10 files changed, 212 insertions, 0 deletions
diff --git a/3554/CH11/EX11.1/Ex11_1.sce b/3554/CH11/EX11.1/Ex11_1.sce
new file mode 100644
index 000000000..973ab4783
--- /dev/null
+++ b/3554/CH11/EX11.1/Ex11_1.sce
@@ -0,0 +1,17 @@
+// Exa 11.1
+
+clc;
+clear all;
+
+// Given data
+
+// Wheatstone's bridge  circuit
+R1=10; // k Ohms
+R2=15; // k Ohms
+R3=40; // k Ohms
+
+// Solution
+// From the equation (11.4) of balanced bridge we have
+
+Rx=R2*R3/R1; // Unknown resistance Rx
+printf(' The unknown resistance Rx is = %d k Ohms \n',Rx);
diff --git a/3554/CH11/EX11.10/Ex11_10.sce b/3554/CH11/EX11.10/Ex11_10.sce
new file mode 100644
index 000000000..458d8f06a
--- /dev/null
+++ b/3554/CH11/EX11.10/Ex11_10.sce
@@ -0,0 +1,23 @@
+// Exa 11.10
+
+clc;
+clear all;
+
+// Given data
+
+// Wien's bridge
+R1=3.1; // k Ohms
+C1=5.2; // micro farads
+R2=25; // k Ohms
+f=2.5; // kHz
+R4=100;// k Ohms
+
+// Solution
+
+w=2*%pi*f; // Angular frequency
+// Substituting the value of C3 from Eq. 11.22(page no. 330) in Eq.11.21(pagr no. 330) to get value of R3 as follows
+R3=R4/R2 *(R1+1/(w^2*R1*C1^2));
+// Also we can get C3 from Eq. 11.22(page no. 330)
+C3=1/(w^2*C1*R1*R3);
+printf(' The parallel resistance of %.1f K ohms and capacitance of %.1f pf\n     causes a Wien bridge to null with values of given component values. \n',R3,C3*10^6);
+
diff --git a/3554/CH11/EX11.2/Ex11_2.sce b/3554/CH11/EX11.2/Ex11_2.sce
new file mode 100644
index 000000000..aa19bc4ee
--- /dev/null
+++ b/3554/CH11/EX11.2/Ex11_2.sce
@@ -0,0 +1,24 @@
+// Exa 11.2
+
+clc;
+clear all;
+
+// Given data
+//Refering fig. 11.5 - Unbalanced Wheatstone bridge
+
+R1=1; // in k Ohms
+R2= 2.5; // in k Ohms
+R3=3.5; // in k Ohms
+R4=10; // in k Ohms
+V= 6; // Applied Voltage(V)
+Rg=0.3; // Galvanometer resistance in k Ohms
+
+// Solution
+
+// Eth=Ea-Eb ; \\ Thevenin's equivalent voltage
+Eth=V*(R4/(R2+R4) - R3/(R1+R3));
+Rth=(R1*R3/(R1+R3)) + (R2*R4/(R2+R4)) ;
+// Refering the equivalent circuit connected along with the galvanometer as shown in fig. 11.6
+Ig=Eth/(Rth+Rg) ; // Current through galvanometer
+printf(' The current through galvanometer is = %.2f micro Amp \n',round(Ig*10^3));
+//The answer vary due to round off error
diff --git a/3554/CH11/EX11.3/Ex11_3.sce b/3554/CH11/EX11.3/Ex11_3.sce
new file mode 100644
index 000000000..14520fe33
--- /dev/null
+++ b/3554/CH11/EX11.3/Ex11_3.sce
@@ -0,0 +1,19 @@
+// Exa 11.3
+
+clc;
+clear all;
+	
+// Given data
+
+// Refering Fig. 11.9(page no.311) - slightly unbalanced Wheatstone bridge
+R= 700; // in Ohms
+Dell_R= 35; // in Ohms
+E=10; // Supplied voltage(V)
+Rg=125;//Internal resistance of galvanometer(Ohms)
+    
+// Solution
+
+Eth= E*Dell_R/(4*R) ; // Thevenin's equivalent voltage(V)
+Rth=R; // Thevenin's equivalent resistance(Ohms)
+Ig= Eth/(Rth+Rg); // Current through galvanometer(Amp)
+printf(' The current through galvanometer by the approximation method is %.1f micro Amp \n',Ig*10^6);
diff --git a/3554/CH11/EX11.4/Ex11_4.sce b/3554/CH11/EX11.4/Ex11_4.sce
new file mode 100644
index 000000000..240568637
--- /dev/null
+++ b/3554/CH11/EX11.4/Ex11_4.sce
@@ -0,0 +1,19 @@
+// Exa 11.4
+
+clc;
+clear all;
+
+// Given data
+// Refering Fig. 11.12(page no.315)- Kelvin's bridge
+
+Ra_b=1000;// The ratio of Ra to Rb
+R1= 5; // in Ohms
+
+// Solution
+
+// We have R1=0.5*R2
+R2=R1/0.5;
+
+//From the eqation for Kelvin'd bridge- Rx*Ra=Rb*R2
+Rx=R2*(1/1000); // Unknown resistance
+printf(' The value of Rx = %.2f Ohm \n ',Rx);
diff --git a/3554/CH11/EX11.5/Ex11_5.sce b/3554/CH11/EX11.5/Ex11_5.sce
new file mode 100644
index 000000000..8d39131b8
--- /dev/null
+++ b/3554/CH11/EX11.5/Ex11_5.sce
@@ -0,0 +1,24 @@
+// Exa 11.5
+
+clc;
+clear all;
+
+// Given data
+// Refering circuit in Fig. 11.15(a) and graph in 11.15(b) on page no.317
+
+R1=5; // k Ohms
+R2=5; //k Ohms
+R3= 5; // k Ohms
+E=6; // Applied voltage(V)
+
+// Solution
+
+// The value of Rv when bridge is balanced is calculated as
+Rv=R2*R3/R1;
+printf(' The value of Rv = %d K Ohms \n' , Rv);
+disp(" From the graph, the temperature at which bridge is balanced is = 80      degree celsius");
+disp(" From the graph, the resistance Rv for balancing bridge at 60 degree celcius comes out to be 4.5 k Ohms ");
+// Therefore
+Rv1=4.5; // Resistance Rv at 60 degree celcius(K ohms)
+es=E*(R3/(R1+R3) - Rv1/(R2+Rv1) ); // Error signal
+printf(' The amplitude of error signal at 60 degree celsius is = %.3f V \n',es);
diff --git a/3554/CH11/EX11.6/Ex11_6.sce b/3554/CH11/EX11.6/Ex11_6.sce
new file mode 100644
index 000000000..fb39d7b71
--- /dev/null
+++ b/3554/CH11/EX11.6/Ex11_6.sce
@@ -0,0 +1,20 @@
+// Exa 11.6
+
+clc;
+clear all;
+
+// Given data
+f=2; // kHz
+C3=100; // micro farads
+R1=10; // k Ohms
+R2=50; // k Ohms
+R3=100; // k Ohms
+
+// Solution
+
+// Using equations 11.12(a) and 11.12(b) (page no. 321)to find values of Rx and Cx
+
+Rx=R2*R3/R1;
+Cx=R1/R2 *C3;
+
+printf(' The equivalent circuit consist of  resistance Rx of %d K ohms \n  in series with a capacitor Cx of %d micro farads \n',Rx,Cx);
diff --git a/3554/CH11/EX11.7/Ex11_7.sce b/3554/CH11/EX11.7/Ex11_7.sce
new file mode 100644
index 000000000..f9c3a66ba
--- /dev/null
+++ b/3554/CH11/EX11.7/Ex11_7.sce
@@ -0,0 +1,19 @@
+// Exa 11.7
+
+clc;
+clear all;
+
+// Given data
+
+C1=0.01; // micro farads
+R1=470; // k Ohms
+R2=5.1; // k Ohms
+R3=100; // k Ohms
+
+// Solution
+// Using equation 11.15 given on page no. 324 to find Rx and Lx
+
+Rx=R2*R3/R1; 
+Lx=R2*R3*C1;
+
+printf(' The series equivalent of the unknown impedence consist of series combination\n  of Rx = %.2f k Ohms and Lx= %.1f H \n' , Rx, Lx);
diff --git a/3554/CH11/EX11.8/Ex11_8.sce b/3554/CH11/EX11.8/Ex11_8.sce
new file mode 100644
index 000000000..b9511ba11
--- /dev/null
+++ b/3554/CH11/EX11.8/Ex11_8.sce
@@ -0,0 +1,21 @@
+// Exa 11.8
+
+ clc;
+clear all;
+
+// Given data
+
+w=3000; // Angular frequency in rad/s
+R2=10*10^3; // Ohms
+R1= 2*10^3; // Ohms
+C1=1*10^-6; // farads
+R3=1*10^3; // Ohms
+
+// Solution
+
+// Using equations 11.19 and 11.18 (page no.326)to find values of Rx and Lx
+
+Rx=w^2*R1*R2*R3*C1^2/(1+w^2*R1^2*C1^2);
+Lx=R2*R3*C1/(1+w^2*R1^2*C1^2);
+
+printf(' The series equivalent inductance and resistance of the network consist of\n  Rx of %.2f k Ohms and Lx of %d mH \n',Rx/1000,Lx*10^3);
diff --git a/3554/CH11/EX11.9/Ex11_9.sce b/3554/CH11/EX11.9/Ex11_9.sce
new file mode 100644
index 000000000..1d9f63b6a
--- /dev/null
+++ b/3554/CH11/EX11.9/Ex11_9.sce
@@ -0,0 +1,26 @@
+// Exa 11.9
+
+clc; 
+clear all;
+
+// Given data
+
+// Refering Fig. 11.26(page no.328) - an AC bridge(SCHERING'S BRIDGE)
+
+R1= 1; // k Ohms
+C1=0.5; // micro farads
+R2=2; // k Ohms
+C3=0.5; // micro farads
+f= 1000; // Hz
+
+// Solution
+// Using Equations 11.20(a) and 11.20(b) given on page no. 328 we get value Rx and Cx
+
+Rx=C1/C3*R2;// in k Ohms
+Cx=R1/R2 * C3; // in micro farads
+
+D=2*%pi*f*Cx*10^-6*Rx*10^3; // Dissipation factor
+
+printf(' The unknown capacitance Cx is equal to %.2f micro farads\n ',Cx);
+printf(' The dissipation factor = %.4f \n ',D);
+