From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

---
 181/CH3/EX3.2/example3_2.sce | 39 +++++++++++++++++++++++++++++++++++++++
 1 file changed, 39 insertions(+)
 create mode 100755 181/CH3/EX3.2/example3_2.sce

(limited to '181/CH3/EX3.2/example3_2.sce')

diff --git a/181/CH3/EX3.2/example3_2.sce b/181/CH3/EX3.2/example3_2.sce
new file mode 100755
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--- /dev/null
+++ b/181/CH3/EX3.2/example3_2.sce
@@ -0,0 +1,39 @@
+// 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
\ No newline at end of file
-- 
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