Three loads are connected in parallel across a singlephase s

Three loads are connected in parallel across a single-phase source voltage of 240 V (rms). Load 1 absorbs 12 kW and 6.667 kVAR, Load 2 absorbs 4 kVA at 0.96 PF leading, Load 3 absorbs 15 kW at unity power factor. (a) Calculate the equivalent, Z, of the three parallel loads for two cases (i) series combination of R and X and (ii) parallel combination of R and X.    

Solution

P = Active Power Q = Reactive Power S = Total power FOR LOAD 1 = P = S*cos(theta1) Q = S*sin(theta1) tan(theta1) = Q/P = 6.667/12 = 0.55555 theta1 = arctac(0.55555) 29.056 deg lagging P = V*I*cos(theta1) I = P/(V*cos(theta1) = 57.2A lagging Z1 = V/I = 4.196 ohm with an angle of Series Combination of R & X = Z1 = 4.196cos + j*4.196*sin(29.056) Z1 = 3.67 + j2.038 ohm For Load 2 = theta2 = arccos(0.96) = 16.26 deg leading P = S*cos(theta2) = 4000*0.96 = 3840W I = P/(Vcos(theta2)) = 3840/(240*0.96) = 16.67 A leading Z2= V/I = 14.395 ohm with angle of 16.26 deg Z2 = 13.82 - 4.031j ohm For Load 3 :- P = 15000W unity pf resistive load Z3 = V^2/P = 3.84ohm (A) If R& X are in series Ynet = Y1 + Y2 + Y3 (using admittances easier as they are parallel loads) Ynet = (1/Z1)+(1/Z2)+(1/Z3) Ynet= 0.535 - 0.096j Znet =1/Ynet = 1.809 + 0.33j ohm Znet = 1.838 ohm with an angle of 10.19 deg (B)if R&X are in parallel Ynet = Z1 + (1/Z2) + (1/Z3) Ynet = 3.9971 + 2.33 j Znet = 1/Ynet = 0.186 - 0.109j Znet = 0.216 ohm with angle -30.31 degree