blob: 4a464107924d281f9756a698115f582b571d49de (
plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
|
// Electric Machinery and Transformers
// Irving L kosow
// Prentice Hall of India
// 2nd editiom
// Chapter 14: TRANSFORMERS
// Example 14-32
clear; clc; close; // Clear the work space and console.
// Given data
S_1 = 10 ; // VA rating of small transformer
V = 115 ; // voltage rating of transformer in volt
V_2_1 = 6.3 ; // voltage rating of one part of secondary winding in volt
V_2_2 = 5.0 ; // voltage rating of other part of secondary winding in volt
Z_2_1 = 0.2 ; // impedance of one part of secondary winding in ohm
Z_2_2 = 0.15 ; // impedance of other part of secondary winding in ohm
// Calculations
// case a
V_2 = V_2_1 + V_2_2 ; // voltage across secondary winding in volt
I_2 = S_1 / V_2 ; // Rated secondary current in A when the LV secondaries are
// connected in series-aiding
// case b
I_c = (V_2_1 - V_2_2) / (Z_2_1 + Z_2_2); // Circulating current when LV windings are paralled
percent_overload = (I_c / I_2)*100 ; // percent overload produced
// Display the results
disp("Example 14-32 Solution : ");
printf(" \n a: Both coils must be series-connected and used to account for the ");
printf(" \n full VA rating of the transformer. Hence,the rated current in 5 V ");
printf(" \n and 6.3 V winding is : \n");
printf(" \n I_2 = %.3f A \n\n", I_2);
printf(" \n b: When the windings are paralleled, the net circulating current is ");
printf(" \n the net voltage applied across the total internal impedance of ");
printf(" \n the windings,or :\n");
printf(" \n I_c = %.2f A \n ",I_c);
printf(" \n The percent overload is = %f percent ≃ %.f percent ",percent_overload,percent_overload);
|
