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

---
 2360/CH3/EX3.26/ex3_26.sce | 21 +++++++++++++++++++++
 1 file changed, 21 insertions(+)
 create mode 100755 2360/CH3/EX3.26/ex3_26.sce

(limited to '2360/CH3/EX3.26/ex3_26.sce')

diff --git a/2360/CH3/EX3.26/ex3_26.sce b/2360/CH3/EX3.26/ex3_26.sce
new file mode 100755
index 000000000..c3d524802
--- /dev/null
+++ b/2360/CH3/EX3.26/ex3_26.sce
@@ -0,0 +1,21 @@
+// Exa 3.26
+format('v',7);clc;clear;close;
+// Given data
+Vm = 0.1;//full scale deflection voltage in V
+Rm = 20;//meter resistance in ohm
+Im = Vm/Rm;//current in A
+I1= 10;// in A
+I2= 1;// in A
+I3= 100*10^-3;// in A
+// I1*R1 = Im*(R2+R3+Rm) or   I1*R1 - Im*R2 - Im*R3 = Im*Rm       (i)
+// I2*(R1+R2) = Im*(R3+Rm) or I2*R1 + I2*R2 -Im*R3 = Im*Rm      (ii)
+// I3*(R1+R2+R3) = Im*Rm or   I3*R1 + I3*R2 + I3*R3 = Im*Rm     (iii)
+A= [I1 I2 I3;-Im I2 I3;-Im -Im I3];
+B= [Im*Rm Im*Rm Im*Rm];
+R= B*A^-1;// Solving equation (i), (ii) and (iii) by matrix method
+R1= R(1);// in ohm
+R2= R(2);// in ohm
+R3= R(3);// in ohm
+disp(R1,"The value of R1 in ohm is : ")
+disp(R2,"The value of R2 in ohm is : ")
+disp(R3,"The value of R3 in ohm is : ")
-- 
cgit