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+//CHAPTER 2- STEADY-STATE ANALYSIS OF SINGLE-PHASE A.C. CIRCUIT
+//Example 6
+
+disp("CHAPTER 2");
+disp("EXAMPLE 6");
+
+//VARIABLE INITIALIZATION
+f=50;                            //in Hertz
+I1=20;                           //in Amperes
+pf1=0.75;                        //power factor
+v=230;                           //in Volts
+pf2=0.9;                         //power factor(lagging)
+
+//SOLUTION
+//V.I1.cos(Φ1) = P
+phi1=acos(pf1);
+res1=tan(phi1);                  //result1 = tan(Φ1)
+phi2=acos(pf2);
+res2=tan(phi2);                  //result2 = tan(Φ2)
+Ic=I1*pf1*(res1-res2);
+w=2*%pi*f;                      //w=2.pi.f
+c=Ic/(v*w);
+disp(sprintf("The value of capacitance is %5.2f μF",c*(10^6)));//text book answer is 82.53 mF
+Qc=v*Ic;			// reactive power in kVAr
+disp(sprintf("The reactive power is %6.4f kVAR",Qc/(10^3)));//text book answer is 1.3716
+I2=I1*(pf1/pf2);		//I1.cos(Φ1) = I2.cos(Φ2)
+disp(sprintf("The new supply current is %5.2f A",I2));
+
+//END