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+clc
+//Example 11.9
+printf("Given")
+disp('Power of induction motor=50kW ,power factor is 0.8 lag,Source voltage is 230V')
+disp('The wish of the consumer is to raise the power factor to 0.95 lag')
+//Let S1 be the complex power supplied to the indiction motor
+V=230;Pmag=50*10^3;pf=0.8;
+Pang=(acos(pf)*180)/%pi
+S1mag=Pmag/pf
+S1ph=Pang
+x=S1mag * cos (( Pang * %pi ) /180) ;
+y=S1mag * sin (( Pang * %pi ) /180) ;
+z= complex (x,y)
+disp(z ,'S1=')
+//To achieve a power factor of 0.95
+pf1=0.95
+//Now the total complex power be S
+P1ang=(acos(pf1)*180)/%pi
+Smag=Pmag/pf1
+Sph=P1ang
+a=Smag * cos (( P1ang * %pi ) /180) ;
+b=Smag * sin (( P1ang * %pi ) /180) ;
+c= complex (a,b)
+disp(c,'S=')
+//Let S2 be the complex power drawn by the corrective load
+S2=c-z
+disp(S2,'S2=')
+disp('Let a phase angle of voltage source selected be 0 degree')
+//Let I2 be the current
+I2=-S2/V
+//Let Z2 be the impedance of corrective load
+Z2=V/I2
+disp(Z2,'Z2=')
+