17 files changed, 280 insertions, 0 deletions
diff --git a/1316/CH2/EX2.1/example2_1.sce b/1316/CH2/EX2.1/example2_1.sce
new file mode 100644
index 000000000..a7f5357e7
--- /dev/null
+++ b/1316/CH2/EX2.1/example2_1.sce
@@ -0,0 +1,29 @@
+//Chapter 2
+//Example 2.1
+//Page 55
+
+clear;
+clc;
+
+Ri = 10;
+TF = 0.020;
+Ro = 5;
+T = 50;
+
+printf("The unloaded output of sensor is simply ")
+//Calculation of Vt
+x = TF*T;
+printf("Vt = %.1f \n",x)
+printf("Since the amplifier has a gain of 10,the output of the amplifier appears to be ")
+//Calculation of Vout
+y = Ri*x;
+printf("Vout = %.0f V \n",y)
+printf("But this is wrong because of loading !\n ")
+//Correct Analysis
+printf("Here we see that  there will be a voltage dropped across the \n output resistance of the sensor.The actual amplifier input voltage will be given by ")
+//Calculation of Vin
+a = x*(1-((Ro)/(Ro+Ri)));
+printf("Vin = %.2f V.\n",a)
+printf("Thus the output of amplifier is actually Vout = %.1f V \n",Ri*a)
+
+
diff --git a/1316/CH2/EX2.1/resultexample2_1.txt b/1316/CH2/EX2.1/resultexample2_1.txt
new file mode 100644
index 000000000..a2a032eb1
--- /dev/null
+++ b/1316/CH2/EX2.1/resultexample2_1.txt
@@ -0,0 +1,7 @@
+ The unloaded output of sensor is simply Vt = 1.0 
+Since the amplifier has a gain of 10,the output of the amplifier appears to be Vout = 10 V 
+But this is wrong because of loading !
+ Here we see that  there will be a voltage dropped across the 
+ output resistance of the sensor.The actual amplifier input voltage will be given by Vin = 0.67 V.
+Thus the output of amplifier is actually Vout = 6.7 V 
+ 
+\ No newline at end of file
diff --git a/1316/CH2/EX2.10/example2_10.sce b/1316/CH2/EX2.10/example2_10.sce
new file mode 100644
index 000000000..7417d62b6
--- /dev/null
+++ b/1316/CH2/EX2.10/example2_10.sce
@@ -0,0 +1,25 @@
+//Chapter 2
+//Example 2.10
+//Page 68
+
+clear;
+clc;
+
+R1 = 1000;
+R2 = 2000;
+R3 = 1000;
+C1 = 1;
+
+printf("Because the bridge is at null , we have \n ")
+printf("Z2*Z3 = Z1*Zx \n")
+printf("R2(R3-j/wC)=R1(Rx-j/wCx) \n")
+printf("The real and imaginary parts must be indpendently equal,so that \n")
+printf("Rx-(R2*R3/R1)=0 \n")
+//Calculation of value of Rx
+x=(R2*R3/R1)/1000
+printf("Rx = %.0f kilo ohm \n",x)
+//Calculation of value of Cx
+y=(C1*(R1/R2))
+printf("Cx=C(R1/R2) \n Cx = %.1f uF",y)
+
+
diff --git a/1316/CH2/EX2.2/example2_2.sce b/1316/CH2/EX2.2/example2_2.sce
new file mode 100644
index 000000000..239adea00
--- /dev/null
+++ b/1316/CH2/EX2.2/example2_2.sce
@@ -0,0 +1,36 @@
+//Chapter 2
+//Example 2.2
+//Page 58
+
+clear;
+clc;
+
+R1 = 10;
+Vs = 5;
+R_low = 4;
+R_high = 12;
+
+
+printf("a. The solution is given by VsR2/R1+R2.For R2=4k.ohm,we have\n")
+//Calculation of Vd
+x = (Vs*R_low)/(R1+R_low);
+printf("VD = %.2f V\n",x)
+printf("For R2=12k.ohm,we have\n")
+//Calculation of Vd
+y = (Vs*R_high)/(R1+R_high);
+printf("VD = %.2f V\n",y)
+printf("b. Thus the voltage varies from %.2f to %.2f V",x,y)
+printf("c. The range of output impedance is found from the parallel \n combination of R1 and R2 for the minimum and maximum of R2.\n Simple parallel resistance computation shows that this will be from ")
+//Calculation of parallel resistances
+a = (R1*R_low)/(R1+R_low);
+b = (R1*R_high)/(R1+R_high);
+printf("%.2f to %.2f k.ohm",a,b)
+printf("d. The power dissipated by the sensor can be determined most easily from V^2/R2,as the voltage across R2 has been calculated.The power dissipated varies from ")
+c = (Vs^2)/R_high;
+d = (Vs^2)/R_low;
+printf("%.4f to %.4f W",c,d)
+
+
+
+
+
diff --git a/1316/CH2/EX2.2/resultexample2_2.txt b/1316/CH2/EX2.2/resultexample2_2.txt
new file mode 100644
index 000000000..a4bdafdfc
--- /dev/null
+++ b/1316/CH2/EX2.2/resultexample2_2.txt
@@ -0,0 +1,7 @@
+a. The solution is given by VsR2/R1+R2.For R2=4k.ohm,we have
+VD = 1.43 V
+For R2=12k.ohm,we have
+VD = 2.73 V
+b. Thus the voltage varies from 1.43 to 2.73 Vc. The range of output impedance is found from the parallel 
+ combination of R1 and R2 for the minimum and maximum of R2.
+ Simple parallel resistance computation shows that this will be from 2.86 to 5.45 k.ohmd. The power dissipated by the sensor can be determined most easily from V^2/R2,as the voltage across R2 has been calculated.The power dissipated varies from 2.0833 to 6.2500 W 
+\ No newline at end of file
diff --git a/1316/CH2/EX2.3/example2_3.sce b/1316/CH2/EX2.3/example2_3.sce
new file mode 100644
index 000000000..edf7d6df7
--- /dev/null
+++ b/1316/CH2/EX2.3/example2_3.sce
@@ -0,0 +1,17 @@
+//Chapter 2
+//Example 2.3
+//Page 60
+
+clear;
+clc;
+
+R1 = 1000;
+R2 = 842;
+R3 = 500;
+
+printf("Because bridge is nulled we find R4 using equation:\n R1R4=R3R2 \n")
+//Calculation of R4
+R4=R3*R2/R1;
+printf("%.0f ohm",R4)
+
+
diff --git a/1316/CH2/EX2.3/resultexample2_3.txt b/1316/CH2/EX2.3/resultexample2_3.txt
new file mode 100644
index 000000000..82b35d588
--- /dev/null
+++ b/1316/CH2/EX2.3/resultexample2_3.txt
@@ -0,0 +1,3 @@
+ Because bridge is nulled we find R4 using equation:
+ R1R4=R3R2 
+421 ohm 
+\ No newline at end of file
diff --git a/1316/CH2/EX2.4/example2_4.sce b/1316/CH2/EX2.4/example2_4.sce
new file mode 100644
index 000000000..f1b08d7a8
--- /dev/null
+++ b/1316/CH2/EX2.4/example2_4.sce
@@ -0,0 +1,18 @@
+//Chapter 2
+//Example 2.4
+//Page 60
+
+clear;
+clc;
+
+R1 = 120;
+R2 = 120;
+R3 = 120;
+R4 = 121;
+V=10;
+printf("Assuming detector impedance to be very high , we find the offset as \n")
+//Calculation of Delta Vd
+Del_Vd = V*(((R3*R2)-(R1*R4))/((R1+R3)*(R2+R4)))
+printf("Delta Vd = %f V",Del_Vd)
+
+
diff --git a/1316/CH2/EX2.4/resultexample2_4.txt b/1316/CH2/EX2.4/resultexample2_4.txt
new file mode 100644
index 000000000..c9b08d046
--- /dev/null
+++ b/1316/CH2/EX2.4/resultexample2_4.txt
@@ -0,0 +1,2 @@
+ Assuming detector impedance to be very high , we find the offset as 
+Delta Vd = -0.020747 V  
diff --git a/1316/CH2/EX2.5/example2_5.sce b/1316/CH2/EX2.5/example2_5.sce
new file mode 100644
index 000000000..7d3a3d660
--- /dev/null
+++ b/1316/CH2/EX2.5/example2_5.sce
@@ -0,0 +1,31 @@
+//Chapter 2
+//Example 2.5
+//Page 61
+
+clear;
+clc;
+
+R1 = 2000.00;
+R2 = 2000.00;
+R3 = 2000.00;
+R4 = 2050.00;
+V = 5.00;
+Rg = 50.0;
+
+//Calculation of Offset Current
+printf("From equation the offset voltage is Vth. \n")
+
+//Calculation of Vth
+x = V *(((R3*R2)-(R1*R4))/((R1+R3)*(R2+R4)))
+printf("Vth = %f V  \n",x)
+
+//Calculation of Thevenin Resistance
+printf("We next find the bridge Thevenin Resistance form equation as : ")
+y = ((R1*R2)/(R1+R2))+((R3*R4)/(R3+R4))
+printf("Rth = %f ohm \n",y)
+
+//Calculation of Current
+printf("Finally the current is given by the equation ")
+z = x/(y + Rg)
+printf("Ig = %f Ampere \n",z)
+printf("The negative sign on current simply means that current flows fromright to left.")
diff --git a/1316/CH2/EX2.5/resultexample2_5.txt b/1316/CH2/EX2.5/resultexample2_5.txt
new file mode 100644
index 000000000..d8e003697
--- /dev/null
+++ b/1316/CH2/EX2.5/resultexample2_5.txt
@@ -0,0 +1,5 @@
+ From equation the offset voltage is Vth. 
+Vth = -0.030864 V  
+We next find the bridge Thevenin Resistance form equation as : Rth = 2012.345679 ohm 
+Finally the current is given by the equation Ig = -0.000015 Ampere 
+The negative sign on current simply means that current flows fromright to left. 
+\ No newline at end of file
diff --git a/1316/CH2/EX2.7/example2_7.sce b/1316/CH2/EX2.7/example2_7.sce
new file mode 100644
index 000000000..4f13b9be6
--- /dev/null
+++ b/1316/CH2/EX2.7/example2_7.sce
@@ -0,0 +1,34 @@
+//Chapter 2
+//Example 2.7
+//Page 65
+
+clear;
+clc;
+
+R1 = 10000.00;
+R2 = 10000.00;
+R3 = 1000;
+R4 = 950.00;
+R5 = 50.00;
+Del_R3 = 1;
+V =10.00;
+
+//Calculation of Current required
+printf("First,for the nominal resistance values given, \n the bridge is set at anull with I=0,because\n")
+
+//Calculation of Va
+x = (V * R3)/(R1 + R3)
+printf("Va = %f V \n",x)
+
+//Calculation of Vb
+printf("With I=0 we get \n")
+y = ((V) * (R4+R5))/(R2 + R4 + R5)
+printf("Vb = %f V \n",y)
+
+//Calculation of Va for change in R3 
+printf("When R3 increases by 1 ohm to 1001 ohm, Va becomes")
+z = (V * (R3+1000))/(R1 + (R3+1000))
+printf("Va = %f V \n",z)
+printf("which shows that the voltage at b must increase by 0.0008 V to renull the bridge.\nThis can be proved by current,from equation and Delta V=0,found from \n")
+I = 0.0008/50;
+printf("I = %f A",I)
diff --git a/1316/CH2/EX2.7/resultexample2_7.txt b/1316/CH2/EX2.7/resultexample2_7.txt
new file mode 100644
index 000000000..041ea5956
--- /dev/null
+++ b/1316/CH2/EX2.7/resultexample2_7.txt
@@ -0,0 +1,9 @@
+ First,for the nominal resistance values given, 
+ the bridge is set at anull with I=0,because
+Va = 0.909091 V 
+With I=0 we get 
+Vb = 0.909091 V 
+When R3 increases by 1 ohm to 1001 ohm, Va becomesVa = 1.666667 V 
+which shows that the voltage at b must increase by 0.0008 V to renull the bridge.
+This can be proved by current,from equation and Delta V=0,found from 
+I = 0.000016 A 
+\ No newline at end of file
diff --git a/1316/CH2/EX2.8/example2_8.sce b/1316/CH2/EX2.8/example2_8.sce
new file mode 100644
index 000000000..38bc4c129
--- /dev/null
+++ b/1316/CH2/EX2.8/example2_8.sce
@@ -0,0 +1,22 @@
+//Chapter 2
+//Example 2.8
+//Page 66
+
+clear;
+clc;
+
+R1 = 1000.00;
+R2 = 1000.00;
+R3 = 605.00;
+R4 = 500.00;
+V =10.00;
+
+//Finding Unknown Potential
+printf("Here simply we use the equation to solve for Vx \n")
+
+//Calculation of Vx
+x = -[(V*R3)/(R3+R1)]+[(V*R4)/(R4+R2)]
+printf("Vx = %f V \n",x)
+
+
+printf("The negative sign tells us the polarity of the voltage , Vx.Since Vx numerically \n substracts from Va,we see that its positive terminal must be connected to pointa in figure \n")
diff --git a/1316/CH2/EX2.8/resultexample2_8.txt b/1316/CH2/EX2.8/resultexample2_8.txt
new file mode 100644
index 000000000..02116a092
--- /dev/null
+++ b/1316/CH2/EX2.8/resultexample2_8.txt
@@ -0,0 +1,4 @@
+ Here simply we use the equation to solve for Vx 
+Vx = -0.436137 V 
+The negative sign tells us the polarity of the voltage , Vx.Since Vx numerically 
+ substracts from Va,we see that its positive terminal must be connected to pointa in figure 
+\ No newline at end of file
diff --git a/1316/CH2/EX2.9/example2_9.sce b/1316/CH2/EX2.9/example2_9.sce
new file mode 100644
index 000000000..b540a958a
--- /dev/null
+++ b/1316/CH2/EX2.9/example2_9.sce
@@ -0,0 +1,24 @@
+//Chapter 2
+//Example 2.9
+//Page 67
+
+clear;
+clc;
+
+R1 = 5000;
+R2 = 5000;
+R3 = 1000;
+R4 = 990;
+R5 = 10;
+Vx =10;
+Pot = 0.012;
+
+//Finding Current necessary to null the bridge
+printf("First an examination of the resistancs show \n that the bridge is nulled when I = 0 and Vx = 0\n from the equation (Vx)+((R3*V)/(R1+R3))-(V*(R4+R5)/(R2+R4+R5))-(I*R5)=0")
+x=(R3*Vx)/(R1+R3);
+printf("(R3*V)/(R1+R3))=%.3f V\n",x)
+y=(Vx*(R4+R5)/(R2+R4+R5))
+printf("(V*(R4+R5)/(R2+R4+R5))=%.3f V",y)
+z = Pot/R5*1000;
+printf("\n Thus,we can use equation Vx-IR5=0: \n 12mV-10I=0 \n Thus I = %0.1f mA",z)
+
diff --git a/1316/CH2/EX2.9/resultexample2_9.txt b/1316/CH2/EX2.9/resultexample2_9.txt
new file mode 100644
index 000000000..99fb723ab
--- /dev/null
+++ b/1316/CH2/EX2.9/resultexample2_9.txt
@@ -0,0 +1,7 @@
+ First an examination of the resistancs show 
+ that the bridge is nulled when I = 0 and Vx = 0
+ from the equation (Vx)+((R3*V)/(R1+R3))-(V*(R4+R5)/(R2+R4+R5))-(I*R5)=0(R3*V)/(R1+R3))=1.667 V
+(V*(R4+R5)/(R2+R4+R5))=1.667 V
+ Thus,we can use equation Vx-IR5=0: 
+ 12mV-10I=0 
+ Thus I = 1.2 mA 
+\ No newline at end of file