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diff --git a/174/CH5/EX5.1/example5_1.sce b/174/CH5/EX5.1/example5_1.sce
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+// To find deflection caused by the given unbalance

+// Modern Electronic Instrumentation And Measurement Techniques

+// By Albert D. Helfrick, William D. Cooper

+// First Edition Second Impression, 2009

+// Dorling Kindersly Pvt. Ltd. India

+// Example 5-1 in Page 101

+

+

+clear; clc; close;

+

+// Given data

+// Resistances of the 4 arms in ohm

+R_1 = 1000;

+R_2 = 100;

+R_3 = 200;

+R_4 = 2005;

+

+E = 5; // battery EMF in volt

+S_I = 10*(10^-3)/(10^-6); //Current sensitivity in m/A

+R_g = 100; //Internal resistance of galvanometer in ohm

+

+//Calculations

+

+//Calculations are made wrt fig 5-3 in page 103

+//Bridge balance occurs if arm BC has a resistance of 2000 ohm. The diagram shows arm BC has as a resistance of 2005 ohm

+

+//To calculate the current in the galvanometer, the ckt is thevenised wrt terminals B and D.

+//The potenttial from B to D, with the galvanometer removed is the Thevenin voltage

+

+// E_TH = E_AD - E_AB 

+

+E_TH = E * ((R_2/(R_2+R_3)) - (R_1/ (R_1+R_4)));

+R_TH = ((R_2 * R_3/(R_2+R_3)) + (R_1 * R_4/ (R_1+R_4)));

+

+//When the galvanometer is now connected to the output terminals, The current through the galvanometer is

+

+I_g = E_TH /(R_TH +R_g);

+d = I_g * S_I;

+printf("The deflection of the galvanometer = %0.2f mm",(d*1000));

+

+//Result

+// The deflection of the galvanometer = 33.26 mm 

+

+

+

diff --git a/174/CH5/EX5.1/example5_1.txt b/174/CH5/EX5.1/example5_1.txt
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+The deflection of the galvanometer = 33.26 mm 

diff --git a/174/CH5/EX5.2/example5_2.sce b/174/CH5/EX5.2/example5_2.sce
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index 000000000..9ade38e92
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+++ b/174/CH5/EX5.2/example5_2.sce
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+// To check the capability of detecting unbalance

+// Modern Electronic Instrumentation And Measurement Techniques

+// By Albert D. Helfrick, William D. Cooper

+// First Edition Second Impression, 2009

+// Dorling Kindersly Pvt. Ltd. India

+// Example 5-2 in Page 102

+

+

+clear; clc; close;

+

+// Given data

+// Resistances of the 4 arms in ohm

+R_1 = 1000;

+R_2 = 100;

+R_3 = 200;

+R_4 = 2005;

+

+E = 5; // battery EMF in volt

+S_I = 1*(10^-3)/(10^-6); //Current sensitivity in m/A

+R_g = 500; //Internal resistance of galvanometer in ohm

+

+

+

+

+//Calculations

+

+//Calculations are made wrt fig 5-3 in page 103

+//Bridge balance occurs if arm BC has a resistance of 2000 ohm. The diagram shows arm BC has as a resistance of 2005 ohm

+

+//To calculate the current in the galvanometer, the ckt is thevenised wrt terminals B and D.

+//The potenttial from B to D, with the galvanometer removed is the Thevenin voltage

+

+// E_TH = E_AD - E_AB 

+

+E_TH = E * ((R_2/(R_2+R_3)) - (R_1/ (R_1+R_4)));

+R_TH = ((R_2 * R_3/(R_2+R_3)) + (R_1 * R_4/ (R_1+R_4)));

+

+//When the galvanometer is now connected to the output terminals, The current through the galvanometer is

+

+I_g = E_TH /(R_TH +R_g);

+d = I_g * S_I;

+printf("The deflection of the galvanometer = %0.3f mm",d*1000);

+disp('Given that galvanometer is capable of detecting a deflection of 1mm');

+disp('Hence looking at the result,it can be seen that this galvanometer produces a deflection that can be easily observed');

+

+//Result

+// The deflection of the galvanometer = 2.247 mm 

+// Given that galvanometer is capable of detecting a deflection of 1mm   

+ 

+// Hence looking at the result,it can be seen that this galvanometer produces a deflection that can be easily observed   

+ 

+

+

+

diff --git a/174/CH5/EX5.2/example5_2.txt b/174/CH5/EX5.2/example5_2.txt
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+ The deflection of the galvanometer = 2.247 mm 

+ Given that galvanometer is capable of detecting a deflection of 1mm   

+ 

+ Hence looking at the result,it can be seen that this galvanometer produces a deflection that can be easily observed   

+ 

diff --git a/174/CH5/EX5.3/example5_3.sce b/174/CH5/EX5.3/example5_3.sce
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+++ b/174/CH5/EX5.3/example5_3.sce
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+// To find the unknown impedence

+// Modern Electronic Instrumentation And Measurement Techniques

+// By Albert D. Helfrick, William D. Cooper

+// First Edition Second Impression, 2009

+// Dorling Kindersly Pvt. Ltd. India

+// Example 5-3 in Page 111

+

+

+clear; clc; close;

+

+// Given data

+// The given polar forms in textbook is represented in rect form

+Z_1 = 17.36482 +%i *98.48078;

+Z_2 = 250;

+Z_3 = 346.4102 +%i *200;

+

+//Calculations

+//The first condition for bridge balance is Z_1*Z_4 = Z_2*Z_3

+mod_Z_4 = (abs(Z_2) *abs(Z_3)/abs(Z_1));

+

+//The second condition for bridge balance requires that sum of the phase angles of opposite arms be equal

+theta_Z_4 = (atan(imag(Z_2),real(Z_2)) +atan(imag(Z_3),real(Z_3)) -atan(imag(Z_1),real(Z_1)))*180/%pi;

+

+printf("The impedence of the unknown arm = %d ohm /_ %d deg\n",mod_Z_4,theta_Z_4);

+printf("Here the magnitude of impedence is 1000 and phase angle is 50 in degrees\n");

+printf("The above value indicates that we are dealing with a capacitive element, possibly consisting of a series combination of a resistor and capacitance");

+

+//Result

+// The impedence of the unknown arm = 1000 ohm /_ -50 deg

+// Here the magnitude of impedence is 1000 and phase angle is 50 in degrees

+// The above value indicates that we are dealing with a capacitive element, possibly consisting of a series combination of a resistor and capacitance 

+ 

+

+

+

diff --git a/174/CH5/EX5.3/example5_3.txt b/174/CH5/EX5.3/example5_3.txt
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+The impedence of the unknown arm = 1000 ohm /_ -50 deg

+Here the magnitude of impedence is 1000 and phase angle is 50 in degrees

+The above value indicates that we are dealing with a capacitive element, possibly consisting of a series combination of a resistor and capacitance 

diff --git a/174/CH5/EX5.4/example5_4.sce b/174/CH5/EX5.4/example5_4.sce
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index 000000000..e57f9c242
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+++ b/174/CH5/EX5.4/example5_4.sce
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+// To find the unknown impedence

+// Modern Electronic Instrumentation And Measurement Techniques

+// By Albert D. Helfrick, William D. Cooper

+// First Edition Second Impression, 2009

+// Dorling Kindersly Pvt. Ltd. India

+// Example 5-4 in Page 112

+

+

+clear; clc; close;

+

+// Given data

+// The notations are wrt to the figure 5-10 in page 109

+

+//Arm AB

+R_1 = 450;

+//Arm BC

+R_2 = 300;

+C = 0.265 *(10^-6);

+//Arm DA

+R_3 = 200;

+L = 15.9*(10^-3);

+f = 1000;

+

+//Calculations

+w = 2*%pi*f;

+Z_1 = 450;

+Z_2 = R_2 - %i *floor(1/(w*C));

+Z_3 = R_3 + %i*ceil(w*L);

+

+Z_4 = Z_1*Z_3/Z_2;

+printf("The impedence of the unknown arm = %di ohm\n",imag(Z_4));

+printf("The result indicates that Z_4 is a pure inductance with an inductive reactance of 150 ohm at a frequency of 1 khz.\n")

+

+L_ans = imag(Z_4)/w;

+printf("The inductance present in the arm CD = %0.1fm H",L_ans*1000);

+

+//Result

+// The impedence of the unknown arm = 150i ohm

+// The result indicates that Z_4 is a pure inductance with an inductive reactance of 150 ohm at a frequency of 1 khz.

+// The inductance present in the arm CD = 23.9m H 

+

+

+

+

+

diff --git a/174/CH5/EX5.4/example5_4.txt b/174/CH5/EX5.4/example5_4.txt
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+The impedence of the unknown arm = 150i ohm

+The result indicates that Z_4 is a pure inductance with an inductive reactance of 150 ohm at a frequency of 1 khz.

+The inductance present in the arm CD = 23.9m H 

diff --git a/174/CH5/EX5.5/example5_5.sce b/174/CH5/EX5.5/example5_5.sce
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+// To balance the unbalanced bridge

+// Modern Electronic Instrumentation And Measurement Techniques

+// By Albert D. Helfrick, William D. Cooper

+// First Edition Second Impression, 2009

+// Dorling Kindersly Pvt. Ltd. India

+// Example 5-5 in Page 119

+

+

+clear; clc; close;

+

+// Given data

+Z_1 = -1000*%i;

+Z_2 = 500;

+Z_3 = 1000;

+Z_4 = 100+500*%i;

+

+// The balance is not possible with this condition as theta_1+theta_4 will be slightly negative than theta_2+theta3

+// Balance can be achieved by 2 methods:

+disp('First option is to modify Z_1 so that its phase angle is decreased to less than 90deg by placing a resistor in parallel with the capacitor.')

+// The resistance R_1 can be determined by the standard approach

+

+//Calculations

+Y_1 = Z_4/(Z_2*Z_3);

+//Also,

+// Y_1 = (1/R) + %i/1000;

+// equating both the equations and solving for R_1

+

+R_1 = 1/(Y_1-(%i/1000 ));

+printf("The value of the resistor R_1 in parallel with capacitor = %d ohm\n",R_1);

+

+// It should be noted that the addition of R_1 upsets the first balance condition as the magnitude of Z_1 is changed

+// Hence the variable R_3 should be adjusted to compensate this effect

+

+disp('The second option is to modify the phase angle of arm 2 or arm 3 by adding series capacitor');

+Z_3_1 = Z_1 *Z_4/Z_2;

+// substituting for the component values and solving for X_C yeilds

+

+X_C = abs(1000- Z_3_1)/-%i;

+printf("The value of the reactance of the capacitor used, X_C = %d ohm",imag(X_C));

+

+

+//In this case the magnitude of the Z_3 is increased so that the first balance condition is changed

+//A small adjustment of R_3 is necessary to restore balance

+

+//Result

+// First option is to modify Z_1 so that its phase angle is decreased to less than 90deg by placing a resistor in parallel with the capacitor.   

+// The value of the resistor R_1 in parallel with capacitor = 5000 ohm

+ 

+// The second option is to modify the phase angle of arm 2 or arm 3 by adding series capacitor   

+// The value of the reactance of the capacitor used, X_C = 200 ohm 

+

+

+

+

diff --git a/174/CH5/EX5.5/example5_5.txt b/174/CH5/EX5.5/example5_5.txt
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+First option is to modify Z_1 so that its phase angle is decreased to less than 90deg by placing a resistor in parallel with the capacitor.   

+The value of the resistor R_1 in parallel with capacitor = 5000 ohm

+ 

+ The second option is to modify the phase angle of arm 2 or arm 3 by adding series capacitor   

+The value of the reactance of the capacitor used, X_C = 200 ohm