initial commit / add all books

70 files changed, 1365 insertions, 0 deletions

diff --git a/3369/CH12/EX12.1/Ex12_1.sce b/3369/CH12/EX12.1/Ex12_1.sce
new file mode 100755
index 000000000..78ef92ed8
--- /dev/null
+++ b/3369/CH12/EX12.1/Ex12_1.sce
@@ -0,0 +1,29 @@
+//Chapter 12, Exmaple 1, page 403
+//Calculate radial thickness of insulating layer
+clc
+clear
+//based on equation 12.15 and v1alues of E1 and E2 
+E1 = 40 // kV/cm
+E2 = 25 // kV/cm
+ep1 = 6 // permittives of the material
+ep2 = 4 //permittives of the material
+d1 = 4 // cm
+d2 = 10 // cm
+r1 = 2 // cm
+r2 = (E1*ep1*2)/(E2*ep2)
+inner = r2-(d1/2)
+outer = (d2/2)-r2
+//based on equation 12.16
+V1peak = E1*r1*log(r2/r1) // inner dielectric
+V2peak = E2*r2*log(d2/(2*r2)) // outter dielectric
+Vcab = V1peak+V2peak // Peak volatge of cable
+rms = Vcab/sqrt(2)
+printf("\n Radius = %f cm ",r2)
+printf("\n Inner radial thickness = %f cm ",inner)
+printf("\n Outer radial thickness = %f cm",outer)
+printf("\n Vpeak of outer dielectric = %f kV", V1peak)
+printf("\n Vpeak of inner dielectric = %f kV", V2peak)
+printf("\n Peak voltage of cable = %f kV", Vcab)
+printf("\n Safe opearating voltage = %f kV", rms)
+
+// Answers may vary due to round off error.
diff --git a/3369/CH12/EX12.10/Ex12_10.sce b/3369/CH12/EX12.10/Ex12_10.sce
new file mode 100755
index 000000000..7ffcb2f60
--- /dev/null
+++ b/3369/CH12/EX12.10/Ex12_10.sce
@@ -0,0 +1,18 @@
+//Chapter 12,Example 10, page 412
+//Determine the maximum stress 
+clear
+clc
+a = 1 //cm
+r1 = 2 // cm
+b = 2.65 // cm
+er1 = 4.5
+er2 = 3.6
+V = 53.8 // kV
+ba = 5.3/2 // b/a
+alpha = 1.325
+E1max = V/(log(r1)+(er1/er2)*log(alpha))
+E2max = V/((r1*(er2/er1)*log(r1))+log(alpha))
+printf("\n E1max = %f kV/cm",E1max)
+printf("\n E2max = %f kV/cm",E2max) // answer vary from the text
+
+// Answer vary from the text due to round off 
diff --git a/3369/CH12/EX12.2/Ex12_2.sce b/3369/CH12/EX12.2/Ex12_2.sce
new file mode 100755
index 000000000..4cc866adf
--- /dev/null
+++ b/3369/CH12/EX12.2/Ex12_2.sce
@@ -0,0 +1,11 @@
+//Chapter 12, Exmaple 2, page 404
+//Calculate optimum value of r
+clear
+clc
+//Based on equation 12.17
+V1 = 100 // kV
+V2 = 55 // kV
+r = V1*sqrt(2)/V2
+printf("\n Radius = %f cm ",r)
+
+// Answers may vary due to round off error
diff --git a/3369/CH12/EX12.3/Ex12_3.sce b/3369/CH12/EX12.3/Ex12_3.sce
new file mode 100755
index 000000000..0ad841adb
--- /dev/null
+++ b/3369/CH12/EX12.3/Ex12_3.sce
@@ -0,0 +1,11 @@
+//Chapter 12, Exmaple 3, page 406
+//Calculate resistivity
+clear
+clc
+l = 10^4 // cable length in m
+Rr = 3/1.5 // R/r ratio
+ins = 0.5*10**6 // insulation in ohms
+p = 2*%pi*l*ins/log(Rr)
+printf("\n Resistivity of insulation material = %e ohm/m ",p)
+
+// Answers may vary due to round off error  
diff --git a/3369/CH12/EX12.4/Ex12_4.sce b/3369/CH12/EX12.4/Ex12_4.sce
new file mode 100755
index 000000000..22c60b7b0
--- /dev/null
+++ b/3369/CH12/EX12.4/Ex12_4.sce
@@ -0,0 +1,13 @@
+//Chapter 12, Exmaple 4, page 406
+//Calculate resistivity
+clear
+clc
+// Baased on Equation 12.1*10**2
+c4 = 0.5*10**2/10 // micro F
+Ic = 2*10**4*2*%pi*5*50*10**-6/sqrt(3)
+C = (sqrt(3)*10000*Ic)*(10**-9*10**6)
+printf("\n C4 = %f mircoF ",c4)
+printf("\n Line charging current = %f A ",Ic)
+printf("\n Charging = %f kVA ",C)
+
+// Answers may vary due to round off error
diff --git a/3369/CH12/EX12.5/Ex12_5.sce b/3369/CH12/EX12.5/Ex12_5.sce
new file mode 100755
index 000000000..bb084db23
--- /dev/null
+++ b/3369/CH12/EX12.5/Ex12_5.sce
@@ -0,0 +1,18 @@
+//Chapter 12,Example 5, page 408
+//Calculate capasitance and kVAr 
+clear
+clc
+//(a) Using the notations used in FiVgs. 12.15 and 12.16
+C2 = 0.75/3 // microF/km
+C3 = (0.6*3-2*C2)/2 // microF/km
+C4 = (C2+C3)/2 // microF/km
+printf("\n C2 = %f mircoF/Km ",C2)
+printf("\n C3 = %f mircoF/Km ",C3)
+printf("\n C4 = %f mircoF/Km ",C4)
+//(b)Capacitance of 10 km between 2 cores
+V = 33*10**3
+w = 2*%pi*50
+C = 2*V^2*w*C4*10*10**-9
+printf("\n Carging = %f kVAr ",C)
+
+// Answers may vary due to round of errors.
diff --git a/3369/CH12/EX12.6/Ex12_6.sce b/3369/CH12/EX12.6/Ex12_6.sce
new file mode 100755
index 000000000..c55cf9976
--- /dev/null
+++ b/3369/CH12/EX12.6/Ex12_6.sce
@@ -0,0 +1,30 @@
+//Chapter 12,Example 6, page 409
+//Determine the efective electrical parameters 
+clear
+clc
+rc = 0.0875*(1+0.004*50) // conductor resistance in ohm/km
+Rc = 0.105*85 // ohm
+w = 2*%pi*50
+Rsh = 23.2*10**-6*85*10**5/(%pi*(3^2-2.5^2)) // Resistance of sheath
+D = 8
+rsh = 1/2*(2.5+3)
+Xm = w*2*log(D/rsh)*10**-7*85000
+Ref = Rc + Xm^2*Rsh/(Rsh^2+Xm^2) // Effective AC resistance 
+Xc = 11.1// reactance with sheaths open-circuit
+Xef = Xc-(Xm^2/(Rsh^2+Xm^2)) //Effective reactance per cable
+SlCl = Rsh*Xm^2/(Rc*(Rsh^2+Xm^2)) // Sheath loss/conductor loss
+I = 400 // A
+emf = Xm*I // emf induced per sheath
+printf("\n Conductor resistance = %f ohm",rc)
+printf("\n Conductor resistance for the whole leangth (Rc) = %f ohm",Rc)
+printf("\n Resistance of sheath (Rsh) = %f ohm/Km ",Rsh)
+printf("\n Conductor to sheath mutual inductive reactance (Xm)= %f ohm/m ",Xm)
+printf("\n Effective AC resistance(Ref) = %f ohm ",Ref)
+printf("\n Reactance with sheaths open-circuit(Xc) = %f ohm ",Xc)
+printf("\n Effective reactance per cable(Xef) = %f ohm ",Xef)
+printf("\n Sheath loss/conductor loss = %f ",SlCl)
+printf("\n emf induced per sheath(emf) = %f V",emf)
+
+
+
+
diff --git a/3369/CH12/EX12.7/Ex12_7.sce b/3369/CH12/EX12.7/Ex12_7.sce
new file mode 100755
index 000000000..d21a8d60e
--- /dev/null
+++ b/3369/CH12/EX12.7/Ex12_7.sce
@@ -0,0 +1,12 @@
+//Chapter 12,Example 7, page 410
+//Determine the induced sheath voltage 
+clear
+clc
+D = 15 // cm
+rsh = 5.5/2 // Sheath diameter converted to radius in cm
+I = 250 // A
+E = 2*10^-7*314*I*log(D/rsh)*10^3
+printf("\n Induced sheath voltage per Km = %f V/km",E)
+printf("\n If the sheaths are bonded at one end, the voltage between them at the other end = = %f V/km",E*sqrt(3))
+
+// Answers may vary due to round off errors. 
diff --git a/3369/CH12/EX12.8/Ex12_8.sce b/3369/CH12/EX12.8/Ex12_8.sce
new file mode 100755
index 000000000..51caa1ce9
--- /dev/null
+++ b/3369/CH12/EX12.8/Ex12_8.sce
@@ -0,0 +1,27 @@
+//Chapter 12,Example 8, page 411
+//Determine the maximum stress 
+clear
+clc
+ba = 5.3/2 // b/a
+alpha = nthroot(ba,3)
+r1 = 1.385 // cm
+r2 = 1.92 // cm
+r = 2.65 // cm
+V = 66*sqrt(2)/sqrt(3)
+V2 = V/(1+(1/alpha)+(1/alpha^2))
+V1 = (1+1/alpha)*V2
+//calculating maximim and minimum stress without sheaths
+Emax0 = V/1*log(r/1)
+Emin0 = V/(r*log(r))
+//calculating max and min stress with the sheaths
+Emax = Emax0*3/(1+(alpha)+(alpha^2))
+Emin = Emax/alpha
+printf("\n Peak voltage of the conductor V = %f kV",V)
+printf("\n V1 = %f kV",V1)
+printf("\n V2 = %f kV",V2)
+printf("\n Maximum stress without sheaths = %f kV/cm",Emax0)
+printf("\n Minimum stress without sheaths = %f kV/cm",Emin0)
+printf("\n Maximum stress with sheaths = %f kV/cm",Emax)
+printf("\n Minimum stress with sheaths = %f kV/cm",Emin)
+
+// Answers vary due to round off errors.
diff --git a/3369/CH12/EX12.9/Ex12_9.sce b/3369/CH12/EX12.9/Ex12_9.sce
new file mode 100755
index 000000000..2253390d9
--- /dev/null
+++ b/3369/CH12/EX12.9/Ex12_9.sce
@@ -0,0 +1,26 @@
+//Chapter 12,Example 9, page 412
+//Determine the maximum stress 
+clear
+clc
+Emax = 47.5 // kV
+b = 2.65 // cm
+a = 1 // cm
+ba = 0.55*3 // 1/3(b-a)
+r1 = 1.55 // cm
+r2 = 2.1 // cm2Vr = 2.65 // cm  
+V = 53.8 // kV
+alpha = nthroot(ba,3)
+// based on the example 12_8 
+//calculating VEmax1, Emax2, Emax3 
+x = 1/(a*log(r1/a))
+y = 1/(r1*log(r2/r1))
+z = 1/(r2*log(b/r2))
+VV1 = Emax/x
+V1V2 = Emax/y
+V2 = Emax/z
+V1 = V2+(Emax/y)
+printf("\n Emax = %f kV/cm",Emax)
+printf("\n V1 = %f kV/cm",V1)
+printf("\n V2 = %f kV/cm",V2)
+
+// Answers may vary due to round off error.
diff --git a/3369/CH14/EX14.4.2.1/Ex14_1.sce b/3369/CH14/EX14.4.2.1/Ex14_1.sce
new file mode 100755
index 000000000..187a4d2cd
--- /dev/null
+++ b/3369/CH14/EX14.4.2.1/Ex14_1.sce
@@ -0,0 +1,14 @@
+//Chapter 14,Example 1, page 453
+//Determine the time to crest 
+clear
+clc
+I = 400 // mH of inductance
+L = 500*10^-3 // mH
+C = 1.5*10^-6 // micro F
+
+f = 1/(2*%pi*sqrt(L*C))  
+t = 10**6/(4*f) // calulation done in the text is wrong
+printf("\n f1 = %f Hz",f)
+printf("\n Time to crest = %f micro seconds",t)
+
+// Answer may vary due to round off error.
diff --git a/3369/CH16/EX16.1/Ex16_1.sce b/3369/CH16/EX16.1/Ex16_1.sce
new file mode 100755
index 000000000..fa190e90d
--- /dev/null
+++ b/3369/CH16/EX16.1/Ex16_1.sce
@@ -0,0 +1,31 @@
+//Chapter 16,Example 1, page 556
+//Determine the (a)ripple voltage (b)voltage drop (c)Average output volatge (d)ripple factor 
+clear
+clc
+I1 = 5*10^-3 // A
+C2 = 0.05*10^-6 // F
+C1 = 0.01*10^-6 // F
+Vs = 100 // kV
+f = 50 // Hz 
+// (a) Ripple voltage
+printf("\n Part (a)")
+delV = I1/(C2*f)
+printf("\n Ripple Voltage = %f V", delV)
+// (b) Voltage drop 
+printf("\n Part (b)")
+Vd = I1/f*((1/C1)+(1/(2*C2)))
+printf("\n Voltage drop = %f V", Vd)
+// (c) Average output voltage 
+printf("\n Part (c)")
+Vav = 2*Vs*sqrt(2)-Vd*10^-3
+printf("\n Average output voltage = %f kV", Vav)
+// (d) Ripple factor 
+printf("\n Part (d)")
+RF = Vd*10^-3/(2*Vs*sqrt(2)) 
+printf("\n Ripple Factor in percentage = %f", RF*100)
+
+
+
+
+
+
diff --git a/3369/CH16/EX16.10/Ex16_10.sce b/3369/CH16/EX16.10/Ex16_10.sce
new file mode 100755
index 000000000..486af2393
--- /dev/null
+++ b/3369/CH16/EX16.10/Ex16_10.sce
@@ -0,0 +1,24 @@
+//Chapter 16,Example 10,page 564
+//Determine the from and tail times
+clear
+clc
+n = 12
+C1 = 0.125*10^-6/n // micro F
+C2 = 0.001*10^-6 // micro F
+R1 = 70*n // ohm
+R2 = 400*n // ohm
+// beased on figure 16.15
+theta = sqrt(C1*C2*R1*R2)
+neta = 1/(1+R1/R2+C2/C1)
+a = R2*C1/(2*theta*neta) // alpha
+T2 = 7*theta*10^6
+T1 = T2/25
+printf("\n R1 = %f Ohm", R1) 
+printf("\n R2 = %f Ohm", R2) 
+printf("\n Theta = %f microS",theta*10^6)
+printf("\n Neta = %f",neta)
+printf("\n Alpha = %f ",a)
+printf("\n T1 = %f microS", T1) 
+printf("\n T2 = %f microS", T2) 
+
+// Answers greatly vary due to round off error
diff --git a/3369/CH16/EX16.11/Ex16_11.sce b/3369/CH16/EX16.11/Ex16_11.sce
new file mode 100755
index 000000000..19a900067
--- /dev/null
+++ b/3369/CH16/EX16.11/Ex16_11.sce
@@ -0,0 +1,17 @@
+//Chapter 16,Example 11,page 564
+//Determine the equation generated by impulse
+clear
+clc
+w = 0.02*10^6 // s^-1 obtained by solving eq 16.47 iteratively
+R = sqrt(4-(sqrt(8*8*4)*0.02)^2) // solved the simplified equation
+L = 8*10^-6
+V = 25*10^3
+// In equation 16.46
+y = R/(2*L)
+// Deriving the equation
+a = V/(w*L)
+printf("\n R = %e ohm",R)
+printf("\n y = %e s^-1",y)
+printf("\n I(t) = %e * exp(%et) * sin(%et) A",a,-y,w) 
+
+// Answers may vary due to round off error
diff --git a/3369/CH16/EX16.2/Ex16_2.sce b/3369/CH16/EX16.2/Ex16_2.sce
new file mode 100755
index 000000000..6503ada92
--- /dev/null
+++ b/3369/CH16/EX16.2/Ex16_2.sce
@@ -0,0 +1,28 @@
+//Chapter 16,Example 2, page 556
+//Determine the (a)ripple voltage (b)voltage drop (c)Average output volatge (d)ripple factor 
+clear
+clc
+I1 = 5*10^-3 // A
+C3 = 0.10*10^-6 // F
+C2 = 0.05*10^-6 // F
+C1 = 0.01*10^-6 // F
+Vs = 100 // kV
+f = 50 // Hz 
+// (a) Ripple voltage
+printf("\n Part (a)")
+delV = I1/f*((2/C1)+(1/C3))
+printf("\n Ripple Voltage = %f kV", delV*10^-3)
+// (b) Voltage drop 
+printf("\n Part (b)")
+Vd = I1/f*((1/C2)+(1/C1)+(1/(2*C3)))
+printf("\n Voltage drop = %f kV", Vd*10^-3)
+// (c) Average output voltage 
+printf("\n Part (c)")
+Vav = 3*Vs*sqrt(2)-Vd*10^-3
+printf("\n Average output voltage = %f kV", Vav)
+// (d) Ripple factor 
+printf("\n Part (d)")
+RF = Vd*10^-3/(3*Vs*sqrt(2)) 
+printf("\n Ripple Factor in percentage = %f", RF*100)
+
+// Answers may vary due to round off error
diff --git a/3369/CH16/EX16.3/Ex16_3.sce b/3369/CH16/EX16.3/Ex16_3.sce
new file mode 100755
index 000000000..aa26bd30e
--- /dev/null
+++ b/3369/CH16/EX16.3/Ex16_3.sce
@@ -0,0 +1,32 @@
+//Chapter 16,Example 3, page 557
+//Determine the (a)ripple voltage (b)voltage drop (c)Average output volatge (d)ripple factor (e)optimum number of stages
+clear
+clc
+I1 = 5*10^-3 // A
+C = 0.15*10^-6 // F
+Vs = 200 // kV
+f = 50 // Hz 
+n = 12
+// (a) Ripple voltage
+printf("\n Part (a)")
+delV = I1*n*(n+1)/(f*C*2)
+printf("\n Ripple Voltage = %f kV", delV*10^-3)
+// (b) Voltage drop 
+printf("\n Part (b)")
+a = I1/(f*C)
+Vd = a*((2/3*n^3)+(n^2/2)-(n/6)+(n*(n+1)/4))
+printf("\n Voltage drop = %f kV", Vd*10^-3)
+// (c) Average output voltage 
+printf("\n Part (c)")
+Vav = 2*n*Vs*sqrt(2)-Vd*10^-3
+printf("\n Average output voltage = %f kV", Vav)
+// (d) Ripple factor 
+printf("\n Part (d)")
+RF = Vd*10^-3/(2*n*Vs*sqrt(2)) 
+printf("\n Ripple Factor in percentage = %f", RF*100)
+// (e) Optimum number of stages
+printf("\n Part (e)")
+nopt = sqrt(Vs*sqrt(2)*10^3*f*C/I1) 
+printf("\n Optimum number of stages = %d stages", nopt)
+
+// Answers may vary due to round off error
diff --git a/3369/CH16/EX16.4/Ex16_4.sce b/3369/CH16/EX16.4/Ex16_4.sce
new file mode 100755
index 000000000..3ae18c2a6
--- /dev/null
+++ b/3369/CH16/EX16.4/Ex16_4.sce
@@ -0,0 +1,29 @@
+//Chapter 16,Example 4, page 558
+//Determine the input voltage and power
+clear
+clc
+Vc = 500*10^3 // V
+A = 4 // A
+Xl = 8/100 // in percentage 
+kV = 250
+Xc = Vc/A // Reactance of the cable
+XL = Xl*(kV**2/100)*10**3 // Leakage reactance of the transformer
+Radd = Xc-XL // Additional series reactance
+Ind = Radd/(2*%pi*XL) // Inductance of required series inductor
+R = 3.5/100*(kV**2/100)*10**3  // Total circuit resistance
+Imax = 100/250 // maximum current that can be supplied by the transformer
+Vex = Imax*R // Exciting voltage of transformer secondary
+Vin = Vex*220/kV // Input voltage of transformer primary
+P = Vin*100/220 // Input power of the transformer
+printf("\n Reactance of the cable = %f k ohm", Xc*10^-3)
+printf("\n Leakage reactance of the transformer = %f k ohm", XL*10^-3)
+printf("\n Additional series reactance = %f k ohm", Radd*10^-3)
+printf("\n Inductance of required series inductor = %f H", Ind*10^3)
+printf("\n Total circuit resistance = %f k ohm", R*10^-3)
+printf("\n maximum current that can be supplied by the transformer = %f A", Imax)
+printf("\n Exciting voltage of transformer secondary = %f kV", Vex*10^-3)
+printf("\n Input voltage of transformer primary = %f V", Vin*10^-3)
+printf("\n Input power of the transformer = %f kW", P*10^-3)
+
+// Answers may vary due to round off error
+
diff --git a/3369/CH16/EX16.5/Ex16_5.sce b/3369/CH16/EX16.5/Ex16_5.sce
new file mode 100755
index 000000000..a72530af7
--- /dev/null
+++ b/3369/CH16/EX16.5/Ex16_5.sce
@@ -0,0 +1,14 @@
+//Chapter 16,Example 5,page 559
+//Determine the charging current and potential difference
+clear
+clc
+ps = 0.5*10**-6 // C/m^2
+u = 10 // m/s
+w = 0.1 // m
+I = ps*u*w 
+Rl = 10^14 // ohm
+V = I*Rl*10^-6
+printf("\n Charging current= %f micro A", I*10^6)
+printf("\n Potential difference = %f MV", V)
+
+// Answers may vary due to round off error
diff --git a/3369/CH16/EX16.6/Ex16_6.sce b/3369/CH16/EX16.6/Ex16_6.sce
new file mode 100755
index 000000000..e546e1992
--- /dev/null
+++ b/3369/CH16/EX16.6/Ex16_6.sce
@@ -0,0 +1,33 @@
+//Chapter 16,Example 6,page 560
+//Determine the wave generated
+clear
+clc
+// With refrence to table 16.1
+C1 = 0.125*10^-6 // F
+C2 = 1*10^-9 // F
+R1 = 360 // ohm
+R2 = 544 // ohm
+V0 = 100 // kV
+theta = sqrt(C1*C2*R1*R2)
+neta = 1/(1+(1+R1/R2)*C2/C1)
+alpha = R2*C1/(2*theta*neta)
+printf("\n Theta = %f micro S",theta*10^6)
+printf("\n Neta = %f",neta)
+printf("\n Alpha = %f ",alpha)
+// Coresponding to alpha the following can be deduced from Fig 16.12
+T2 = 10.1*theta*10^6
+T1 = T2/45
+imp = T1/T2 // generated lighting impulse
+// From equations 16.41 and 16.42
+a1 = (alpha-sqrt(alpha^2-1))*10^-6/(theta) 
+a2 = (alpha+sqrt(alpha^2-1))*10^-6/theta 
+printf("\n T1 = %f microS", T1)
+printf("\n T2 = %f microS", T2)
+printf("\n Generated lighting impulse = %e wave", imp)
+printf("\n alpha1 = %f microS", a1)
+printf("\n alpha2 = %f microS", a2)
+// According to equation 16.40
+et = neta*(alpha*V0)/sqrt(alpha^2-1)
+printf("\n e(t) = %f * (e^%ft - f^%ft)",et,-a1,-a2) // Equation of the wave form generated by the impulese
+
+//Answers may vary due to round off error 
diff --git a/3369/CH16/EX16.7/Ex16_7.sce b/3369/CH16/EX16.7/Ex16_7.sce
new file mode 100755
index 000000000..2e4321658
--- /dev/null
+++ b/3369/CH16/EX16.7/Ex16_7.sce
@@ -0,0 +1,30 @@
+//Chapter 16,Example 6,page 561
+//Determine the wave generated
+clear
+clc
+C1 = 0.125*10^-6 // F
+C2 = 1*10^-9 // F
+R1 = 360 // ohm
+R2 = 544 // ohm
+V0 = 100 // kV
+theta = sqrt(C1*C2*R1*R2)
+neta = 1/(1+R1/R2+C2/C1)
+alpha = R2*C1/(2*theta*neta)
+printf("\n Theta = %f micro S",theta*10^6)
+printf("\n Neta = %f",neta)
+printf("\n Alpha = %f ",alpha)
+// Coresponding to alpha the following can be deduced from Fig 16.12
+T2 = 16.25*theta*10^6
+T1 = T2/120
+// From equations 16.41 and 16.42
+a1 = (alpha-sqrt(alpha^2-1))*10^-6/(theta) 
+a2 = (alpha+sqrt(alpha^2-1))*10^-6/theta 
+printf("\n T1 = %f microS", T1) // Answer given in the text is wrong
+printf("\n T2 = %f microS", T2) 
+printf("\n alpha1 = %f microS", a1)
+printf("\n alpha2 = %f microS", a2)
+// According to equation 16.40
+et = neta*(alpha*V0)/sqrt(alpha^2-1)
+printf("\n e(t) = %f * (e^%ft - f^%ft)",et,-a1,-a2) // Equation of the wave form generated by the impulese
+
+//Answers may vary due to round off error 
diff --git a/3369/CH16/EX16.8/Ex16_8.sce b/3369/CH16/EX16.8/Ex16_8.sce
new file mode 100755
index 000000000..94475426e
--- /dev/null
+++ b/3369/CH16/EX16.8/Ex16_8.sce
@@ -0,0 +1,24 @@
+//Chapter 16,Example 8,page 562
+//Determine the circuit efficiency
+clear
+clc
+C1 = 0.125*10^-6 // F
+C2 = 1*10^-9 // F
+T2 = 2500
+T1 = 250
+// Bsaed on Figure 16.12
+T2T1 = T2/T1
+a = 4 // alpha
+theta = T2/6
+// From table 16.1
+X = (1/a^2)*(1+C2/C1)
+R1 = (a*theta*10^-6/C2)*(1-sqrt(1-X))
+R2 = (a*theta*10^-6/(C1+C2))*(1+sqrt(1-X))
+neta = 1/(1+(1+R1/R2)*C2/C1)
+printf("\n Theta = %f micro S", theta)
+printf("\n X = %f ", X)
+printf("\n R1 = %f k Ohm", R1*10^-3) 
+printf("\n R2 = %f k Ohm", R2*10^-3) 
+printf("\n neta = %f ", neta)
+
+// Answers may vary due to round off error
diff --git a/3369/CH16/EX16.9/Ex16_9.sce b/3369/CH16/EX16.9/Ex16_9.sce
new file mode 100755
index 000000000..6b63423ea
--- /dev/null
+++ b/3369/CH16/EX16.9/Ex16_9.sce
@@ -0,0 +1,33 @@
+//Chapter 16,Example 9,page 563
+//Determine the maximum output voltage and energy rating
+clear
+clc
+n = 8
+C1 = 0.16/n // micro F
+C2 = 0.001 // micro F
+T2 = 50
+T1 = 1.2
+// beased on figure 16.12
+a = 6.4 // alpha
+theta = T2/9.5
+X = (1/a^2)*(1+C2/C1)
+R1 = (a*theta*10^-6/C2)*(1-sqrt(1-X))
+R2 = (a*theta*10^-6/(C1+C2))*(1+sqrt(1-X))
+R1n = R1/n
+R2n = R2/n
+V0 = n*120 
+neta = 1/(1+(1+R1/R2)*C2/C1)
+V = neta*V0
+E = 1/2*C1*V0^2 
+printf("\n Theta = %f micro S", theta)
+printf("\n X = %f ", X)
+printf("\n V0 = %f ", V0)
+printf("\n R1 = %f Ohm", R1*10^6) 
+printf("\n R2 = %f Ohm", R2*10^6) 
+printf("\n R1/n = %d Ohm", R1n*10^6) 
+printf("\n R2/n = %d Ohm", R2n*10^6) 
+printf("\n neta = %f ", neta)
+printf("\n Maximum output voltage = %f kV", V) 
+printf("\n Energy rating = %f J", E)
+
+// Answers greatly vary due to round off error
diff --git a/3369/CH19/EX19.1/Ex19_1.sce b/3369/CH19/EX19.1/Ex19_1.sce
new file mode 100755
index 000000000..844d9c8e2
--- /dev/null
+++ b/3369/CH19/EX19.1/Ex19_1.sce
@@ -0,0 +1,13 @@
+//Chapter 19,Example 1,page 665
+//Determine the sepration between the particles
+clear
+clc
+// Based on the equations 19.6, 19.7, 19.8, 19.9 and 19.10
+E = 8*10^5 // V/m
+qm = 10*10^-6 // C/kg, qm = q/m
+y = -1 // m
+t = (1*2/9.8)
+x = 1/2*qm*E*t
+printf("\n The seperation between the particles = %f m",2*x)
+
+// Answers may vary due to round off error
diff --git a/3369/CH19/EX19.10/Ex19_10.sce b/3369/CH19/EX19.10/Ex19_10.sce
new file mode 100755
index 000000000..37b45184d
--- /dev/null
+++ b/3369/CH19/EX19.10/Ex19_10.sce
@@ -0,0 +1,13 @@
+//Chapter 19,Example 10,page 679
+//Determine the minimum voltage required for gnerating drops witha charge of 50 pC per drop
+clear
+clc
+q = 50*10^-12
+a = 25*10^-6
+b = 750*10^-6
+E0 = 8.84*10^-12
+r = 50*10^-6 
+V = (3*q*b^2*log(b/a))/(7*%pi*E0*r^3)
+printf("\n The minimum voltage required for gnerating drops witha charge of 50 pC per drop = %f kV",V*10^-6)
+
+// Answers may vary due to round off error
diff --git a/3369/CH19/EX19.2/Ex19_2.sce b/3369/CH19/EX19.2/Ex19_2.sce
new file mode 100755
index 000000000..0fcf57844
--- /dev/null
+++ b/3369/CH19/EX19.2/Ex19_2.sce
@@ -0,0 +1,10 @@
+//Chapter 19,Example 2,page 667
+//Determine the pumping pressure
+clear
+clc
+p0 = 30*10^-3 // C/m^3
+V = 30*10^3 // V
+P = p0*V
+printf("\n The pumping pressure P = %f N/m^2",P)
+
+// Answers may vary due to round off error
diff --git a/3369/CH19/EX19.4/Ex19_4.sce b/3369/CH19/EX19.4/Ex19_4.sce
new file mode 100755
index 000000000..53f6c933a
--- /dev/null
+++ b/3369/CH19/EX19.4/Ex19_4.sce
@@ -0,0 +1,23 @@
+//Chapter 19,Example 4,page 670
+//Determine the vertical displacement of the drop
+clear
+clc
+d = 0.03*10^-3 // m
+p = 2000 // kg/m^3
+q = 100*10^-15 // C
+V0 = 3500 // V
+d2 = 2*10^-3 // m
+L1 = 15*10^-3 // m
+L2 = 12*10^-3 // m
+Vz = 25 // m/s
+
+m = 4/3*%pi*(1/2*d)^3*p
+t0 = L1/Vz
+Vx0 = q*V0*t0/(m*d2)
+x0 = 1/2*Vx0*t0
+t1 = (L1+L2)/Vz
+x1 = x0+Vx0*(t1-t0)
+
+printf("\n The vertical displacement of the drop = %e m",x1)
+ 
+// Answers may vary due to round off error
diff --git a/3369/CH19/EX19.5/Ex19_5.sce b/3369/CH19/EX19.5/Ex19_5.sce
new file mode 100755
index 000000000..bd790a316
--- /dev/null
+++ b/3369/CH19/EX19.5/Ex19_5.sce
@@ -0,0 +1,19 @@
+//Chapter 19,Example 5,page 672
+//Determine the electric stress and charge density
+clear
+clc
+a = 25*10^-6 // m
+b = 75*10^-6 // m
+Er = 2.8
+ps = 25*10^-6 // C/m^3
+E0 = 8.84*10^-12
+
+Ea = (b*ps)/(ps*E0+b*Er*E0)
+Eb = (a*ps)/(ps*E0+b*Er*E0) // the negative noation is removed to obtain positive answer as in the book 
+psc = E0*Eb
+
+printf("\n Ea = %e V/m",Ea)
+printf("\n Eb = %e V/m",Eb)
+printf("\n Charge density = %e C/m^2",psc)
+
+// Answers may vary due to round off error
diff --git a/3369/CH19/EX19.6/Ex19_6.sce b/3369/CH19/EX19.6/Ex19_6.sce
new file mode 100755
index 000000000..16b58e5f6
--- /dev/null
+++ b/3369/CH19/EX19.6/Ex19_6.sce
@@ -0,0 +1,12 @@
+//Chapter 19,Example 6,page 675
+//Determine the current density
+clear
+clc
+E0 = 8.84*10^-12
+Us = 1.5*10^-3*10^-4
+V = 100
+d3 = 10^-6 // d^3
+J = 4*E0*Us*V^2/d3
+printf("\n Current density = %e A/m^2",J)
+
+// Answer may vary due to round off error
diff --git a/3369/CH19/EX19.7/Ex19_7.sce b/3369/CH19/EX19.7/Ex19_7.sce
new file mode 100755
index 000000000..2794dd095
--- /dev/null
+++ b/3369/CH19/EX19.7/Ex19_7.sce
@@ -0,0 +1,11 @@
+//Chapter 19,Example 7,page 676
+//Determine the thickness of dust layer
+clear
+clc
+Edb = 3*10^6
+E0 = 8.84*10^-12
+p0 = 15*10^-3
+d = Edb*E0/p0
+printf("\n Thickness of the dust layer = %e m",d)
+
+// Answers may vary due to round off errors
diff --git a/3369/CH19/EX19.8/Ex19_8.sce b/3369/CH19/EX19.8/Ex19_8.sce
new file mode 100755
index 000000000..2dba7fd25
--- /dev/null
+++ b/3369/CH19/EX19.8/Ex19_8.sce
@@ -0,0 +1,14 @@
+//Chapter 19,Example 8,page 676
+//Determine the velocity of the ejected ions and propolsion force
+clear
+clc
+mi = 133*1.67*10^-27 // kg
+qi = 1.6*10^-19 // C
+Va = 3500 // V
+I = 0.2 // A
+vi = sqrt(2*qi*Va/mi)
+F = vi*mi*I/qi
+printf("\n Ion velocity = %e m/s",vi)
+printf("\n Populsion force = %e N",F)
+
+// Answers may vary due to round off errors
diff --git a/3369/CH19/EX19.9/Ex19_9.sce b/3369/CH19/EX19.9/Ex19_9.sce
new file mode 100755
index 000000000..c7db3bfcc
--- /dev/null
+++ b/3369/CH19/EX19.9/Ex19_9.sce
@@ -0,0 +1,18 @@
+//Chapter 19,Example 9,page 677
+//Determine the position of the particle 
+clear
+clc
+V = 120*10^3 // applied voltage in V
+d = 0.6 //  space b/w the plates in m
+vd = 1.2 // vertical dimention in m 
+qm = 10*10^-6 // charge to mass C/kg 
+y = 4.9
+
+t0 = sqrt(vd/y)
+// based on eq 19.51 and 19.52
+dx2 = qm*V/d
+x = t0^2
+printf("\n Velocity = %d m/s2",dx2)
+printf("\n Position of the particle = %f m",x)
+
+// Answer may vary due to round off error
diff --git a/3369/CH2/EX2.11/Ex2_11.sce b/3369/CH2/EX2.11/Ex2_11.sce
new file mode 100644
index 000000000..67670ce53
--- /dev/null
+++ b/3369/CH2/EX2.11/Ex2_11.sce
@@ -0,0 +1,20 @@
+//Chapter 2, Exmaple 11, page 75
+//Calculate the potential within the mesh
+clc
+clear
+//Based on figure 2.38(b)
+//equations are obtained using Eq.2.46
+A1 = 1/2*(0.54+0.16)
+A2 = 1/2*(0.91+0.14)
+S = [0.5571 -0.4571 -0.1;-0.4751 0.828 0.3667;-0.1 0.667 0.4667]
+//By obtaining the elements of the global stiffness matrix(Sadiku,1994)
+//and by emplying the Eq.2.49(a)
+S1 = [1.25 -0.014;-0.014 0.8381]
+S2 = [-0.7786 -0.4571;-0.4571 -0.3667]
+Phi13 = [0 ;10]
+val1 = S2*Phi13
+Phi24 = S1\val1
+disp(-Phi24,"The values of Phi2 and Phi4 are:")
+
+//Answers may vary due to round of error  
+
diff --git a/3369/CH2/EX2.5/Ex2_5.sce b/3369/CH2/EX2.5/Ex2_5.sce
new file mode 100755
index 000000000..e3a24bd96
--- /dev/null
+++ b/3369/CH2/EX2.5/Ex2_5.sce
@@ -0,0 +1,14 @@
+//Chapter 2, Example 5, page 65
+//Calculate the maximum field at the sphere surface
+clc
+clear
+//Calulating Field at surface E based on figure 2.31 and table 2.3
+Q1 = 0.25
+e0 = 8.85418*10**-12 //Epselon nought
+RV1= ((1/0.25**2)+(0.067/(0.25-0.067)**2)+(0.0048/(0.25-0.067)**2))
+RV2= ((0.25+0.01795+0.00128)/(0.75-0.067)**2)
+RV= RV1+RV2
+E = (Q1*RV)/(4*%pi*e0)
+printf("Maximum field = %e V/m per volt",E)
+
+//Answers vary due to round off error
diff --git a/3369/CH2/EX2.6/Ex2_6.sce b/3369/CH2/EX2.6/Ex2_6.sce
new file mode 100755
index 000000000..9548d382f
--- /dev/null
+++ b/3369/CH2/EX2.6/Ex2_6.sce
@@ -0,0 +1,62 @@
+//Chapter 2, Exmaple 6, page 66
+clc
+clear
+//calculation based on figure 2.32
+
+//(a)Charge on each bundle
+printf("Part a\n")
+req = sqrt(0.0175*0.45)
+printf("Equivalent radius = %e m \n", req)
+V = 400*10**3 //Voltage
+H = 12 //bundle height in m
+d = 9 //pole to pole spacing in m
+e0 = 8.85418*10**-12 //Epselon nought
+Hd = sqrt((2*H)^2+d^2)//2*H^2 + d^2
+Q = V*2*%pi*e0/(log((2*H/req))-log((Hd/d)))
+q = Q/2
+printf("Charge per bundle = %e uC/m \n",Q) //micro C/m
+printf("Charge per sunconducter = %e uC/m \n",q) //micro C/m
+
+//(b part i)Maximim & average surface feild
+printf("\nPart b")
+printf("\nSub part 1\n")
+r = 0.0175 //subconductor radius
+R = 0.45 //conductor to subconductor spacing
+MF = (q/(2*%pi*e0))*((1/r)+(1/R)) // maximum feild
+printf("Maximum feild = %e kV/m \n",MF)
+MSF = (q/(2*%pi*e0))*((1/r)-(1/R)) // maximum surface feild
+printf("Maximum feild = %e kV/m \n",MSF)
+ASF = (q/(2*%pi*e0))*(1/r) // Average surface feild
+printf("Maximum feild = %e kV/m \n",ASF)
+
+//(b part ii) Considering the two sunconductors on the left
+printf("\nSub part 2\n")
+//field at the outer point of subconductor #1 
+drO1 = 1/(d+r)
+dRrO1 = 1/(d+R+r)
+EO1 =  MF -((q/(2*%pi*e0))*(drO1+dRrO1))
+printf("EO1 = %e kV/m \n",EO1)
+//field at the outer point of subconductor #2 
+drO2 = 1/(d-r)
+dRrO2 = 1/(d-R-r)
+EO2 =  MF -((q/(2*%pi*e0))*(dRrO2+drO2))
+printf("EO2 = %e kV/m \n",EO2)
+
+//field at the inner point of subconductor #1 
+drI1 = 1/(d-r)
+dRrI1 = 1/(d+R-r)
+EI1 =  MSF -((q/(2*%pi*e0))*(drI1+dRrI1))
+printf("EI1 = %e kV/m \n",EI1)
+//field at the inner point of subconductor #2 
+drI2 = 1/(d+r)
+dRrI2 = 1/(d-R+r)
+EI2 =  MSF -((q/(2*%pi*e0))*(dRrI2+drI2)) 
+printf("EI2 = %e kV/m \n",EI2)
+
+//(part c)Average of the maximim gradient
+printf("\nPart c\n")
+Eavg = (EO1+EO2)/2
+printf("The average of the maximum gradient = %e kV/m \n",Eavg)
+
+
+//Answers might vary due to round off error
diff --git a/3369/CH2/EX2.7/Ex2_7.sce b/3369/CH2/EX2.7/Ex2_7.sce
new file mode 100755
index 000000000..6522de255
--- /dev/null
+++ b/3369/CH2/EX2.7/Ex2_7.sce
@@ -0,0 +1,12 @@
+//Chapter 2, Exmaple 7, page 69
+//Electric feild induced at x
+clc
+clear
+e0 = 8.85418*10**-12 //Epselon nought
+q = 1 // C/m
+C = (q/(2*%pi*e0))
+//Based on figure 2.33
+E = C-(C*(1/3+1/7))+(C*(1+1/5+1/9))+(C*(1/5+1/9))-(C*(1/3+1/7))
+printf("Electric Feild = %e V/m \n",E)
+
+//Answers might vary due to round off error
diff --git a/3369/CH2/EX2.8/Ex2_8.sce b/3369/CH2/EX2.8/Ex2_8.sce
new file mode 100755
index 000000000..887ff754e
--- /dev/null
+++ b/3369/CH2/EX2.8/Ex2_8.sce
@@ -0,0 +1,26 @@
+//Chapter 2, Exmaple 8, page 70
+//Calculate the volume of the insulator
+clc
+clear
+//Thinkness of graded design
+V = 150*sqrt(2)
+Ebd = 50
+T = V/Ebd
+printf("\nThickness of graded design= %e cm \n",T)
+//Based on figure 2.24
+r = 2 // radius of the conductor
+l = 10 //length of graded cylinder; The textbook uses 10 instead of 20
+zr = l*(T+r)
+printf("Curve = %e cm^2 \n",zr)
+//Volume of graded design V1
+V1 = 4*%pi*zr*(zr-r)
+printf("V1 = %e cm^3 \n",V1) //Unit is wrong in the textbook
+//Thickness of regular design as obtained form Eq.2.77
+pow = V/(2*Ebd)
+t = 2*(%e^pow-1)
+printf("Thickness of regular design = %e cm \n",t)
+//Volume of regular design V2
+V2 = %pi*((2+t)^2-4)
+printf("V2 = %e cm^3 \n",V2)//unit not mentioned in textbook
+ 
+//Answers may vary due to round off error
diff --git a/3369/CH3/EX3.1/Ex3_1.sce b/3369/CH3/EX3.1/Ex3_1.sce
new file mode 100755
index 000000000..9ac8a49b8
--- /dev/null
+++ b/3369/CH3/EX3.1/Ex3_1.sce
@@ -0,0 +1,16 @@
+//Chapter 3, Exmaple 1, page 103
+//Movement of oxygen molecule
+clc
+clear
+//using equation 3.3
+R = 3814 // J/Kg.mol.K
+T = 300 // K
+M = 32 // mol^-1
+V2 = 3*R*(T/M)
+V = sqrt(V2)
+printf("Velocity of Oxygen (O2)= %d m^2/s^2\n",V2)
+//Since Oxygen is a diatomic gas
+printf("Velocity of Oxygen (O)= %d m/s",V)
+//Velocity of oxygen is about 300 m/s
+
+//Answer given in the textbook is wrong
diff --git a/3369/CH3/EX3.10/Ex3_10.sce b/3369/CH3/EX3.10/Ex3_10.sce
new file mode 100755
index 000000000..96fbf7aca
--- /dev/null
+++ b/3369/CH3/EX3.10/Ex3_10.sce
@@ -0,0 +1,16 @@
+//Chapter 3, Exmaple 10, page 106
+//KE and velocity of photoelectron
+clc
+clear
+h = 4.15*10**-15
+c = 3*10**8
+l = 200*10**-10
+BE = 13.6 // Binding energy
+PE = h*c/l
+KE = PE-BE // Kinetic energy of photoelectron
+Ve = sqrt((2*KE*1.6*10**-19)/9.11*10**-31)*10**31
+printf("\nPhoton energy eV = %e ",PE)
+printf("\nKinetic energy eV = %e ",KE)
+printf("\nVelocity m/s = %e ",Ve)
+
+//Answer may vary due to round off error
diff --git a/3369/CH3/EX3.11/Ex3_11.sce b/3369/CH3/EX3.11/Ex3_11.sce
new file mode 100755
index 000000000..a4e989765
--- /dev/null
+++ b/3369/CH3/EX3.11/Ex3_11.sce
@@ -0,0 +1,11 @@
+//Chapter 3, Exmaple 11, page 107
+//Find the absorption coefficient
+clc
+clear
+// Using equation 3.20
+x = 20
+I0 = 6
+Mu = -1/x*log(1/I0)
+printf("\nLiquid photon absorption coefficient cm^-1 = %e ",Mu)
+
+//Answer may vary due to round off error 
diff --git a/3369/CH3/EX3.12/Ex3_12.sce b/3369/CH3/EX3.12/Ex3_12.sce
new file mode 100755
index 000000000..1e6e0a8fd
--- /dev/null
+++ b/3369/CH3/EX3.12/Ex3_12.sce
@@ -0,0 +1,11 @@
+//Chapter 3, Exmaple 12, page 107
+//Binding energy
+clc
+clear
+h = 4.15*10**-15
+c = 3*10**8
+Imax = 1000*10**-10
+We = h*c/Imax
+printf("\nBinding Energy = %e eV ",We)
+
+//Answer may vary due to round off errorS
diff --git a/3369/CH3/EX3.14/Ex3_14.sce b/3369/CH3/EX3.14/Ex3_14.sce
new file mode 100755
index 000000000..854f67a5e
--- /dev/null
+++ b/3369/CH3/EX3.14/Ex3_14.sce
@@ -0,0 +1,12 @@
+//Chapter 3, Exmaple 14, page 108
+//Diameter of argon atom
+clc
+clear
+//As derived from example 13
+N = (1.01*10**5/760)/(1.38*10**-23*273)
+printf("\nN = %e atoms/m^3 ",N)
+//Use equation 3.10
+ra = sqrt((85*10^2/(%pi*3.527*10**22))) 
+printf("\nra = %e m ",ra)
+
+//Answer may vary due to round off error
diff --git a/3369/CH3/EX3.15/Ex3_15.sce b/3369/CH3/EX3.15/Ex3_15.sce
new file mode 100644
index 000000000..6fb064290
--- /dev/null
+++ b/3369/CH3/EX3.15/Ex3_15.sce
@@ -0,0 +1,13 @@
+//Chapter 3, Exmaple 15, page 109
+//Mobility of electrons
+clc
+clear
+Ie = 3
+d = 0.8
+A = 8*10**-4
+Vne = 20*10**17 //V*ne
+e = 1.6*10**-19
+ke = (Ie*d)/(A*Vne*e)
+printf("\Mobility of electrons = %d m^2/s*V ",ke)
+
+//Answer may vary from the text
diff --git a/3369/CH3/EX3.17/Ex3_17.sce b/3369/CH3/EX3.17/Ex3_17.sce
new file mode 100755
index 000000000..fc69b6308
--- /dev/null
+++ b/3369/CH3/EX3.17/Ex3_17.sce
@@ -0,0 +1,11 @@
+//Chapter 3, Exmaple 17, page 110
+//Determine the ion density
+clc
+clear
+//Based on equation 3.50 and 3.52
+nplus = 10**11*%e**(-1.6*10**-19*5*0.02/(1.38*10**-23*293))
+nminus = 10**11*%e**(-1.6*10**-19*5*-0.02/(1.38*10**-23*293)) //textbook uses 0.02 inseatead of -0.02. In the program I have used -0.02
+printf("\n+(0.02) = %e ions/m^3 ",nplus)
+printf("\n+(-0.02) = %e ions/m^3 ",nminus)
+
+//answers may vary due to round off error
diff --git a/3369/CH3/EX3.18/Ex3_18.sce b/3369/CH3/EX3.18/Ex3_18.sce
new file mode 100755
index 000000000..6bcdcaba8
--- /dev/null
+++ b/3369/CH3/EX3.18/Ex3_18.sce
@@ -0,0 +1,17 @@
+//Chapter 3, Exmaple 18, page 110
+//Determine the diameter
+clc
+clear
+//Based on the equation 3.40
+k = 1.38*10**-23
+T = 293
+z2z1 = 0.05
+e = 1.6*10**-19
+E = 250
+r1 = 0.09*10**-6
+r1r2 = (6*k*T*z2z1)/(e*E)
+r2 = sqrt(r1+r1r2)
+printf("\n r1^2-r2^2 = %e ",r1r2)
+printf("\n r2 = %e m ",r2)
+
+//answers may vary due to round off error
diff --git a/3369/CH3/EX3.19/Ex3_19.sce b/3369/CH3/EX3.19/Ex3_19.sce
new file mode 100755
index 000000000..06ffba79d
--- /dev/null
+++ b/3369/CH3/EX3.19/Ex3_19.sce
@@ -0,0 +1,13 @@
+//Chapter 3, Exmaple 19, page 111
+//Determine mean free path and ionization
+clc
+clear
+//(a)Mean free path
+//Based on equation 3.14 and 3.15
+lambda = 1/(9003*0.5)
+//(b)Ionization potential
+Vi = 256584/9003
+printf("\n lambda = %e m ",lambda)
+printf("\n Vi = %f V ",Vi)
+
+//answers may vary due to round off error
diff --git a/3369/CH3/EX3.2/Ex3_2.sce b/3369/CH3/EX3.2/Ex3_2.sce
new file mode 100755
index 000000000..0b4f51c8f
--- /dev/null
+++ b/3369/CH3/EX3.2/Ex3_2.sce
@@ -0,0 +1,12 @@
+//Chapter 3, Exmaple 2, page 104
+//Kinetic energy of oxygen molecule
+clc
+clear
+//from Eq.3.2
+G = (2*10**-3/32)*(8314*298*1.01*10**5)*10**-10
+printf("\nG = %e m^3\n",G) // Answer is is wrong in the text 
+//From equation 3.1
+mv2 = 3/2*1.01*10**5 // 1/2*m0*v^2
+KE = mv2*G//total transalational K.E
+printf("K.E = %f J\n",KE)
+//Answer may varry due to round off error
diff --git a/3369/CH3/EX3.3/Ex3_3.sce b/3369/CH3/EX3.3/Ex3_3.sce
new file mode 100755
index 000000000..d182cf325
--- /dev/null
+++ b/3369/CH3/EX3.3/Ex3_3.sce
@@ -0,0 +1,17 @@
+//Chapter 3, Exmaple 3, page 104
+//Maximum pressure in the chamber
+clc
+clear
+//Making use of equation 3.10
+N1 = (4*%pi*1.7*1.7*0.10*10^-10*10^-10)
+N = 1/N1
+//Using equation 3.2
+R = 8314 // J/Kg*mol*K 
+M = 28 // Mol^-1
+N = 220*10**-8 // Kg
+T = 300 // K
+p =  N/M*R*T
+printf("\nN = %e ",N1) // answer mentioned in the tectbook is wrong
+printf("\nPressure = %f N/m^2",p)
+
+//Answer vary due to round off error
diff --git a/3369/CH3/EX3.4/Ex3_4.sce b/3369/CH3/EX3.4/Ex3_4.sce
new file mode 100755
index 000000000..26f96dbc7
--- /dev/null
+++ b/3369/CH3/EX3.4/Ex3_4.sce
@@ -0,0 +1,12 @@
+//Chapter 3, Exmaple 4, page 105
+//Temperature & Average K.E of He atom
+clc
+clear
+m0 = 1
+v2 = 1.6*10**-19 // V^2
+KE = m0*v2
+//Using equation 3.3
+T = 2*KE/(3*1.38*10**-23) 
+printf("\nK.E = %e J",KE)
+printf("\nTemperature = %e K",T)
+  
diff --git a/3369/CH3/EX3.5/Ex3_5.sce b/3369/CH3/EX3.5/Ex3_5.sce
new file mode 100755
index 000000000..cc49f0bbe
--- /dev/null
+++ b/3369/CH3/EX3.5/Ex3_5.sce
@@ -0,0 +1,9 @@
+//Chapter 3, Exmaple 5, page 105
+//Volume of Helium 
+clc
+clear
+// Using equation 3.2
+G = (1*8314*273)/(2.016*1.01*10**5)
+printf("\nVolume of He = %f m^3",G)
+
+//Answer may vary due to round off error.
diff --git a/3369/CH3/EX3.6/Ex3_6.sce b/3369/CH3/EX3.6/Ex3_6.sce
new file mode 100755
index 000000000..702cde8e4
--- /dev/null
+++ b/3369/CH3/EX3.6/Ex3_6.sce
@@ -0,0 +1,12 @@
+//Chapter 3, Exmaple 6, page 105
+//Determine mean free path 
+clc
+clear
+//(a) Mean free path
+na = %e^-1
+//(b) 5 times mean free path
+nb = %e^-5
+printf("\n Mean free path = %f*n0 ",na)
+printf("\n 5 times mean free path = %f*n0 ",nb)
+
+//Answer may vary due to round of error 
diff --git a/3369/CH3/EX3.7/Ex3_7.sce b/3369/CH3/EX3.7/Ex3_7.sce
new file mode 100755
index 000000000..b1cb00182
--- /dev/null
+++ b/3369/CH3/EX3.7/Ex3_7.sce
@@ -0,0 +1,13 @@
+//Chapter 3, Exmaple 7, page 105
+//Mean square velocity of Helium 
+clc
+clear
+//based on equation 3.2 and 3.3 we derive the gas density
+N = 178*10**-3 // kg/m^3
+// calculating mean square velocity
+v2 = (3*1.01*10**5)/N
+printf("\nV^2 = %e m^2/s^2",v2)
+v = sqrt(v2)
+printf("\nMean square velocity = %f m/s",v)
+
+//Answer may vary due to round off error
diff --git a/3369/CH3/EX3.8/Ex3_8.sce b/3369/CH3/EX3.8/Ex3_8.sce
new file mode 100755
index 000000000..417bd8547
--- /dev/null
+++ b/3369/CH3/EX3.8/Ex3_8.sce
@@ -0,0 +1,11 @@
+//Chapter 3, Exmaple 8, page 106
+//Energy of free electron
+clc
+clear
+//Using equation 3.3
+mv2 = (3/2*1.38*10**-21*293) // 1/2*m*v^2
+E = mv2*10**38/1.6*10**-19
+printf("\n1/2*m*v^2 = %e J",mv2)
+printf("\nEnergy of free electron = %f eV",E)
+
+//Answers may vary due to round off error
diff --git a/3369/CH3/EX3.9/Ex3_9.sce b/3369/CH3/EX3.9/Ex3_9.sce
new file mode 100755
index 000000000..6d23d456f
--- /dev/null
+++ b/3369/CH3/EX3.9/Ex3_9.sce
@@ -0,0 +1,14 @@
+//Chapter 3, Exmaple 9, page 106
+//Average separation and volume occupied by one atom
+clc
+clear
+NA = 6.0244*10**23
+NoA = NA*0.075 // Number of atoms/cm^3
+V = 1/NoA // Average volume occupied by one atom
+S = nthroot(V,3) //  Average separation between atoms
+printf("\nNumber od atoms per cm^3 = %e ",NoA)
+printf("\nAverage vloume occupied by one atom = %e cm^3",V)
+printf("\nAverage separation between atoms = %e cm",S)
+
+
+//Answers may vary due to round off error
diff --git a/3369/CH4/EX4.1/Ex4_1.sce b/3369/CH4/EX4.1/Ex4_1.sce
new file mode 100755
index 000000000..4a9e91ba7
--- /dev/null
+++ b/3369/CH4/EX4.1/Ex4_1.sce
@@ -0,0 +1,17 @@
+//Chapter 4, Exmaple 1, page 139
+//Claculate alpha and No. of electrons emmited
+clc
+clear
+//Claculate (a)alpha
+d2 = 0.01
+d1 = 0.005
+I2 = 2.7*10**-7
+I1 = 2.7*10**-8
+alpha = 1/(d2-d1)*log(I2/I1)
+//(b)number of electrons emmited from cathode per second
+I0 = I1*%e**(-alpha*d1)
+n0 = I0/(1.6*10**-19)
+printf("\n Part (a)\n alpha = %f m^-1",alpha)
+printf("\n Part (b)\n I0 = %e ",I0)
+printf("\n No of electrons emitted = %e electrons/s",n0)
+//Answer may vary due to round off error
diff --git a/3369/CH4/EX4.10/Ex4_10.sce b/3369/CH4/EX4.10/Ex4_10.sce
new file mode 100755
index 000000000..ab65e2563
--- /dev/null
+++ b/3369/CH4/EX4.10/Ex4_10.sce
@@ -0,0 +1,28 @@
+//Chapter 4, Exmaple 10, page 144
+//Claculate (a)Raether's criterion (b)Meek and Lobe's criterion
+clc
+clear
+//(a)Raether's criterion
+// as assumed by Raether and based equation 3.3, 3.50, 4.22 and 4.23
+d = 0.001 // m
+alpha = 10792.2 // m^-1
+p = 101.3 //kPa^-1
+ap = 106.54 // alpha/p Unit: m^-1*kPa^-1
+T = 11253.7 // m^-1*kPa^-1
+B = 273840 // V/m*kPa
+Ep = 58764.81 // E/p Unit:V/m*kPa
+
+ad = 17.7 + log(d)
+E = Ep*p
+Vs = E*d*10^-3 //Voltage breakdown
+printf("\n E = %e V/m",E)
+printf("\n Voltage breakdown = %f kV",Vs)
+
+//(b)Meek and Loeb's criterion
+//Using equation 4.11 and based on 4.24 & 4,25 
+//+ we get Er = 468*10^4 V/m
+Er = 468*10^4 // V/m
+Vs2 = Er*0.001*10^-3
+printf("\n Voltage breakdown = %f kV",Vs2)
+
+// Answers may vary due to round of error
diff --git a/3369/CH4/EX4.11/Ex4_11.sce b/3369/CH4/EX4.11/Ex4_11.sce
new file mode 100755
index 000000000..8efd2abd0
--- /dev/null
+++ b/3369/CH4/EX4.11/Ex4_11.sce
@@ -0,0 +1,13 @@
+//Chapter 4, Exmaple 11, page 146
+//Claculate the first Townsend's ionization coefficient
+clc
+clear
+t = 0.2*10**-6 // transit time of electrons in seconds
+d = 0.05 // m
+ve = d/t
+TC = 35*10**-9 // Time constant
+a = 1/(ve*TC)
+printf("\n Electron drift velocity = %e m/s",ve)
+printf("\n alpha = %e m^-1",a)
+
+// Answers may vary due to round of error
diff --git a/3369/CH4/EX4.12/Ex4_12.sce b/3369/CH4/EX4.12/Ex4_12.sce
new file mode 100755
index 000000000..6e591561e
--- /dev/null
+++ b/3369/CH4/EX4.12/Ex4_12.sce
@@ -0,0 +1,19 @@
+//Chapter 4, Exmaple 12, page 146
+//Travel time and maximum frequency
+clc
+clear
+//(a)Determine the travel time
+Ea = 200*sqrt(2)*10**3/0.1
+x = 1.4*10**-4*2828.4*10**3/(2*%pi*50)
+d = 0.1
+printf("\n Ea = %e V/m",Ea)
+printf("\n x = %f*sin(3.14*t)",x)
+//obtaining t from x
+t = asin(d/x)/3.14
+printf("\n t = %f ms",t) // answer mentioned in the text is wrong
+//(b)Determine the maximum frequency
+k = 1.4*10**-4
+fmax = k*Ea/(2*%pi*d)
+printf("\n fmax = %f Hz",fmax)
+
+//Answer may vary due to round off error
diff --git a/3369/CH4/EX4.2/Ex4_2.sce b/3369/CH4/EX4.2/Ex4_2.sce
new file mode 100755
index 000000000..4c6c80ee8
--- /dev/null
+++ b/3369/CH4/EX4.2/Ex4_2.sce
@@ -0,0 +1,18 @@
+//Chapter 4, Exmaple 2, page 140
+//Claculate electrode space
+clc
+clear
+//based on the values of example 1
+d2 = 0.01
+d1 = 0.005
+I2 = 2.7*10**-7
+I1 = 2.7*10**-8
+a = 1/(d2-d1)*log(I2/I1) // alpha
+//10^9 = %e^a(a*d) 
+//multiplying log on bith sides log(10^9) = a*d
+ad = log(10^9)
+printf("\n a*d = %f ",ad)
+d = ad/a
+printf("\n electrode space = %f m",d) 
+
+//Answers may vary due to round off error
diff --git a/3369/CH4/EX4.3/Ex4_3.sce b/3369/CH4/EX4.3/Ex4_3.sce
new file mode 100755
index 000000000..5858a2127
--- /dev/null
+++ b/3369/CH4/EX4.3/Ex4_3.sce
@@ -0,0 +1,13 @@
+//Chapter 4, Exmaple 3, page 140
+//Claculate size of developed avalanche
+clc
+clear
+a = 4*10**4
+b = 15*10**5
+//Rewriting equation 4.2
+x0=0;x1=0.0005;
+X=integrate('a-b*sqrt(x)','x',x0,x1);
+As = exp(X) // Avelanche size
+printf("\n Avalanche size = %f m",As)
+ 
+//Answers may vary due to round of error
diff --git a/3369/CH4/EX4.4/Ex4_4.sce b/3369/CH4/EX4.4/Ex4_4.sce
new file mode 100755
index 000000000..1db17855f
--- /dev/null
+++ b/3369/CH4/EX4.4/Ex4_4.sce
@@ -0,0 +1,13 @@
+//Chapter 4, Exmaple 4, page 141
+//Claculate distance to produce avalanche
+clc
+clear
+//Rewrite equation 4.2
+//using the values of a and b from previous example
+//convert integartion to quaderatic equation form
+x=poly(0,"x");
+p=59.97-4*10**4*x+7.5*10**5*x^2 // making the polinomial equation
+r= roots(p) //obtaining the roots
+printf("\n %f m or %f m away from the cathode",r(1),r(2))
+
+//Answer may vary due to round of error.
diff --git a/3369/CH4/EX4.5/Ex4_5.sce b/3369/CH4/EX4.5/Ex4_5.sce
new file mode 100755
index 000000000..e78b58fc0
--- /dev/null
+++ b/3369/CH4/EX4.5/Ex4_5.sce
@@ -0,0 +1,11 @@
+//Chapter 4, Exmaple 5, page 141
+//Claculate minimum distance to produce avalanche of size 10^19
+clc
+clear
+//Rewriting equation 4.2 and converting it into quadratic equation
+x=poly(0,"x");
+p=43.75-4*10**4*x+7.5*10**5*x^2 // making the polinomial equation
+r= roots(p) //obtaining the roots
+printf("\n Minimum distance = %f m",r(2)) // other root is disregarded
+
+//Answer may vary due to round of error.
diff --git a/3369/CH4/EX4.7/Ex4_7.sce b/3369/CH4/EX4.7/Ex4_7.sce
new file mode 100755
index 000000000..b388e41e3
--- /dev/null
+++ b/3369/CH4/EX4.7/Ex4_7.sce
@@ -0,0 +1,17 @@
+//Chapter 4, Exmaple 7, page 142
+//Claculate secondary coefficient
+clc
+clear
+//Using equation 3.15
+E = 9*10**3/0.002
+T = 11253.7 // m^-7*kPa^-1
+B = 273840 // V/mkPa
+p = 101.3 // kPa or 1 atm
+d = 0.002 // m
+alpha = p*T*exp(-B*p/E)
+Y = 1/(exp(alpha*d)-1)
+printf("\n E = %e V/m",E)
+printf("\n Alpha = %f m^-1",alpha)
+printf("\n Total secondary coefficient of ionization = %f ",Y)
+
+//Answer may vary due to round off error
diff --git a/3369/CH4/EX4.8/Ex4_8.sce b/3369/CH4/EX4.8/Ex4_8.sce
new file mode 100755
index 000000000..e4c3f69cd
--- /dev/null
+++ b/3369/CH4/EX4.8/Ex4_8.sce
@@ -0,0 +1,31 @@
+//Chapter 4, Exmaple 8, page 143
+//Claculate first and secondary ionization coefficient
+clc
+clear
+//(a)first ionization coefficient
+//Using equation 4.7a
+d1 = 0.005
+a1d1 = log(1.22)
+a1 = a1d1/d1
+
+d2 = 0.01504
+a2d2 = log(1.82)
+a2 = a2d2/d2
+
+d3 = 0.019 // wrong value used in the text
+a3d3 = log(2.22)
+a3 = a3d3/d3
+
+printf("\n Alpha 1 = %f m^-1",a1)
+printf("\n Alpha 2 = %f m^-1",a2)
+printf("\n Alpha 3 = %f m^-1",a3)
+printf("\n From the above results we can understand that ionization mechanism must be acting at d3 ")
+
+//secondary ionization coefficient
+I = 2.22
+e = exp(a1*d3)
+Y = (I-e)/(I*(e-1))
+printf("\n secondary ionization coefficient = %f ",Y)
+
+//Answer may vary due to round off error.
+
diff --git a/3369/CH4/EX4.9/Ex4_9.sce b/3369/CH4/EX4.9/Ex4_9.sce
new file mode 100755
index 000000000..121fb59c9
--- /dev/null
+++ b/3369/CH4/EX4.9/Ex4_9.sce
@@ -0,0 +1,18 @@
+//Chapter 4, Exmaple 9, page 144
+//Claculate distance and voltage
+clc
+clear
+a = 39.8 // alpha
+Y = 0.0354 // corfficient
+p = 0.133 // kPa
+Ep = 12000 // E/P , unit : V/m*kPa
+
+d = (1/a)*(log(1/Y + 1)) // distance
+E = Ep*p
+V = E*d
+
+printf("\n Distance = %f m",d)
+printf("\n E = %f V/m",E)
+printf("\n Volatge  = %f V",V)
+
+//Answers may vary due to round off error
diff --git a/3369/CH5/EX5.10/Ex5_10.sce b/3369/CH5/EX5.10/Ex5_10.sce
new file mode 100755
index 000000000..105e09d23
--- /dev/null
+++ b/3369/CH5/EX5.10/Ex5_10.sce
@@ -0,0 +1,30 @@
+//Chapter 5, Exmaple 10, page 180
+//Calculate corona power loss
+clc
+clear
+p = 75 // pressure
+t = 35 // temprature
+m1 = 0.92
+m2 = 0.95
+t = 5*5*8.66 // the three side of the trangle in m
+Deq = nthroot(t,3) 
+dt = (3.92*p)/(273+t) //Relative air density
+E0 = 30*dt*(1 + 0.3*sqrt(dt))*m1*m2
+En = 27.501 // kVpeak/cm
+Vph = (275*10^3)/sqrt(3)
+V0peak = E0*log(Deq*10**2)
+V0 = En*log(Deq*10**2)
+V0ratio = 275/V0
+printf("\n Reative air density %f ",dt)
+printf("\n Corona onset field = %f kVpeak/cm",E0)
+printf("\n V0peak = %f kVpeak",V0peak)
+printf("\n V0 = %f kV",V0)
+printf("\n Ration of V0 = %f ",V0ratio)
+K = 0.05 // K factor
+Pc = (3.73*K*50*Vph^2)/(Deq*10**2)^2
+Cc = Pc*10^3/Vph
+printf("\n Corona power loss Pc = %f kW/km",Pc*10**-5)
+printf("\n Corona current = %f mA/Km",Cc*10^-2)
+
+//Answer vary due to round off error
+//Some of the answers provided in the textbook are wrong
diff --git a/3369/CH5/EX5.11/Ex5_11.sce b/3369/CH5/EX5.11/Ex5_11.sce
new file mode 100755
index 000000000..217606fe7
--- /dev/null
+++ b/3369/CH5/EX5.11/Ex5_11.sce
@@ -0,0 +1,23 @@
+//Chapter 5, Exmaple 11, page 180
+//Calculate corona onset voltage and effective corona envelope
+clc
+clear
+//(a) corona onset voltage
+r = 3.175 // cm
+h = 13 // m
+m= 0.9 // m1 and m2
+dt = 1 // Relative air density
+E0 = 30*dt*(1 + 0.3/sqrt(r))*m*m
+V0 = 20*r*log(2*h*10^2/r)
+printf("\n E0 = %f kVpeak/cm or 20 kV/cm",E0)
+printf("\n V0 = %f kV",V0)
+printf("\n V0 (line to line) = %f kV",V0*sqrt(3))
+
+//(b)Corona envelope at 2.5 p.u
+V = 2.5*525 // line to line voltage * 2.5
+printf("\n Voltage (line to line) = %f kV",V)
+//Solving the equations in trila and error method
+printf("\n Envelope radius = 5 cm")
+
+// Answers may vary due to round off error.
+ 
diff --git a/3369/CH5/EX5.2/Ex5_2.sce b/3369/CH5/EX5.2/Ex5_2.sce
new file mode 100755
index 000000000..5740be366
--- /dev/null
+++ b/3369/CH5/EX5.2/Ex5_2.sce
@@ -0,0 +1,29 @@
+//Chapter 5, Exmaple 2, page 173
+//Calculate breakdown voltage
+clc
+clear
+//(a)Based on equation 4.13
+p = 101.3 // kPa
+Ep = 2400.4/0.027
+E = p*Ep
+d = 1*10**-3 // 1 mm
+Vs1 = E*d
+printf("\n Part (a): based on equation 4.13")
+printf("\n Breakdown voltage = %f V or %f kV",Vs1,Vs1*10^-3)
+
+//(b)Corrsponding to an avelanche size of 10^8
+p = 101.3 // kPa
+Cp = Ep*0.027*p
+Vs2 = (18.42 + (Cp*10**-3))/0.027
+printf("\n Part (b):Corrsponding to an avelanche size of 10^8")
+printf("\n Breakdown voltage = %f V or %f kV",Vs2,Vs2*10^-3)
+
+//(b)According to criteria expressed by Equations 5.4 and 5.5
+p = 101.3 // kPa
+Vs3a = 9.4
+Vs3b = 9.2
+printf("\n Part (c):According to criteria expressed by Equations 5.4 and 5.5")
+printf("\n Breakdown voltage = %f kV or %f kV",Vs3a,Vs3b)
+
+//Answer may vary due to round off error
+
diff --git a/3369/CH5/EX5.3/Ex5_3.sce b/3369/CH5/EX5.3/Ex5_3.sce
new file mode 100755
index 000000000..641d2cfa5
--- /dev/null
+++ b/3369/CH5/EX5.3/Ex5_3.sce
@@ -0,0 +1,31 @@
+//Chapter 5, Exmaple 3, page 174
+//Calculate breakdown voltage at atm pressure 3 and 5
+clc
+clear
+//(a)Based on equation 5.14
+p = 101.3 // kPa
+Ep = 2400.4/0.027
+E = p*Ep
+d = 1*10**-3 // 1 mm
+Vs13 = E*d*3 // at 3 atm
+Vs15 = E*d*5 // at 5 atm
+printf("\n Part (a): based on equation 5.14")
+printf("\n Breakdown voltage = %f kV or %f kV",Vs13*10^-3,Vs15*10^-3)
+
+//(b)According to eqution 5.13
+p = 101.3 // kPa
+Cp3 = Ep*0.027*p*3 // at 3 atm
+Vs23 = (18.42 + (Cp3*10**-3))/0.027
+Cp5 = Ep*0.027*p*5 // at 5 atm
+Vs25 = (18.42 + (Cp5*10**-3))/0.027
+printf("\n Part (b):According to eqution 5.13")
+printf("\n Breakdown voltage = %f V or %f kV",Vs23*10^-3,Vs25*10^-3)
+
+//(b)According to criteria expressed by Equations 5.4 and 5.5
+p = 101.3 // kPa
+Vs3a = 27.73 // at 3 atm
+Vs3b = 45.5 // at 5 atm
+printf("\n Part (c):According to criteria expressed by Equations 5.4 and 5.5")
+printf("\n Breakdown voltage = %f kV or %f kV",Vs3a,Vs3b)
+
+//Answer may vary due to round off error
diff --git a/3369/CH5/EX5.8/Ex5_8.sce b/3369/CH5/EX5.8/Ex5_8.sce
new file mode 100755
index 000000000..637191715
--- /dev/null
+++ b/3369/CH5/EX5.8/Ex5_8.sce
@@ -0,0 +1,43 @@
+//Chapter 5, Exmaple 8, page 179
+//Calculate corona onset voltage
+clc
+clear
+s = 4 // cm
+r = 1 // cm
+D = 5*10^2 // cm
+dt = 1
+E0 = 30*dt*(1 + 0.3*sqrt(dt*r))
+printf("\n E0 = %f kVpeak/cm",E0)
+//using equations (5.18), the positive and negative corona
+En = 27.501 // kVpeak/cm
+//part a
+Vp1 = 6.2*E0
+Vn1 = 6.2*En
+printf("\n Part (a)")
+printf("\n The postive corona = %f kVpeak",Vp1)
+printf("\n The negative corona = %f kV",Vn1)
+//part b
+Vp2 = 8.32*E0
+Vn2 = 8.32*En
+printf("\n Part (b)")
+printf("\n The postive corona = %f kVpeak",Vp2)
+printf("\n The negative corona = %f kV",Vn2)
+//part c
+Vp3 = 9.97*E0
+Vn3 = 9.97*En
+printf("\n Part (c)")
+printf("\n The postive corona = %f kVpeak",Vp3)
+printf("\n The negative corona = %f kV",Vn3)
+//part d
+Vp4 = 11.39*E0
+Vn4 = 11.39*En
+printf("\n Part (d)")
+printf("\n The postive corona = %f kVpeak",Vp4)
+printf("\n The negative corona = %f kV",Vn4)
+
+//Answer CONSIDERABLY vary due to round off error.
+
+
+
+
+
diff --git a/3369/CH5/EX5.9/Ex5_9.sce b/3369/CH5/EX5.9/Ex5_9.sce
new file mode 100755
index 000000000..484d0ab2c
--- /dev/null
+++ b/3369/CH5/EX5.9/Ex5_9.sce
@@ -0,0 +1,20 @@
+//Chapter 5, Exmaple 9, page 180
+//Calculate corona onset voltage
+clc
+clear
+t = 5*5*8.66 // the three side of the trangle in m
+Deq = nthroot(t,3) 
+dt = 1 //delta = 1 at standard temperature and pressure
+r = 1 //radius of the conductor
+En = 27.501 // kVpeak/cm
+E0 = 30*dt*(1 + 0.3*sqrt(dt*r))
+V0peak = E0*log(Deq*10**2)
+V0 = En*log(Deq*10**2)
+
+printf("\n Mean geometric distance between the conductors %f m",Deq)
+printf("\n E0 = %f kVpeak/cm",E0)
+printf("\n V0peak = %f kVpeak",V0peak)
+printf("\n V0 = %f kV",V0)
+
+//Answers may vary due to round off error
+