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diff --git a/174/CH12/EX12.1/example12_1.sce b/174/CH12/EX12.1/example12_1.sce
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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