diff options
| author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
| commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
| tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3556/CH12/EX12.3/Ex12_3.sce | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
| download | Scilab-TBC-Uploads-7f60ea012dd2524dae921a2a35adbf7ef21f2bb6.tar.gz Scilab-TBC-Uploads-7f60ea012dd2524dae921a2a35adbf7ef21f2bb6.tar.bz2 Scilab-TBC-Uploads-7f60ea012dd2524dae921a2a35adbf7ef21f2bb6.zip | |
initial commit / add all books
Diffstat (limited to '3556/CH12/EX12.3/Ex12_3.sce')
| -rw-r--r-- | 3556/CH12/EX12.3/Ex12_3.sce | 61 |
1 files changed, 61 insertions, 0 deletions
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)
+
+
+
+
+
+
|