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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+// Exa 3.23
+format('v',5);clc;clear;close;
+// Given data
+I = 100;// in mA
+Im = 1;// in mA
+Rm = 25;// in ohm
+// m = I/Im = 1 + Rm/Rsh;
+Rsh = Rm/((I/Im) - 1);// in ohm
+del_t = 10;// in °C
+Alpha_c = 0.004;
+Alpha_m = 0.00015;
+// When temperature increase by 10 °C
+R_m = Rm * ( 1 + (Alpha_c*del_t) );// in ohm
+R_sh = Rsh * (1 + (Alpha_m*del_t));// in ohm
+// When I= 100 mA then
+I_m = (R_sh/(R_sh+R_m))*I;// in mA
+// But Im required for full scale deflection
+PerEerror= ((I_m-Im)/Im)*100;// in %
+disp("Part (i) ");
+disp(PerEerror,"The percentage error in % is");
+Rx = 75;// in ohm
+Rtotal = Rm+Rx;// in ohm
+Rsh = Rtotal/((I/Im) - 1);// in ohm
+//R_total =R_m+R_x;
+R_total = R_m + (Rx*(1+(Alpha_m*del_t)));// in ohm
+R_sh = Rsh * (1+( Alpha_m*del_t ));// in ohm
+I_m = (R_sh/(R_sh+R_total))*I;// in mA
+PerEerror = ((I_m-Im)/Im)*100;//percentage error in %
+disp("Part (ii) ");
+disp(PerEerror,"The percentage error in % is");