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/CH12/EX12.3/Ex12_3.sce | 61 +++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 61 insertions(+)
 create mode 100644 3556/CH12/EX12.3/Ex12_3.sce

(limited to '3556/CH12/EX12.3/Ex12_3.sce')

diff --git a/3556/CH12/EX12.3/Ex12_3.sce b/3556/CH12/EX12.3/Ex12_3.sce
new file mode 100644
index 000000000..21a402f2a
--- /dev/null
+++ b/3556/CH12/EX12.3/Ex12_3.sce
@@ -0,0 +1,61 @@
+clc
+// Fundamental of Electric Circuit 
+// Charles K. Alexander and  Matthew N.O Sadiku  
+// Mc Graw Hill of New York 
+// 5th Edition 
+
+// Part 2     :  AC Circuits 
+// Chapter 12 :  Three Phase Circuit 
+// Example 12 - 3
+
+clear; clc; close; 
+//
+// Given data
+Van_mag   = 100.0000;                                
+Van_angle =  10.0000;                                
+Zdelta = complex(8,4);                               
+//
+// Calculations Line Voltage (Vab)
+Vab_mag   = (3)^0.5 * Van_mag                        
+Vab_angle =  Van_angle + 30.0000;                    
+// Calculatioan Phase Current AB 
+I_AB_mag   = Vab_mag/norm(Zdelta)                   
+Zdelta_real = real(Zdelta)                           
+Zdelta_imag = imag(Zdelta)                           
+Zdelta_angle = atand(Zdelta_imag,Zdelta_real)        
+I_AB_angle  = Vab_angle - Zdelta_angle               
+// Calculations Phase Current BC
+I_BC_mag = I_AB_mag;                                 
+I_BC_angle = -120.0000 + I_AB_angle;                 
+// Calculations Phase Current CA 
+I_CA_mag = I_AB_mag;                                 
+I_CA_angle = 120.0000 + I_AB_angle;                  
+// Calculations Line Current a 
+I_a_mag = (3)^0.5 * I_AB_mag;                        
+I_a_angle = -30.0000 + I_AB_angle;                   
+// Calculations Line Current b 
+I_b_mag = I_a_mag;                                   
+I_b_angle = -120.0000 + I_a_angle;                   
+// Calculations Line Current c 
+I_c_mag = I_a_mag;                                   
+I_c_angle = 120.0000 + I_a_angle;                   
+//
+disp("Example 12-3 Solution : ");
+printf(" \n I_AB_mag   = Magnitude of Phase Currents AB      = %.3f A",I_AB_mag)
+printf(" \n I_AB_angle = Angle of Phase Current AB           = %.3f Degree",I_AB_angle)
+printf(" \n I_BC_mag   = Magnitude of Phase Currents BC      = %.3f A",I_BC_mag)
+printf(" \n I_BC_angle = Angle of Phase Current BC           = %.3f Degree",I_BC_angle)
+printf(" \n I_CA_mag   = Magnitude of Phase Currents CA      = %.3f A",I_CA_mag)
+printf(" \n I_CA_angle = Angle of Phase Current CA           = %.3f Degree",I_CA_angle)
+printf(" \n I_a_mag    = Magnitude of Line Currents a        = %.3f A",I_a_mag)
+printf(" \n I_a_angle  = Angle of Line Current a             = %.3f Degree",I_a_angle)
+printf(" \n I_b_mag    = Magnitude of Line Current b         = %.3f A",I_b_mag)
+printf(" \n I_b_angle  = Angle of Line Current b             = %.3f Degree",I_b_angle)
+printf(" \n I_c_mag    = Magnitude of Line Currents c        = %.3f A",I_c_mag)
+printf(" \n I_c_angle  = Angle of Line Current c             = %.3f Degree",I_c_angle)
+
+
+
+
+
+
-- 
cgit