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/CH4/EX4.8/Ex4_8.sce | 46 ++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 46 insertions(+)
 create mode 100644 3556/CH4/EX4.8/Ex4_8.sce

(limited to '3556/CH4/EX4.8')

diff --git a/3556/CH4/EX4.8/Ex4_8.sce b/3556/CH4/EX4.8/Ex4_8.sce
new file mode 100644
index 000000000..f79b0df3c
--- /dev/null
+++ b/3556/CH4/EX4.8/Ex4_8.sce
@@ -0,0 +1,46 @@
+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 4 :  Circuit Theorems
+// Example 4 - 
+8
+clear; clc; close; 
+//
+// Given data
+R4   =   4.00000;
+R12  =  12.0000;
+R1   =   1.0000;
+Vs   =  32.0000;  
+Is   =  -2.0000;
+RL1  =   6.0000;
+RL2  =  16.0000;
+RL3  =  36.0000;
+//
+// Calculations Rth
+Rth1  = (R4*R12)/(R4+R12);
+Rth   = R1 + Rth1;
+// Calculations Vth 
+I1  = (Vs + (R12*Is))/(R4+R12)
+I2  = Is;
+Vth = R12*(I1-I2)
+// Calculation IL1 for RL = 6 Ohm 
+IL1 = Vth/(Rth + RL1)
+// Calculation IL2 for RL = 16 Ohm 
+IL2 = Vth/(Rth + RL2)
+// Calculation IL3 for RL = 36 Ohm 
+IL3 = Vth/(Rth + RL3)
+// Display the result
+disp("Example 4-8 Solution : ");
+printf(" \n Vth    = Voltage Thevenin                  = %.3f Volt",Vth)
+printf(" \n Rth    = Resistance Thevenin               = %.3f Ohm",Rth)
+printf(" \n IL1    = Load Current For RL 6 Ohm         = %.3f A",IL1)
+printf(" \n IL2    = Load Current For RL 16 Ohm        = %.3f A",IL2)
+printf(" \n IL3    = Load Current For RL 36 Ohm        = %.3f A",IL3)
+
+
+
+
-- 
cgit