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authorpriyanka2015-06-24 15:03:17 +0530
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-rwxr-xr-x174/CH5/EX5.1/example5_1.sce45
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-rwxr-xr-x174/CH5/EX5.2/example5_2.sce54
-rwxr-xr-x174/CH5/EX5.2/example5_2.txt5
-rwxr-xr-x174/CH5/EX5.3/example5_3.sce35
-rwxr-xr-x174/CH5/EX5.3/example5_3.txt3
-rwxr-xr-x174/CH5/EX5.4/example5_4.sce45
-rwxr-xr-x174/CH5/EX5.4/example5_4.txt3
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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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+// 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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index 000000000..6798b11c8
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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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+// 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