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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
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