22 files changed, 700 insertions, 0 deletions
diff --git a/1319/CH5/EX5.1/5_1.sce b/1319/CH5/EX5.1/5_1.sce
new file mode 100644
index 000000000..0781b2be2
--- /dev/null
+++ b/1319/CH5/EX5.1/5_1.sce
@@ -0,0 +1,19 @@
+// To find flux density in the core and induced emf in the secondary winding
+
+clc;
+clear;
+
+E1=500;
+A=60*(10^-4);
+f=50;
+N1=400;
+N2=1000;
+
+// E=4.44*f*N*Bm*A Induced EMF equation
+
+Bm=E1/(4.44*f*N1*A);
+
+E2=4.44*f*N2*Bm*A;
+
+printf('a) The peak value of the flux density in the core = %f tesla \n',Bm)
+printf('b) The voltage induced in the secondary winding = %f V \n',E2)
diff --git a/1319/CH5/EX5.10/5_10.sce b/1319/CH5/EX5.10/5_10.sce
new file mode 100644
index 000000000..a9aef1fc1
--- /dev/null
+++ b/1319/CH5/EX5.10/5_10.sce
@@ -0,0 +1,30 @@
+//Calculate efficiency on unity pf at different cases
+
+clc;
+clear;
+
+Pi=25*(10^3);
+
+E1=2000;
+E2=200;
+
+Pil=350;
+Pc=400;
+
+// Full load efficiency
+
+nfl=Pi*100/(Pi+Pil+Pc);
+
+//Half Load efficiency
+
+Pihl=Pi/2;// Half Load
+nhl=Pihl*100/(Pihl+Pil+(Pc/4));
+
+// Load at which maximum efficiency occurs
+
+Piml=sqrt(Pil/Pc)*Pi;
+Pcm=Pc*((Piml/Pi)^2);
+
+printf('a) Efficiency at full load = %f percent \n',nfl)
+printf('b) Efficiency at half load = %f percent \n',nhl)
+printf('c) Maximum Efficiency will occur at %f KVA and the losses are each %d watt. \n',(Piml/1000),Pcm)
diff --git a/1319/CH5/EX5.11/5_11.sce b/1319/CH5/EX5.11/5_11.sce
new file mode 100644
index 000000000..a772cbc98
--- /dev/null
+++ b/1319/CH5/EX5.11/5_11.sce
@@ -0,0 +1,23 @@
+//Calcualte efficiencies at various loads
+
+clc;
+clear;
+
+P=100*(10^3);// Power Input
+Pc=1000;// Copper Loss
+Pil=1000;// Iron Loss
+pf=0.8;
+
+deff('y=unity(x)','y=(P*100*x)/((P*x)+Pil+(Pc*(x^2)))')// Unit Power Factor
+deff('y=pfactor(x)','y=(P*100*x*pf)/((P*pf*x)+Pil+(Pc*(x^2)))')// 0.8 p.f
+
+printf('a) Unity power factor efficiencies at \n \n')
+printf('i) Half of full load = %f percent \n',unity(1/2))
+printf('ii) Full load = %f percent \n',unity(1))
+printf('iii) (5/4) of full load = %f percent \n \n',unity(5/4))
+
+
+printf('b) 0.8 power factor efficiencies at \n \n')
+printf('i) Half of full load = %f percent \n',pfactor(1/2))
+printf('ii) Full load = %f percent \n',pfactor(1))
+printf('iii) (5/4) of full load = %f percent \n',pfactor(5/4))
diff --git a/1319/CH5/EX5.12/5_12.sce b/1319/CH5/EX5.12/5_12.sce
new file mode 100644
index 000000000..4f431998d
--- /dev/null
+++ b/1319/CH5/EX5.12/5_12.sce
@@ -0,0 +1,40 @@
+// To determine all day efficiency
+
+clc;
+clear;
+
+p=15*(10^3);
+t1=12;
+t2=6;
+t3=6;
+
+pf1=0.5;
+pf2=0.8;
+pf3=0.9;
+
+x=poly([0 1],'x','c');
+
+nm=0.98; // Max Efficiency
+
+y=(nm*(p+(2*x)))-p;
+
+x=roots(y); // To find the iron loss or copper loss at unity p.f for maximum efficiency
+
+Pil=x; // Iron loss
+
+Pc=x; // Copper Loss at unity p.f for maximum efficiency
+
+deff('a=culoss(b,c)','a=b*Pc*((c/(p/1000))^2)');
+
+Pc1=culoss(12,(2/pf1)); // Total Copper Loss for 12hrs - 2 kW at p.f 0.5
+Pc2=culoss(6,(12/pf2)); // Total Copper Loss for 6hrs - 12 kW at p.f 0.8
+Pc3=culoss(6,(18/pf3)); // Total Copper Loss for 6hrs - 18 kW at p.f 0.9
+
+Po=((12*2)+(6*12)+(6*18))*(10^3);// Power Output
+
+eff=Po*100/(Po+(Pc1+Pc2+Pc3)+(24*Pil));
+
+// Note the iron loss has to be considered to calculate the Efficiency, Text Error
+
+printf('The all day effciency = %f percent \n',eff)
+
diff --git a/1319/CH5/EX5.13/5_13.sce b/1319/CH5/EX5.13/5_13.sce
new file mode 100644
index 000000000..28b8960f2
--- /dev/null
+++ b/1319/CH5/EX5.13/5_13.sce
@@ -0,0 +1,13 @@
+// Calculating Efficiency using Sumpner test
+
+clc;
+clear;
+
+P=200*(10^3);
+
+W1= 4*(10^3); // Total iron loss for both the transformers
+W2= 6*(10^3); // Total copper loss for both the transformers
+
+n=P*100/(P+(W1/2)+(W2/2));// Efficiency
+
+printf('The Efficiency of each transformer at full load = %f percent \n',n)
diff --git a/1319/CH5/EX5.14/5_14.sce b/1319/CH5/EX5.14/5_14.sce
new file mode 100644
index 000000000..9aa16254c
--- /dev/null
+++ b/1319/CH5/EX5.14/5_14.sce
@@ -0,0 +1,12 @@
+// To determine the ratio of weights of copper
+
+clc;
+clear;
+
+n=3; //transformation ratio
+
+// Ratio of weights of copper in an ato transformer and a two winding transformer
+
+roc=(1-(1/n));
+
+printf('Ratio of weights of copper in an ato transformer and a two winding transformer = %f \n',roc)
diff --git a/1319/CH5/EX5.15/5_15.sce b/1319/CH5/EX5.15/5_15.sce
new file mode 100644
index 000000000..4f29e6c32
--- /dev/null
+++ b/1319/CH5/EX5.15/5_15.sce
@@ -0,0 +1,20 @@
+// To find voltage ratio and output
+
+clc;
+clear;
+
+E1=11500;
+E2=2300;
+
+n1=(E1+E2)/E1; // Voltage ratio of 13.8 kV/11.5 kV auto transformer
+
+Pi=100*(10^3);
+
+P1=Pi*n1/(n1-1);
+
+n2=(E1+E2)/E2; // Voltage ratio of 13.8 kV/2.3 kV auto transformer
+
+P2=Pi*n2/(n2-1);
+
+printf('The transformation ratio of the auto transformer is %g and is rated %g / %g kV, %g KVA \n',n1,(E1+E2)/1000,E1/1000,P1/1000)
+printf('The transformation ratio of the auto transformer is %g and is rated %g / %g kV, %g KVA \n',n2,(E1+E2)/1000,E2/1000,P2/1000)
diff --git a/1319/CH5/EX5.16/5_16.sce b/1319/CH5/EX5.16/5_16.sce
new file mode 100644
index 000000000..4a4144ca6
--- /dev/null
+++ b/1319/CH5/EX5.16/5_16.sce
@@ -0,0 +1,31 @@
+// Determine primary and secondary voltages and current
+
+clc;
+clear;
+
+R1=100;
+R2=40;
+
+P=2; // Power
+
+r=sqrt(R2/R1); // n2/n1 Turns ratio
+
+if(r<1)
+    printf(' The turns ratio is 1 : %g \n',(1/r));
+else
+    printf('The turns ratio is %g : 1 \n',r);
+end
+
+V1=sqrt(P*(R1));
+V2=sqrt(P*(R2));
+
+I1=V1/R1;
+I2=V2/R2;
+
+printf('\n Voltages are as follows \n')
+printf('The primary voltage = %g V \n',V1)
+printf('The secondary voltage = %g V \n',V2)
+printf('\n Currents are as follows \n')
+printf('The primary current = %g A \n',I1)
+printf('The secondary current = %g A \n',I2)
+
diff --git a/1319/CH5/EX5.17/5_17.sce b/1319/CH5/EX5.17/5_17.sce
new file mode 100644
index 000000000..0ea28b0aa
--- /dev/null
+++ b/1319/CH5/EX5.17/5_17.sce
@@ -0,0 +1,18 @@
+// Equivalent resistance and leakage reactance wrt primary
+
+clc;
+clear;
+
+P=1200;
+V=60;
+I1=100;
+R1eq=P/(I1^2);
+
+Zeq=V/I1;
+
+X1eq=sqrt((Zeq^2)-(R1eq^2));
+
+// Secondary short circuited there the parameters calculated are wrt to primary itself
+
+printf('Equivalent Resistance of the transformer w.r.t primary = %g ohms \n',R1eq)
+printf('Leakage Reactance of the transformer w.r.t primary = %g ohms \n',X1eq)// Text Book Error Please note
diff --git a/1319/CH5/EX5.18/5_18.sce b/1319/CH5/EX5.18/5_18.sce
new file mode 100644
index 000000000..700f9fcb4
--- /dev/null
+++ b/1319/CH5/EX5.18/5_18.sce
@@ -0,0 +1,33 @@
+// To determine Input current and voltage during SC test
+
+clc;
+clear;
+
+Vh=6600;
+Vl=250;
+V=400;
+
+a=Vh/Vl; // Turns ratio
+
+Rh=0.21;
+Rl=2.72*(10^-4);
+
+Xh=1;
+Xl=1.3*(10^-3);
+
+Rt=Rh+Rl*(a^2); // Equivalent resistance w.r.t the primary
+Xt=Xh+Xl*(a^2); // Equivalent reactance w.r.t the primary
+
+ZHeq= sqrt((Rt^2)+(Xt^2));
+
+Ih=V/ZHeq; // Current on high voltage side
+
+Pi=(Ih^2)*Rt; // Power input
+
+printf('W.R.T High Voltage side the equivalent resistance is %g ohms and the equivalent reactance is %g ohms \n',Rt,Xt)
+
+printf('The current on the high voltage side is %g A \n',Ih)
+
+printf('Power Input on the high voltage side is %g kW \n',Pi/1000)
+
+
diff --git a/1319/CH5/EX5.19/5_19.sce b/1319/CH5/EX5.19/5_19.sce
new file mode 100644
index 000000000..839c9fa8c
--- /dev/null
+++ b/1319/CH5/EX5.19/5_19.sce
@@ -0,0 +1,35 @@
+// To determine the load for max efficiency at two power factors
+
+clc;
+clear;
+
+P=100*(10^3); // Power Input
+
+E1=1000;
+E2=10000;
+
+Pil=1200;
+
+I2=P/E2; // Full load current on the HV side
+
+Isc=6; // Current for 500W copper loss in HV winding
+Psc=500; // Copper Loss for 6A in HV winding
+
+Pc=((I2/Isc)^2)*Psc; // Copper loss at full load.
+
+Pmax=sqrt(Pil/Pc)*P; // Is a factor of square root of the ratio of Iron loss and Copper loss at full load.
+
+deff('x=eff(y,z)','x=(P*y*z)*100/((P*y*z)+Pil+(Pc*(z^2)))')// Function to find the eifficiency for a given power factor(y) and load(z).
+
+printf('a) The Efficiency at various loads for unity power factor are as follows. \n')
+printf('i) At 25 percent load = %f percent \n',eff(1,0.25))
+printf('ii) At 50 percent load = %g percent \n',eff(1,0.5))
+printf('iii) At 100 percent load = %g percent \n',eff(1,1))
+
+printf('\n b) The Efficiency at various loads for 0.8 power factor are as follows. \n')
+printf('i) At 25 percent load = %g percent \n',eff(0.8,0.25))
+printf('ii) At 50 percent load = %g percent \n',eff(0.8,0.5))
+printf('iii) At 100 percent load = %g percent \n \n',eff(0.8,1))
+
+printf('The Load at which efficiency is maximum = %g kVA \n',(Pmax/1000))
+
diff --git a/1319/CH5/EX5.2/5_2.sce b/1319/CH5/EX5.2/5_2.sce
new file mode 100644
index 000000000..0d1f27f26
--- /dev/null
+++ b/1319/CH5/EX5.2/5_2.sce
@@ -0,0 +1,34 @@
+// To calculate the number of turns per limb on the high and low voltage sides
+
+clc;
+clear;
+
+f=50;
+A=400*(10^-4);
+Bm=1;
+V1=3000;
+V2=220;
+
+l=2; // Number of limbs
+
+//Neglecting the series voltage drop
+
+// Induced EMF equation
+a=V1/(4.44*f*A*Bm);
+
+b=V2*a/V1;
+
+if(modulo(round(a),2)==0) // No. of turns is a whole even number as it has 2 limbs
+    N1=round(a);
+else
+    N1=round(a)+1;
+end
+
+if(modulo(round(b),2)==0) // No. of turns is a whole even number as it has 2 limbs
+    N2=round(b);
+else
+    N2=round(b)+1;
+end
+
+printf('The number of turns in the high voltage side per limb = %d \n',N1/l)
+printf('The number of turns in the low voltage side per limb = %d \n',N2/l)
diff --git a/1319/CH5/EX5.20/5_20.sce b/1319/CH5/EX5.20/5_20.sce
new file mode 100644
index 000000000..85b233b04
--- /dev/null
+++ b/1319/CH5/EX5.20/5_20.sce
@@ -0,0 +1,19 @@
+//To determine the max regulation and the pf at which it occurs
+
+clc;
+clear;
+
+Vr=2.5;
+Vx=5;
+
+printf('The expression for voltage requlation is y= %g cos(phi) + %g sin(phi) \n',Vr,Vx )
+
+printf('Differenciating w.r.t phi and equating it to zero, we get the power factor angle \n')
+
+printf('We get tan(phi)=> Vr/Vx => 5/2.5 => 2 \n \n')
+
+phi=atand(Vx/Vr); // power factor angle
+
+y= Vr*cosd(phi)+Vx*sind(phi); // Max Volatge regulation
+
+printf('The maximum regulation is %g percent \n and the power factor at which it occurs is %g degrees \n',y,phi)
diff --git a/1319/CH5/EX5.21/5_21.jpg b/1319/CH5/EX5.21/5_21.jpg
new file mode 100644
index 000000000..690e660fa
--- /dev/null
+++ b/1319/CH5/EX5.21/5_21.jpg
diff --git a/1319/CH5/EX5.21/5_21.sce b/1319/CH5/EX5.21/5_21.sce
new file mode 100644
index 000000000..43248add5
--- /dev/null
+++ b/1319/CH5/EX5.21/5_21.sce
@@ -0,0 +1,95 @@
+//To calculate secondary terminal voltage and full load efficiency at unity pf
+
+clc;
+clear;
+
+P=4*(10^3);
+E1=200;
+E2=400;
+
+// O.C Test
+V=200;
+Pil=70; // Iron Loss
+Ioc=0.8;
+
+R0=(V^2)/Pil;
+
+Iw=V/R0;
+Im=sqrt((Ioc^2)-(Iw^2));
+
+X0=V/Im;
+
+// S.C Test
+Vsc=17.5;
+Isc=9;
+Psc=50;
+
+R2eq=Psc/(Isc^2);
+
+Z2eq=Vsc/Isc;
+
+X2eq=sqrt((Z2eq^2)-(R2eq^2));
+
+Is=P/E2; // Full load current
+
+Pc=((Is/Isc)^2)*Psc;
+
+fleff=(P*100)/(P+Pil+Pc);// Full load efficiency
+
+printf('i) The Full load efficiency at unity power factor = %g percent \n \n',fleff)
+
+// Secondary Terminal voltages cosidering full load secondary current as reference
+
+Vs=poly([0 1],'Vs','c');
+
+Vz=Is*(R2eq+(X2eq*%i));
+
+// Using the characteristic equation in polar form, 'Is' as reference
+// E = V/_theta + Is/_0 *(Z/_phi)
+
+// Function to evalulate the right side of the equation in complex form
+deff('a=stv(b)','a=Vs*(complex(cosd(b),sind(b)))+Vz')
+
+case1=stv(acosd(1));
+case2=stv(acosd(0.8));
+case3=stv(-acosd(0.8));
+
+// Funtion to calculate the characteristic equation of Vs
+deff('x=svol(y)','x=(real(y)^2)+(imag(y)^2)-(E2^2)')
+
+cs1=svol(case1);
+cs2=svol(case2);
+cs3=svol(case3);
+
+// Roots of the characteristic equations
+
+r1=roots(cs1);
+r2=roots(cs2);
+r3=roots(cs3);
+
+
+// To find the positive roots
+if(imag(sqrt(r1(1))))
+    Vs1=r1(2);
+else
+    Vs1=r1(1);
+end
+
+if(imag(sqrt(r2(1))))
+    Vs2=r2(2);
+else
+    Vs2=r2(1);
+end
+
+if(imag(sqrt(r3(1))))
+    Vs3=r3(2);
+else
+    Vs3=r3(1);
+end
+
+printf('ii) The Secondary terminal voltages for various power factors are as follows \n')
+printf('a) At Unity power factor, Vs = %g V \n',Vs1)
+printf('b) At 0.8 power factor(Lagging), Vs = %g V \n',Vs2)
+printf('c) At 0.8 power factor(Leading), Vs = %g V \n',Vs3)
+
+    
diff --git a/1319/CH5/EX5.3/5_3.sce b/1319/CH5/EX5.3/5_3.sce
new file mode 100644
index 000000000..6592dda92
--- /dev/null
+++ b/1319/CH5/EX5.3/5_3.sce
@@ -0,0 +1,22 @@
+// To calculate resistance of primary interms of secondary and vice versa
+
+clc;
+clear;
+
+N1=90;
+N2=180;
+
+R2=0.233;
+R1=0.067;
+
+n=N2/N1; // Transformation ratio
+
+R1w2=(n^2)*R1;
+R2w1=R2/(n^2);
+
+Rt=R1+R2w1; // Total resistance in terms of primary
+
+printf('a) Resistance of primary in terms of the secondary = %f ohms \n',R1w2)
+printf('b) Resistance of secondary in terms of the primary = %f ohms \n',R2w1)
+printf('c) Total resistance of the transformer in terms of the primary winding =%f ohms \n',Rt)
+
diff --git a/1319/CH5/EX5.4/5_4.sce b/1319/CH5/EX5.4/5_4.sce
new file mode 100644
index 000000000..886ab0e11
--- /dev/null
+++ b/1319/CH5/EX5.4/5_4.sce
@@ -0,0 +1,32 @@
+// Total resitance and total copper loss at full load
+
+clc;
+clear;
+
+P=40*(10^3);
+E1=2000;
+E2=250;
+
+n=E2/E1; //Transformation ratio
+
+R1=1.15;
+R2=0.0155;
+
+R1w2=R1*(n^2);
+R2w1=R2/(n^2);
+
+Rt=R2+R1w2;
+
+// Full load currents
+I1=P/E1;
+I2=P/E2;
+
+Pc1=(I1^2)*R1; // Primary Loss
+Pc2=(I2^2)*R2; // Secondary Loss
+
+Pc= Pc1+Pc2; // Total Copper loss at full load
+
+printf('a) The total resitance in terms of the secondary winding = %f ohms \n',Rt)
+printf('b) Total copper loss on full load = %f watts',Pc)
+
+
diff --git a/1319/CH5/EX5.5/5_5.sce b/1319/CH5/EX5.5/5_5.sce
new file mode 100644
index 000000000..e641c98e8
--- /dev/null
+++ b/1319/CH5/EX5.5/5_5.sce
@@ -0,0 +1,61 @@
+//Voltage regulation at 0.8 pf lagging
+
+clc;
+clear;
+
+E1=1100;
+E2=110;
+
+P=5*(10^3);
+
+I=P/E1; // Primary full load current
+
+I2=P/E2;// Secondary full load current
+
+V=33;
+
+pf=0.8; // Power Factor lagging, so the angle is positive
+
+theta=acosd(pf);// Power factor angle
+
+Pc=85;
+
+R=Pc/(I^2);
+
+Z=V/I;
+
+V1=E1;
+
+X=sqrt((Z^2)-(R^2));
+
+// Using equation 5.22 to determine V2
+
+V2=poly([0 1],'V2','c');
+
+x=(V2^2)+(2*V2*I*R*pf)+(2*V2*I*X*sind(theta))+((I^2)*((R^2)+(X^2)))-(V1^2);
+
+r=roots(x);
+
+a1=sqrt(r(1));
+a2=sqrt(r(2));
+
+if(imag(a1))
+    V2=r(2);
+else
+    if(imag(a2))
+        V2=r(1);
+    else
+        disp('Error')
+    end
+end
+
+reg=(V1-V2)/V2; // Voltage regulation
+
+regper=reg*100;// Voltage regulation percent
+
+disp(x,'The characteristic equation to find V2 equated to zero is')
+
+disp(regper,'The percentage voltage regulation for a load at 0.8 pf lagging is')
+    
+
+
diff --git a/1319/CH5/EX5.6/5_6.sce b/1319/CH5/EX5.6/5_6.sce
new file mode 100644
index 000000000..252c38319
--- /dev/null
+++ b/1319/CH5/EX5.6/5_6.sce
@@ -0,0 +1,31 @@
+// Regulation at laggiing leading and unity power factors
+
+clc;
+clear;
+
+ol=0.01;// Ohmic loss is 1% of the output
+
+// Output = V*I; Ohmic loss =(I^2)*R
+
+//(I*R)/V = 0.01
+
+rd=0.05; // Reactance drop is 5% of the output voltage
+
+// Power Factors
+pf1=0.8;// lag
+pf2=1; // unity
+pf3=0.8;// lead
+
+deff('y=angle(x)','y=acosd(x)');// Function to find out the angle
+
+// Angles
+t1=angle(pf1);// Positive sign as it is lagging
+t2=angle(pf2);
+t3=-angle(pf3); // Minus sign as it is leading
+
+deff('a=vr(b)','a=((ol*cosd(b))+(rd*sind(b)))*100');// Function to find out voltage regulation
+
+printf('The voltage regulation percentages is as follows \n')
+printf('a) For 0.8 p.f lag = %f percent \n',vr(t1))
+printf('b) For unity p.f = %f percent \n',vr(t2))
+printf('c) For 0.8 p.f lead = %f percent \n',vr(t3)) 
diff --git a/1319/CH5/EX5.7/5_7.sce b/1319/CH5/EX5.7/5_7.sce
new file mode 100644
index 000000000..d96f5696f
--- /dev/null
+++ b/1319/CH5/EX5.7/5_7.sce
@@ -0,0 +1,42 @@
+// Calculate the circuit parameters of a transformer using OC and SC tests
+
+clc;
+clear;
+
+E1=200;
+E2=400;
+
+n=E2/E1; // Transformation ratio
+
+// O.C Calculations
+V1=200;
+Ioc=0.7;
+Pi=70;
+
+R0=(V1^2)/Pi;
+
+Iw=V1/R0;
+
+Im=sqrt((Ioc^2)-(Iw^2));
+
+X0=V1/Im;
+
+//S.C Calculations on HT side
+
+Pc=80;
+I=10;
+V=15;
+
+Rth= Pc/(I^2);
+Z=V/I;
+
+Xth=sqrt((Z^2)-(Rth^2));
+
+// Both these value are referred to HT side, but the answer is required to be referred to LT side
+
+Xtl=Xth/(n^2); // Reactance referred to LT side
+Rtl=Rth/(n^2); // Resistance referred to LT side
+
+printf('The Circuit parameters referred to LT side is as follows \n')
+
+printf('Ro = %f ohms \n Xo = %f ohms \n Rt = %f ohms \n Xt = %f ohms \n',R0,X0,Rtl,Xtl)
diff --git a/1319/CH5/EX5.8/5_8.sce b/1319/CH5/EX5.8/5_8.sce
new file mode 100644
index 000000000..40f7123df
--- /dev/null
+++ b/1319/CH5/EX5.8/5_8.sce
@@ -0,0 +1,46 @@
+// To calculate terminal voltage and current and efficiency
+
+clc;
+clear;
+
+n=10; // Transformation ratio
+
+E1=200;
+
+R0=400;
+X0=251*%i;
+
+R1=0.16;
+X1=0.7*%i;
+
+R2=5.96; // As referred to the primary side
+X2=4.44*%i; // As referred to the primary side
+
+I1=E1/(R1+R2+X1+X2);
+
+t1=atand(imag(I1)/real(I1));// Angle for primary current
+
+Iw=E1/R0;
+Im=E1/X0;
+
+Ip=Iw+Im+I1;
+
+Zl=R2+X2;
+
+V2p=I1*Zl;// Secondary terminal voltage referred to primary side
+
+V2=n*V2p;
+
+t2=atand(imag(V2)/real(V2)); // Angle for V2
+
+Po= (abs(I1)^2)*R2; // Output power
+
+Pc=(abs(I1)^2)*R1;// Copper Loss
+
+Pil=(abs(Iw)^2)*R0;// Iron Loss
+
+eff= Po*100/(Po+Pc+Pil)// Efficiency
+
+printf('a) The secondary terminal voltage = %f /_%f V \n',abs(V2),t2)
+printf('b) The primary current = %f /_%f A \n',abs(I1),t1)
+printf('c) The efficiency is %f percent \n',eff)
diff --git a/1319/CH5/EX5.9/5_9.sce b/1319/CH5/EX5.9/5_9.sce
new file mode 100644
index 000000000..b4d14fb83
--- /dev/null
+++ b/1319/CH5/EX5.9/5_9.sce
@@ -0,0 +1,44 @@
+// Regulation at full load p.f 0.8 lag
+
+clc;
+clear;
+
+Pi=500*(10^3);// Power Input
+Meff=97/100;// Max Efficiency
+pf1=1;
+
+E1=3300;
+E2=500;
+
+Po=Pi*pf1*3/4;
+
+// Iron loss = Copper loss at maximum efficiency
+
+x=poly([0 1],'x','c');
+
+Pin=Po+(2*x);
+
+xx=(Pin*Meff)-Po;
+
+x=roots(xx); // Iron Loss = Copper Loss
+
+I2=Po/E2;
+
+R=x/(I2^2);
+
+I2fl=Pi/E2;
+
+Rfl=E2/I2fl;
+
+// Per unit values
+Rpu=R*100/Rfl;
+Zpu=10;
+Xpu=sqrt((Zpu^2)-(Rpu^2));
+
+pf2=0.8; // Lagging
+
+ang=acosd(pf2);// Positive Angle as it is lagging
+
+perreg=(Rpu*cosd(ang))+(Xpu*sind(ang));
+
+printf('The regulation at full load, p.f 0.8 lag = %f percent\n',perreg)