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//CHAPTER 3- THREE-PHASE A.C. CIRCUITS
//Example 11
clc;
disp("CHAPTER 3");
disp("EXAMPLE 11");
//SOLUTION
function power_sum=p1(phi);
power_sum=20*cos(phi); //power_sum=p1+p2=20*cos(phi) and in KiloWatts
endfunction;
function power_diff=p2(phi);
power_diff=(20*sin(phi))/sqrt(3); //power_diff=p1-p2=(20*sin(phi))/sqrt(3) and in KiloWatts
endfunction;
//solution (a): when phi=0
power_sum=20*cos(0); //eq(i)
power_diff=(20*sin(0))/sqrt(3); //eq(ii)
//solving eq(i) and eq(ii) to get values of p1 and p2
A=[1 1;1 -1];
b=[power_sum;power_diff];
x=inv(A)*b;
x1=x(1,:); //to access the 1st row of 2X1 matrix
x2=x(2,:); //to access the 2nd row of 2X1 matrix
disp("Solution (a)");
disp(sprintf("P1 + P2 = %d kW",power_sum));
disp(sprintf("P1 - P2 = %d kW",power_diff));
disp(sprintf("The two wattmeter readings are %d kW and %d kW",x1,x2));
//solution (b): when phi=30 or %pi/6 (lagging)
power_sum=20*cos(%pi/6);
power_diff=(20*sin(%pi/6))/sqrt(3);
A=[1 1;1 -1];
b=[power_sum;power_diff];
x=inv(A)*b;
x1=x(1,:);
x2=x(2,:);
disp("Solution (b)");
disp(sprintf("P1 + P2 = %f kW",power_sum));
disp(sprintf("P1 - P2 = %f kW",power_diff));
disp(sprintf("The two wattmeter readings are %f kW and %f kW",x1,x2));
//solution (c): when phi=60 or %pi/3
power_sum=20*cos(%pi/3);
power_diff=(20*sin(-(%pi/3)))/sqrt(3); //leading
A=[1 1;1 -1];
b=[power_sum;power_diff];
x=inv(A)*b;
x1=x(1,:);
x2=x(2,:);
disp("Solution (c)");
disp(sprintf("P1 + P2 = %f kW",power_sum));
disp(sprintf("P1 - P2 = %f kW",power_diff));
disp(sprintf("The two wattmeter readings are %f kW and %f kW",x1,x2));
//solution (d): when phi=90 or %pi/2
power_sum=20*cos(%pi/2);
power_diff=(20*sin(%pi/2))/sqrt(3); //leading
A=[1 1;1 -1];
b=[power_sum;power_diff];
x=inv(A)*b;
x1=x(1,:);
x2=x(2,:);
disp("Solution (d)");
disp(sprintf("P1 + P2 = %f kW",power_sum));
disp(sprintf("P1 - P2 = %f kW",power_diff));
disp(sprintf("The two wattmeter readings are %f kW and %f kW",x1,x2));
//END
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