initial commit / add all books

11 files changed, 295 insertions, 0 deletions

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
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+++ 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