Question

Instructions: Start each new problem on a new sheet of paper. Staple this sheet and both y our work and . formula sheet. Show procedure for full credit. Neatness and UNITs counts for points. Use only 2 decimals for all calculations. If calculated values are zero with two decimals, express the number in 103, 10*, or 10 with two decimals as well. l. (25 %) The system in Figure 1 has two costumers, one commercial and one residential. The commercial load is an inductive load represented by Zc-6+.22and the residential load is purely resistive represented by Z 6 Ω The transmission line is represented by an inductance whose value is 3.1831 mH. Assuming the source has an instantaneous 60-Hz voltage source v()-120V2cos(wt + 60) V a) (10 %) Determine the current delivered by the source. Report its value in phasor form and its equivalent sinusoidal form. Is the current lagging or leading the voltage? For parts b) and c), assume that a fault down the system puts line out service. With the line out the service, the current source is 18.57 38.20 A and the industrial voltage is 113.62 449.51 V b) (5 %) By using the fault data, determine the complex power observed by the commercial load and its power factor (lagging or leading?) (10 %) A bank of capacitors is placed in parallel with the commercial load to raise its power factor to 1 . Find the net capacitance of the capacitor bank in μF that is required to correct the power factor c) 3 9831 Faut vit) Figure 1. A radial network. The fault is ONLY assumed for parts b) and c).

Answer 1

a) If v(t) = v_M cos (wt + phi), it can be expressed in phasor form of v = v_m angle phi v(t) = 120 squareroot 2 cos (wt + 80) = 120 squareroot 2 angle 60 degree volts F = 60 Hz X_L = j2 pi cL = j 2 pi times 60 times 3.1831 times 10^-3 X_L = j 1.2 Ohm Let v_1 be the node Apply KCL at node v_1: v_1 - 120 squareroot 2 angle 60/j 1.2 + v_1/6 + j 1.2 + v_1/j 1.2 + 6 = 0 v_1 [1/j 1.2 + 1/(6 - j 1.2) + 1/j 1.2 + 6) = 120 squareroot 2 angle 60/j 1.2 v_1 (0.64879 angle -60.3955) = 141.42 angle -30 degree v_1 = 217.97 angle 30.3955 degree v current delivered by the source I_3 = v_s - v_1/X_L = (120 times squareroot 2 angle 60) - 217.97 angle 30.3955 degree/j 1.2 I_s = 91.24 angle 70.42 squareroot Amp rightarrow phasor form sinusoidal form i_s(t) = 91.24 cos (wt + 70.42 degree) i_s(t) = (91.24 cos(2 pi times 60 t + 70.42) i_s(t) = 91.24 cos (376.99 t + 70.42 degree Amp.