From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001
From: prashantsinalkar
Date: Tue, 10 Oct 2017 12:27:19 +0530
Subject: initial commit / add all books

---
 3556/CH2/EX2.16/Ex2_16.sce | 48 ++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 48 insertions(+)
 create mode 100644 3556/CH2/EX2.16/Ex2_16.sce

(limited to '3556/CH2/EX2.16/Ex2_16.sce')

diff --git a/3556/CH2/EX2.16/Ex2_16.sce b/3556/CH2/EX2.16/Ex2_16.sce
new file mode 100644
index 000000000..ba7584f28
--- /dev/null
+++ b/3556/CH2/EX2.16/Ex2_16.sce
@@ -0,0 +1,48 @@
+clc
+// Fundamental of Electric Circuit 
+// Charles K. Alexander and  Matthew N.O Sadiku  
+// Mc Graw Hill of New York 
+// 5th Edition 
+
+// Part 1   :  DC Circuits 
+// Chapter 2:  Basic Laws 
+// Example 2 - 16
+
+clear; clc; close; 
+//
+// Given data
+V    =   9.00;       
+p1   =  20.00;       
+p2   =  15.00;     
+p3   =  10.00;      
+//
+// Calculations 
+// Calculations Ptot
+Ptot = p1 + p2 + p3          
+// Calculations Itot
+Itot = Ptot/V;              
+// Calculations I1
+I1   = p1/V;                 
+// Calculations I2 and I3 
+I2   = Itot - I1;           
+I3   = I2;  //               
+// Calculations R1 
+R1   = p1/(I1^2);         
+// Calculations R2 
+R2   = p2/(I2^2);           
+// Calculations R3 
+R3   = p3/(I3^2);           
+// Display the result
+disp("Example 2-16 Solution : ");
+printf(" \n I   : Total current supplied the battery  = %.3f A ",Itot);
+printf(" \n I1  : Current Through bulb 1              = %.3f A ",I1);
+printf(" \n I2  : Current Through bulb 2              = %.3f A ",I2);
+printf(" \n I3  : Current Through bulb 3              = %.3f A ",I3);
+printf(" \n R1  : Resistance bulb 1                   = %.3f Ohm ",R1);
+printf(" \n R2  : Resistance bulb 2                   = %.3f Ohm ",R2);
+printf(" \n R3  : Resistance bulb 3                   = %.3f Ohm ",R3);
+
+
+
+
+
-- 
cgit