1 files changed, 49 insertions, 0 deletions
diff --git a/1092/CH14/EX14.12/Example14_12.sce b/1092/CH14/EX14.12/Example14_12.sce
new file mode 100755
index 000000000..b66edc909
--- /dev/null
+++ b/1092/CH14/EX14.12/Example14_12.sce
@@ -0,0 +1,49 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-12
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data(from Example 14-11)
+V_1 = 2300 ; // Primary voltage in volt
+V_2 = 230 ; // Secondary voltage in volt
+I_2 = 2174 ; // Secondary current in A
+I_1 = 217.4 ; // Primary current in A
+// calculated values from Example 14-11
+Z_2 = 0.00316 ; // Secondary internal impedance in ohm
+Z_1 = 0.316 ; // Primary internal impedance in ohm
+
+
+// Calculations
+alpha = V_1 / V_2 ; // Transformation ratio
+// case a
+Z_L = V_2 / I_2 ; // Load impedance in ohm
+
+// case b
+Z_p = V_1 / I_1 ; // Primary input impedance in ohm
+
+Zp = (alpha)^2 * Z_L ; // Primary input impedance in ohm
+
+// Display the results
+disp("Example 14-12 Solution : ");
+
+printf(" \n a: Load impedance :\n    Z_L = %.4f ohm \n ", Z_L );
+
+printf(" \n b: Primary input impedance : ");
+printf(" \n    (method 1) :\n    Z_p = %.2f ohm \n ",Z_p );
+printf(" \n    (method 2) :\n    Z_p = %.2f ohm \n ",Zp );
+
+printf(" \n c: The impedance of the load Z_L = %.4f Ω, which is much greater",Z_L);
+printf(" \n    than the internal secondary impedance Z_2 = %.5f Ω .\n ",Z_2);
+printf(" \n    The primary input impedance Z_p = %.2f Ω,which is much greater",Z_p);
+printf(" \n    than the internal primary impedance Z_1 = %.3f Ω .\n",Z_1);
+
+printf(" \n d: It is essential for Z_L to be much greater than Z_2 so that the ");
+printf(" \n    major part of the voltage produced by E_2 is dropped across the ");
+printf(" \n    load impedance Z_L. As Z_L is reduced in proportion to Z_2, the ");
+printf(" \n    load current increases and more voltage is dropped internally ");
+printf(" \n    across Z_2.");