Question

(15 pts.) A voltage source with 115 V rms value, is feeding three loads, which are in parallel to one another. The information about the loads is as follows L1: 200 W, at pf .92 lagging L2: 250 VA, at pf= .81 leading L3: 200 VARs at pf .85 lagging a. (1 point) Show the diagram depicting the given system 5 ะ Jr 2 + b. (6 points) Calculate S, P, Q, and 6 for each of the loads c. (3 points) Draw the power triangles corresponding to each of the loads d. (3 points) Draw the power triangle of the 3 loads combined e. (2 points) Calculate the rms value of the supply current 8. L3

Answer 1

P_1 = 200 w, cos phi_1 = 0.92 (cos) Q (Reactive) p_1 = s_1 cos phi_1 cos phi_1 = 0.92 s_1 = p_1/cos phi_1 phi_1 = cos^-1 (0.92) = 23.0719 phi_1 = p_1/cos phi_1 (sin phi_1) sin phi_1 = 0.3919 = 201/0.92 (0.3919) (P = j phi) - 1 cos sin (p - j phi) - 1 leading = 85.199 VAR s_1 = p_1 + j phi = 200 + j 85.199 VA phi = Tan^-1 (Q/p) = Tan^-1 (85.199)/200 = 23.07 j L_2 s_2: - 250 VA phi_2 = cos^-1 (0.81) = 35.97 p.f = 0.81 leading sin phi_2 = sin phi_2 = 0.586 p_2: - s_2 (cos phi_2) = (250) (0.81) = 202.5 w phi_2 = s_2 sin phi_2 = (250) (0.586) = 146.60 V A R s_2 = 202.5 j 146.60 V A phi_2 = 35.904 L_3 phi_3 = 200 V A R cos phi_3 = 0.85 (losing) phi_3 = 31.788 degree s_3 = phi_3/sin phi_3 sin phi_3 = sin (cos^-1 (0.85)) = 0.5267 = 200/0.5257 = 379.66 V A p_3 = s_3 cos phi_3 = (379.68) (0.85) = 322.7137 w phi_3 = 31.788 s_3 = 322.7137 + j 200 Total power s_T = s_1 + s_2 + s_3 = 200