa = 50.74 ; // Voltage drop across armature resistance in volt
+
+// Calculations
+
+// Torque angle alpha in degrees calculated by different Eqns
+// case a
+alpha1 = acosd( ( E_gp^2 + V_p^2 - E_r^2 ) / ( 2*E_gp*V_p ) ); // Eq.8-12
+
+// case b
+alpha2 = asind( ( E_r * sind(deba) ) / ( E_gp ) ); // Eq.8-13
+
+// case c
+alpha3 = theta - atand( (V_p*sind(theta) + IaXs) / (V_p*cosd(theta) - IaRa) );// Eq.8-14
+
+// Display the results
+disp("Example 8-5 Solution : ");
+printf(" \n a: Using Eq.(8-12) \n    alpha = %.2f degrees \n ", alpha1 );
+
+printf(" \n b: Using Eq.(8-13) \n    alpha = %.2f degrees \n ", alpha2 );
+
+printf(" \n c: Using Eq.(8-14) \n    alpha = %.2f degrees \n ", alpha3 );
+printf(" \n    Slight variation in case c alpha is due to tan inverse value ");
+printf(" \n    which was calulated to be 42.445604 degrees, instead of 42.44 degrees(textbook).")
diff --git a/1092/CH8/EX8.6/Example8_6.sce b/1092/CH8/EX8.6/Example8_6.sce
new file mode 100755
index 000000000..5e3e4e3d1
--- /dev/null
+++ b/1092/CH8/EX8.6/Example8_6.sce
@@ -0,0 +1,59 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-6
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data as per Example 8-4
+// Y-connected synchronous dynamo
+P = 2 ; // No. of poles
+hp = 1000 ; // power rating of the synchronous motor in hp
+V_L = 6000 ; // Line voltage in volt
+f = 60 ; // Frequency in Hz
+R_a = 0.52 ; // Effective armature resistance in ohm
+X_s = 4.2 ; // Synchronous reactance in ohm
+P_t = 811 ; // Input power in kW
+PF = 0.8 ; // Power factor leading
+
+// Calculated values from Example 8-4
+E_gp = 3687 ; // Generated voltage/phase in volt 
+
+I_a = 97.55 ; // Phase armature current in A
+
+phi = (42.45 - 0); // Phase angle between E_gp and I_a in degrees 
+// where 42.45 and 0 are phase angles of E_gp and I_a in degrees respectively.
+
+// Calculations
+// case a 
+P_p = E_gp * I_a * cosd(phi) / 1000; // Mechanical power developed per phase in kW
+
+P_t_a = 3 * P_p ; // Total mechanical power developed in kW
+
+// case b
+P_t_b = P_t_a / 0.746 ; // Internal power developed in hp at rated load
+
+// case c
+S = 120 * f / P ; // Speed of the motor in rpm
+T_int = ( P_t_b * 5252 ) / S ; // Internal torque developed in lb-ft
+
+// case d
+T_ext = ( hp * 5252 ) / 3600 ; // External torque developed in lb-ft
+eta = (T_ext / T_int) * 100 ; // Motor efficiency in percent
+
+// Display the results
+disp("Example 8-6 Solution : ");
+printf(" \n a: Similar to a dc motor, the mechanical power developed in the armature");
+printf(" \n    is the product of the induced EMF per phase, the armature current");
+printf(" \n    per phase, and the cosine of the angle between them.\n");
+printf(" \n    P_p = %.3f kW \n    P_t = %.1f kW \n", P_p, P_t_a );
+
+printf(" \n b: P_t = %.1f hp \n ", P_t_b );
+
+printf(" \n c: T_int = %.f lb-ft \n ", T_int );
+
+printf(" \n d: T_ext = %d lb-ft  \n", T_ext );
+printf(" \n    Motor Efficiency,\n    eta = %.1f percent ", eta );
diff --git a/1092/CH8/EX8.7/Example8_7.sce b/1092/CH8/EX8.7/Example8_7.sce
new file mode 100755
index 000000000..d89d99b8c
--- /dev/null
+++ b/1092/CH8/EX8.7/Example8_7.sce
@@ -0,0 +1,95 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-7
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P_o = 2000 ; // Total power consumed by a factory in kW from the transformer
+cos_theta = 0.6 ; // 0.6 lagging power factor at which power is consumed -
+// - from the transformer
+sin_theta = sqrt(1 - (cos_theta)^2);
+theta = -acosd(0.6); // power factor angle at which power is consumed -
+// - from the transformer in degrees
+
+V_L = 6000 ; // Primary line voltage of a transformer in volt
+
+P = 750 ; // kW expected to be delivered by the dc motor-generator
+
+hp = 1000 ; // hp rating of the motor(induction or synchronous)
+V_L_m = 6000 ; // Line voltage of a synchronous(or induction) motor in volt
+cos_theta_sm = 0.8 ; // 0.8 leading power factor of the synchronous motor
+theta_sm = acosd(0.8); // power factor angle of the synchronous motor in degrees
+
+cos_theta_im = 0.8 ; // 0.8 lagging power factor of the induction motor
+theta_im = -acosd(0.8); // power factor angle of the induction motor in degrees
+
+eta = 0.92 ; // Efficiency of each motor
+
+// Calculations
+// case a : using Induction Motor(IM)
+P_m = ( hp * 746 ) / eta ; // Induction(or synchronous) motor load in W
+I_1 = P_m / ( sqrt(3) * V_L_m * cos_theta_im ); // Lagging current drawn by IM in A
+
+I_1_prime = P_o * 1000 / ( sqrt(3) * V_L * cos_theta ); // Original lagging  -
+// - factory load current in A
+
+// Total load current in A using Induction Motor :
+I_TM = I_1*(cosd(theta_im) + %i*sind(theta_im)) + I_1_prime*(cosd(theta) + %i*sind(theta)) ; 
+I_TM_m = abs(I_TM);//I_TM_m = magnitude of I_TM in A
+I_TM_a = atan(imag(I_TM) /real(I_TM))*180/%pi;//I_TM_a=phase angle of I_TM in degrees
+
+PF_im = cosd(I_TM_a); // Overall PF using induction motor
+
+// case b: using synchronous motor
+I_s1 = P_m / ( sqrt(3) * V_L_m * cos_theta_sm ); // Lagging current drawn by IM in A
+
+// Total load current in A using synchronous motor :
+I_TSM = I_s1*(cosd(theta_sm) + %i*sind(theta_sm)) + I_1_prime*(cosd(theta) + %i*sind(theta)) ; 
+I_TSM_m = abs(I_TSM);//I_TSM_m = magnitude of I_TSM in A
+I_TSM_a = atan(imag(I_TSM) /real(I_TSM))*180/%pi;//I_TSM_a=phase angle of I_TSM in degrees
+
+PF_sm = cosd(I_TSM_a); // Overall PF using Synchronous motor
+
+// case c 
+percent_I_L = ( I_TM_m - I_TSM_m ) / I_TM_m * 100 ; // Percent reduction in  -
+// - total load current in percent
+
+// Display the results
+printf("Note : case a,I1 calculated is around 97.53 A instead of 47.53 A(textbook).\n")
+printf(" Note : case b,Actual I_s1 imaginary part is around 58.52 instead of ");
+printf(" \n        52.52(textbook)so slight variation in I_TSM and percent ")
+printf(" \n        reduction in total load current.\n")
+
+disp("Example 8-7 Solution : ");
+printf(" \n a: Induction(or sunchronous) motor load");
+printf(" \n    P_m = %.f W ",P_m);
+printf(" \n    Lagging current drawn by the IM = I1");
+printf(" \n    I_1 = %.2f <-%.2f A \n",I_1,acosd(cos_theta_sm));
+printf(" \n    I_1 in A = ");disp(I_1*cosd(-36.87)+%i*I_1*sind(-36.87));
+printf(" \n    Original lagging factory load current = I_1_prime");
+printf(" \n    I_1_prime in A = ");disp(I_1_prime*cosd(theta)+%i*I_1_prime*sind(theta));
+printf(" \n    I_1_prime = %.1f <-%.2f A \n",I_1_prime,acosd(cos_theta));
+printf(" \n    Total load current = motor load + factory load");
+printf(" \n    I_TM = I_1 + I_1_prime\n");
+printf(" \n    I_TM in A = ");disp(I_TM);
+printf(" \n    I_TM = %.1f <%.1f A \n ",I_TM_m , I_TM_a );
+printf(" \n    Overall system PF = %.4f lagging \n ", PF_im );
+
+printf(" \n b: Synchronous motor load\n    I_s1 = %.2f <%.2f A\n",I_1,acosd(cos_theta_sm));
+printf(" \n    I_s1 in A = ");disp(I_s1*cosd(36.87)+%i*I_s1*sind(36.87));
+printf(" \n    Total load current : I_TSM = I_s1 + I_1_prime \n");
+printf(" \n    I_TSM in A = ");disp(I_TSM);
+printf(" \n    I_TSM = %.1f <%.1f A \n ",I_TSM_m , I_TSM_a );
+printf(" \n    Overall system PF = %.1f lagging \n ", PF_sm );
+
+printf(" \n c: Percent reduction in total load current = %.1f percent \n",percent_I_L);
+
+printf(" \n d: PF improvement: Using the synchronous motor ( in lieu of the IM)");
+printf(" \n    raises the total system PF from %.4f lagging to %.1f lagging.",PF_im,PF_sm);
+
+
diff --git a/1092/CH8/EX8.8/Example8_8.sce b/1092/CH8/EX8.8/Example8_8.sce
new file mode 100755
index 000000000..c4442f8b2
--- /dev/null
+++ b/1092/CH8/EX8.8/Example8_8.sce
@@ -0,0 +1,41 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-8
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data from Ex.8-3a
+// 3- phase Y-connected synchronous motor
+P = 6 ; // No. of poles
+hp = 50 ; // power rating of the synchronous motor in hp
+V_L = 440 ; // Line voltage in volt
+X_s = 2.4 ; // Synchronous reactance in ohm
+R_a = 0.1 ; // Effective armature resistance in ohm
+alpha = 20 ; // The rotor shift from the synchronous position in
+// electrical degrees.
+E_gp = 240 ; // Generated voltage/phase in volt when the motor is under-excited
+f = 60 ; // Frequency in Hz
+
+// Calculated values from Example 8-3a
+V_p = 254 ; // Phase voltage in volt
+
+// Calculations
+// case a
+// Torque developed per phase Using Eq.(8-17a)
+S = 120 * f / P ; // Speed of the motor in rpm
+T_p = ( 7.04 * E_gp * V_p ) / ( S*X_s) * sind(alpha); 
+
+// case b
+// Total horsepower developed using part a
+Horsepower = ( 3*T_p*S )/5252; 
+
+// Display the results
+disp("Example 8-8 Solution : ");
+printf(" \n From given and calculated data of Ex.8-3a,\n");
+printf(" \n a: T_p = %.2f lb-ft \n ", T_p );
+
+printf(" \n b: Horsepower = %.1f hp ", Horsepower );
diff --git a/1092/CH8/EX8.9/Example8_9.sce b/1092/CH8/EX8.9/Example8_9.sce
new file mode 100755
index 000000000..6eb6e417a
--- /dev/null
+++ b/1092/CH8/EX8.9/Example8_9.sce
@@ -0,0 +1,64 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-9
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P_o = 2000 ; // Total power consumed by a factory in kW 
+cos_theta = 0.6 ; // 0.6  power factor at which power is consumed 
+sin_theta = sqrt( 1 - (cos_theta)^2 );
+V = 6000 ; // Line voltage in volt
+// Synchronous capacitor is used to raise the overall PF to unity
+P_loss_cap = 275 ; // Synchronous capacitor losses in kW
+
+// Calculations
+// case a
+S_o_conjugate = P_o / cos_theta ; // apparent complex power in kW
+jQ_o = S_o_conjugate * sin_theta ; // Original kilovars of lagging load
+
+// case b
+jQ_c = -jQ_o ; // Kilovars of correction needed to bring the PF to unity
+
+// case c
+R = P_loss_cap ; // Synchronous capacitor losses in kW
+S_c_conjugate = R - %i*( abs(jQ_c) ) ; // kVA rating of the synchronous capacitor
+S_c_conjugate_m = abs(S_c_conjugate);//S_c_conjugate_m = magnitude of S_c_conjugate in kVA
+S_c_conjugate_a = atan(imag(S_c_conjugate) /real(S_c_conjugate))*180/%pi;
+//S_c_conjugate_a=phase angle of S_c_conjugate in degrees 
+PF = cosd(S_c_conjugate_a); // Power factor of the synchronous capacitor
+
+// case d
+I_o = S_o_conjugate * 1000 / V ; // Original current drawn from the mains in A
+
+
+// case e
+P_f = P_o + P_loss_cap ; // Total power in kW
+S_f = P_f ; // Total apparent power in kW
+S_f_m = abs(S_f);//S_f_m = magnitude of S_f in A
+S_f_a = atan(imag(S_f) /real(S_f))*180/%pi;//S_f_a=phase angle of S_f in degrees 
+
+I_f = S_f * 1000 / V ; // Final current drawn from the mains after correction in A
+
+// Display the results
+disp("Example 8-9 Solution : ");
+printf(" \n a: S*o = %d kVA \n", S_o_conjugate );
+printf(" \n    +jQo in kvar = ");disp(%i*jQ_o);
+
+printf(" \n b: -jQc in kvar =  " );disp(%i*jQ_c);
+
+printf(" \n c: S*c in kVA = ");disp(S_c_conjugate);
+printf(" \n    S*c = %.f <%.1f kVA \n", S_c_conjugate_m , S_c_conjugate_a );
+printf(" \n    PF = %.3f leading \n",PF );
+
+printf(" \n d: I_o = %.1f A \n ",I_o );
+
+printf(" \n e: S_f in A = ");disp(S_f);
+printf(" \n    S_f = %d <%d kVA \n" , S_f_m , S_f_a );
+printf(" \n    I_f = %.1f A \n ", I_f);
+
+printf(" \n f: See Fig.8-25.");