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A four wire, three phase system has a balanced supply voltage of 208V. Three wattmeters are connected to measure the power dissipated on the three resistors. Calculate for a) Neutral Voltage b) Line currents c) readings of wattmeter d) total power of the load 2002 402 120 Figure 2. Question number 5 You can use the sample problem below as guide answering the problem, please use 3 decimal points for the final answer! Sample Problem solution for guide (the given from this problem is different from the problem above) 208 (a) V₂ = - 120 V (b) Because the load is unbalanced, we have an unbalanced three-phase system. Assuming an abc sequence, 2.520 A 120/0° 48 120-120° =32-120° A 40 120/120 =24120° A 60 -I, I₁ +1₂ +1, 1₁ -2.5+ (3) (-0.5-√³)+(2)(-0.5+j) √√3 1₂ =j²²2₂ =j0.866= 0.866/90° A 1₁ = 2.5 A. 1₂ - 3 A. 1=0.866 A (2.5) (48) 300 W (3) (40) = 360 W (2)³ (60) = 240 W O Hence, P, IR, P₂1R, P₁ = 1; R, (d) P, P,+P, + P, = 900 W 1₁ = 2 A.