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+// Find the diode currents

+// Basic Electronics

+// By Debashis De

+// First Edition, 2010

+// Dorling Kindersley Pvt. Ltd. India

+// Example 3-2 in page 144

+

+clear; clc; close;

+

+// Given data

+R=10*10^3; // Resistance in K-ohms

+

+// Calculation

+printf("(a) R = 10K.Assume both diodes are conducting.We have:\n");

+printf("100 = 10.02*I1 + 10*I2 + 0.2\n 100 = 10.01*I2 + 10*I1 + 0.6\n");

+function y=f(i);

+    y(1)=10.02*i(1)+10*i(2)+0.2-100

+    y(2)=10.015*i(2)+10*i(1)+0.6-100

+endfunction

+ans=fsolve([0.1;0.1],f);

+I1=ans([1]);

+I2=ans([2]);

+printf("I1 = %0.3f A,I2 = %0.3f A\n",I1,I2);

+printf("Solving,we find I2<0.Thus D is not ON\n");

+I1=(100-0.2)/10.02;

+printf("I1 = %0.2e A and I2 = 0\n\n",I1);

+printf("(b) R=1K.Assume both diodes are ON,we have:\n");

+printf("100 = 1.52*I1 + 1.5*I2 + 0.2\n 100 = 1.515*I2 + 1.5*I1 + 0.6\n");

+function y1=g(j);

+    y1(1)=1.52*j(1)+1.5*j(2)+0.2-100

+    y1(2)=1.515*j(2)+1.5*j(1)+0.6-100

+endfunction

+ans1=fsolve([0.1;0.1],g);

+I1=ans1([1]);

+I2=ans1([2]);

+printf("Solving,we find\nI1 = %0.3f A and I2 = %0.3f A.Hence assumption is valid",I1,I2);

+

+// Result

+// Since both currents are positive,assumption is valid for I1 = 39.717 mA and I2 = 26.287 mA
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