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//Book - Power System: Analysis & Design 5th Edition
//Authors - J. Duncan Glover, Mulukutla S. Sarma, and Thomas J. Overbye
//Chapter - 10 ; Example 10.1
//Scilab Version - 6.0.0 ; OS - Windows
clc;
clear;
CTratio=100/5; //CT ratio
Zs=0.082; //Secondary resistance of a 100:5 CT in Ohm
IZB=[5 0.5; 8 0.8; 15 1.5]; //Secondary output current in Amperes and burden resistance in Ohm
E=(Zs+IZB(1,2))*IZB(1,1); //Secondary Excitation voltage in Volts
printf('\nCase: a');
printf('\nThe Secondary excitation voltage is %0.4f Volts',E);
Ie=0.25 //Secondary Excitation current for the secondary voltage from figure !0.8 in Amperes
printf('\nThe Secondary excitation current is %0.4f Amperes',Ie);
I=CTratio*(IZB(1,1)+Ie); //Primary current of the CT in Amperes
printf('\nThe Primary current is %d Amperes',I);
CTerr=Ie*100/(IZB(1,1)+Ie)'; //Error in CT
printf('\nThe error of the CT is %0.4f percentage',CTerr);
E=(Zs+IZB(2,2))*IZB(2,1); //Secondary Excitation voltage in Volts
printf('\n\nCase: b');
printf('\nThe Secondary excitation voltage is %0.4f Volts',E);
Ie=0.4 //Secondary Excitation current for the secondary voltage from figure !0.8 in Amperes
printf('\nThe Secondary excitation current is %0.4f Amperes',Ie);
I=CTratio*(IZB(2,1)+Ie); //Primary current of the CT in Amperes
printf('\nThe Primary current is %d Amperes',I);
CTerr=Ie*100/(IZB(2,1)+Ie)'; //Error in CT
printf('\nThe error of the CT is %0.4f percentage',CTerr);
E=(Zs+IZB(3,2))*IZB(3,1); //Secondary Excitation voltage in Volts
printf('\n\nCase: c');
printf('\nThe Secondary excitation voltage is %0.4f Volts',E);
Ie=20 //Secondary Excitation current for the secondary voltage from figure !0.8 in Amperes
printf('\nThe Secondary excitation current is %0.4f Amperes',Ie);
I=CTratio*(IZB(3,1)+Ie); //Primary current of the CT in Amperes
printf('\nThe Primary current is %d Amperes',I);
CTerr=Ie*100/(IZB(3,1)+Ie)'; //Error in CT
printf('\nThe error of the CT is %0.4f percentage',CTerr);
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