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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+//Chapter 16, Problem 13
+clc;
+eff = 0.8; // effficiency
+f = 50; // in ohm
+Pout = 4800; // in Watt
+pf1 = 0.625 // power factor
+pf2 = 0.95 // power factor
+V = 240; // in Volts
+//calculation:
+Pin = Pout/eff
+Im = Pin/(V*pf1)
+phi1 = acos(pf1)
+phi1d = phi1*180/%pi
+//When a capacitor C is connected in parallel with the motor a current Ic flows which leads V by 90°.
+phi2 = acos(pf2)
+phi2d = phi2*180/%pi
+Imh = Im*cos(phi1)
+//Ih = I*cos(phi2)
+Ih = Imh
+I = Ih/cos(phi2)
+Imv = Im*sin(phi1)
+Iv = I*sin(phi2)
+Ic = Imv - Iv
+C = Ic/(2*%pi*f*V)
+kvar = V*Ic/1000
+printf("\n\n (a)Current taken by the motor, Im = %.0f A",Im)
+printf("\n\n (b)Supply current after p.f. correction, I = %.2f A ",I)
+printf("\n\n (c)Magnitude of the capacitor current Ic = %.0f A",Ic)
+printf("\n\n (d)Capacitance, C = %.0f μF ",(C/1E-6))
+printf("\n\n (d)kvar rating of the capacitor = %.2f kvar ",kvar)