blob: 46b03988a224e7fdc51504f748330fb0e7a84fba (
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
|
//CHAPTER 7- SINGLE PHASE TRANSFORMER
//Example 24
disp("CHAPTER 7");
disp("EXAMPLE 24");
//VARIABLE INITIALIZATION
va=10000;
v1=2500; //primary voltage in Volts
v2=250; //secondary voltage in Volts
R1=4.8;
X1=11.2;
R2=0.048;
X2=0.112;
//SOLUTION
//
R_dash_2=R2*(v1/v2)^2;
R_e1=R1+R_dash_2;
X_dash_2=X2*(v1/v2)^2;
X_e1=X1+X_dash_2;
//
R_dash_1=R1*(v2/v1)^2;
R_e2=R2+R_dash_1;
X_dash_1=X1*(v2/v1)^2;
X_e2=X2+X_dash_1;
//leakage impedence
z0=R_e2+X_e2*%i;
//applied load
Zl=5+3.5*%i;
//total impedence in series
Z=z0+Zl;
magZ=sqrt(real(Z)^2+imag(Z)^2);
magZl=sqrt(real(Zl)^2+imag(Zl)^2);
I2=v2/magZ;
V2=I2*magZl
disp("SOLUTION (a)");
disp(sprintf("The secondary terminal voltage is %f V",V2));
//
//part (b) of the problem cannot be solved mathematically alone.
disp(" ");
//
//END
|
