6 files changed, 163 insertions, 0 deletions
diff --git a/3830/CH6/EX6.1/Ex6_1.sce b/3830/CH6/EX6.1/Ex6_1.sce
new file mode 100644
index 000000000..7a6e42123
--- /dev/null
+++ b/3830/CH6/EX6.1/Ex6_1.sce
@@ -0,0 +1,40 @@
+// Exa 6.1
+
+clc;
+clear;
+
+// Given data
+
+// Refering bridge shown in fig. 6.8 
+R1 = 1000; // Ohms
+R2 = 4000; // Ohms
+R3 = 100; // Ohms
+R4 = 400; // Ohms
+Rg = 100; // Ohms
+Si = 100; // Sensitivity in mm/microAmp
+V = 3; // Voltage applied
+R4_imbalance = 1; // resistance added in R4 to create imbalance
+
+// Solution
+
+printf('The bridge is originally in balance. Therefore, R1/R3 = R2/R4 \n');
+printf('Let there be imbalance in the bridge circuit because of increase in value of R4 value by 1 Ohm \n');
+printf('Therefore, R4 = 400+X Ohms \n');
+printf('Thevenins Resistance  Rth = (100*1000)/(100+1000) + (4000*(400+X))/(4400+X) \n'); // Rth = R1*R3/(R1+R3) + R2*R4/(R2+R4)
+printf('Neglecting X \n');
+// Therefore
+Rth = R1*R3/(R1+R3) + R2*R4/(R2+R4);
+printf('Rth becomes %d ohms \n',round(Rth));
+printf('Eth = [R3/(R1+R3) + R4/(R2+R4)]*E; \n');
+// Applying binomial expansion and neglecting X2 term, X is small 
+// Therefore
+X = R4_imbalance;
+
+Eth =  V*10*X/48400;
+printf('Applying binomial expansion, Eth = %.2f µV \n',round(Eth*10^6));
+Ig = Eth/(Rth+Rg); // Galvanometer current
+D = Ig*Si; // Deflection in mm
+printf('Galvanometer Current Ig = %.3f µA \n', Ig*10^6);
+printf('Galvanometer deflection D = %.2f mm \n',D*10^6);
+
+// The answer provided in the textbook is wrong
diff --git a/3830/CH6/EX6.2/Ex6_2.sce b/3830/CH6/EX6.2/Ex6_2.sce
new file mode 100644
index 000000000..2bd384dcb
--- /dev/null
+++ b/3830/CH6/EX6.2/Ex6_2.sce
@@ -0,0 +1,27 @@
+// Exa 6.2
+
+clc;
+clear;
+
+// Given
+
+//Fig. 6.9 shows wheatstone bridge
+R1 = 1000; // Ohms
+R2 = 100; // Ohms
+R3 = 400; // Ohms
+Rx = 41; // Ohms(Unknown resistance)
+V = 1.5; // Voltage supplied
+Rg = 50; // Galvanometer resistance (ohms)
+Si = 2; // current sensitivity in mm/microAmp
+
+
+// Solution
+
+Rth = (R1*R3/(R1+R3)) + R2*Rx/(R2+Rx);
+Eth = V*(R3/(R1+R3) - Rx/(R2+Rx));
+Ig = Eth/(Rth+Rg);
+d = Ig*Si;
+printf('The thevenins equivalent resistance = %.1f Ohms \n',round(Rth));
+printf(' The thevenins equivalent voltage = %.1f mV \n',abs(Eth*10^3));
+printf(' The current through the galvanometer = %.2f micro Amp \n',abs(Ig*10^6));
+printf(' The deflection produced by the galvanometer caused by the imbalance in the circuit = %.2f mm \n',abs(d*10^6));
diff --git a/3830/CH6/EX6.3/Ex6_3.sce b/3830/CH6/EX6.3/Ex6_3.sce
new file mode 100644
index 000000000..c73deb1f3
--- /dev/null
+++ b/3830/CH6/EX6.3/Ex6_3.sce
@@ -0,0 +1,32 @@
+// Exa 6.3
+
+clc;
+clear;
+
+// Given
+
+//Fig 6.41 shows an AC bridge
+R1 = 800; // Ohms
+C1 = 0.4; // microFarad
+R2 = 500; // Ohms
+C2 = 1.0; // microFarad
+R3 = 1200; // Ohms
+
+
+// Solution
+
+// Z = R + j X;
+// Z1 = 800 + j/(w*C1)
+// Y2 = 1/R2 - j*(w*C2)
+//Z3 = 1200
+
+printf('At balance, Z1/Z4 = Z2/Z3 \n');
+
+printf(' Rearranging the equation, Z4 = Z1*Z3*Y2 \n') ;
+printf(' Equating the real and imaginary parts on both sides, \n');
+Z4 = R1*R3*1/R2;
+w = sqrt(C1*C2);
+printf(' The value of R in arm DA to produce a balance = %d ohms \n',Z4);
+printf(' The value of frequency at balance = %.4f Hz \n',w/(2*%pi));
+
+// The answers given in textbook for R and f are incorrect
diff --git a/3830/CH6/EX6.4/Ex6_4.sce b/3830/CH6/EX6.4/Ex6_4.sce
new file mode 100644
index 000000000..87530ec6a
--- /dev/null
+++ b/3830/CH6/EX6.4/Ex6_4.sce
@@ -0,0 +1,23 @@
+// Exa 6.4
+
+clc;
+clear;
+
+// Given
+// Referring Fig 6.42 to get expression for unknowns Rs and Ls
+
+
+// Solution
+
+printf('It is a bridged-T network. At balance,Z1+Z3+ Z1*Z3/Z2 = 0 \n ');
+printf('Z1 = 1/jwC \n ');
+printf('Z3 = 1/jwC \n ');
+printf('Z2 = R \n ');
+printf('Z4 = Rs+jwLs \n ');
+
+printf('substituting these values in the equation, equating real and imagnary parts, and simplifying,\n ');
+
+printf('1/jwC + 1/jwC - 1/(w^2*C^2*R)+Rs+jwLs = 0 \n ');
+printf('Therefore \n ');
+printf('Rs = 1/(w^2*C^2*R) \n ');
+printf('wLs = 2/wc \n ');
diff --git a/3830/CH6/EX6.5/Ex6_5.sce b/3830/CH6/EX6.5/Ex6_5.sce
new file mode 100644
index 000000000..592acea07
--- /dev/null
+++ b/3830/CH6/EX6.5/Ex6_5.sce
@@ -0,0 +1,26 @@
+// Exa 6.5
+
+clc;
+clear;
+
+// Given
+
+// Referring Fig. 6.43
+
+
+// Solution
+
+printf('This is also a bridged-T network. This circuit is used to compare different coils, Lp and Rp. Using the general equation for a bridged-T netwrok at balance,\n ');
+
+printf('Z1+Z3+ Z1*Z3/Z2+Z4= 0 \n ');
+printf('Z1 = 1/jwC \n ');
+printf('Z3 = 1/jwC \n ');
+printf('Z2 = Rp+ 1/jwLp \n ');
+printf('Z4 = R \n ');
+
+printf('substituting these values in the equation, equating real and imagnary parts, and simplifying \n ');
+
+printf('1/jwC + 1/jwC - 1/(w^2*C^2*R)+Rs+jwLs = 0 \n ');
+printf('Therefore \n ');
+printf('w*Lp = 1/(2*w*C) \n ');
+printf('Rp = 1/(R*(w*C)^2) \n ');
diff --git a/3830/CH6/EX6.6/Ex6_6.sce b/3830/CH6/EX6.6/Ex6_6.sce
new file mode 100644
index 000000000..a24b642c7
--- /dev/null
+++ b/3830/CH6/EX6.6/Ex6_6.sce
@@ -0,0 +1,15 @@
+// Exa 6.6
+
+clc;
+clear;
+
+// Given
+
+// Fig 6.44 shows R.L.C bridge
+
+// Solution
+
+printf('For a given RLC circuit the expressions for as follows :- \n ');
+printf('Resistance, Rx = (R2*R3)/R1 \n ');
+printf('Inductance Lx = R2*R3*C \n ');
+printf('Capacitance Cx = (C*R3)/R1 \n ')