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+// To find percentage error

+// Modern Electronic Instrumentation And Measurement Techniques

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

+// First Edition Second Impression, 2009

+// Dorling Kindersly Pvt. Ltd. India

+// Example 12-1 in Page 360

+

+

+clear; clc; close;

+

+// Given data

+R = 1; //Resistance of the wire in ohm

+R_L = 10*10^3; //Load resistance in ohm

+I_supply = 50*10^-3; //power supply current in A

+V_out = 1; //output of the amplifier in V

+

+//Calculations

+V_L = (V_out+(I_supply*R))*R_L/(2*R+R_L);

+printf("The load voltage calculated = %0.2f\n",V_L);

+

+%error = ceil((V_L -V_out)/V_L*100);

+printf("The percentage error is about %d %%, which is unacceptable in most systems",%error);

+

+//Result

+// The load voltage calculated = 1.05

+// The percentage error is about 5 %, which is unacceptable in most systems 

+

+

+

diff --git a/174/CH12/EX12.1/example12_1.txt b/174/CH12/EX12.1/example12_1.txt
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+The load voltage calculated = 1.05

+The percentage error is about 5 %, which is unacceptable in most systems