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//Electric Power Generation, Transmission and Distribution by S.N.Singh
//Publisher:PHI Learning Private Limited
//Year: 2012 ; Edition - 2
//Example 3.2
//Scilab Version : 6.0.0 ; OS : Windows
clc;
clear;
vs=220; // Supply voltage in Volts
rs=5; // Series resistance in Ohms
rp=2; // Parallel resistance in Ohms
xlp=8*%i; // Parallel inductive reactance in Ohms
xcp=-6*%i; // Parallel capacitive reactance in Ohms
zeq=((rp+xlp)*xcp)/(rp+xlp+xcp); // Equivalent impedance of parallel branch in Ohms
I=vs/(rs+zeq); // Current in the series branch in Ampere
Ps=((I)^2)*rs; // Power in 5 ohm resistor Watts
I1=I*xcp/(rp+xlp+xcp); // Current in branch ab in Ampere
I2=I*(rp+xlp)/(rp+xlp+xcp); // Current in branch cd in Ampere
Pab=(I1^2)*rp; // Power loss in branch ab resistor in Watts
Qab=(I1^2)*xlp; // Power loss in branch ab inductor in VAR
Qcd=(I2^2)*(xcp); // Power loss in branch cd capacitor in VAR
printf('The power loss in 5 ohm resistor is %.2f watts \n',abs(Ps))
printf('The power loss in branch ab resistor is %.2f watts \n',abs(Pab))
printf('The power loss in branch ab induoctor is %.2f VAR \n',abs(Qab))
printf('The power loss in branch cd capacitor is %.2f VAR \n',-abs(Qcd)) //Negative sign since capacitor supplies reactive power
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