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Each phase of a -connected load consists of the series combination of a 350-mH inductor, a 20-F capacitor, and a 170- resistance. This load is fed by a Y-connected source. The source has a positive phase sequence and Va 4000 VRMS . Assume zero line resistance and = 1000 rad/s. Determine the total average power absorbed by the load. Spring 2014, Exam #2, Problem #1 Answer: 2040 W 1 In the circuit shown, the three-phase source is balanced with a positive phase sequence and va 169.7cos 377t V . Each of the inductors is 46 mH and each of the resistors is 10 . Determine the total average power absorbed by the Y load. Spring 2015, Exam #3, Problem #1 Answer: 1080 W 2 For the circuit given, determine the magnitude of the neutral-to-neutral voltage, |VNn| . Spring 2015, Exam #3, Problem #2 Answer: 22 V 3 In the circuit shown, the three-phase source is balanced with a positive phase sequence and Va 86.70 VRMS . Let Z1 Z2 Z3 15 26 j . Determine the average power absorbed by phase impedance Z2 . Spring 2015, Exam #3, Problem #3 Answer: 375 W 4 In the circuit shown, the source is balanced with a positive phase sequence and va 240cos 377t V . Load A is a 40- resistor and a 184-mH inductor in series. Load B is a 51-F capacitor and a 30- resistor in series. Load C is a 16- resistor. Determine the neutral-to-neutral current, iNn(t) . Spring 2015, Exam #3, Problem #4 Answer: 8cos 377t 120 A 5 A Y-source is balanced with a negative phase sequence and Va 2000 VRMS . This source is connected to a Y-load where each phase impedance is 28 37 j . The line impedance between each source phase and each load phase is 2 3 j . Determine the total average power absorbed by the lines. Spring 2015, Exam #3, Problem #5 Answer: 96 W 6 A balanced -connected load with Z P 8 6 j is connected to a balanced Y-connected source with Van 163.30 V and a positive phase sequence. Assume that the line impedances between the source and the load are negligible. Calculate the total average power absorbed by the load. Spring 2015, Final exam, Problem #5 Answer: 9.6 kW 7 In the circuit below, the Y source has a positive phase sequence with Va 1100 VRMS . The load impedances are ZA 50 j80 , ZB j50 , ZC 100 j 25 . Determine the complex power delivered to ZA . Spring 2015, Final exam, Problem #6 Answer: 27658 VA 8 In the circuit below, the resistor markings unfortunately have been omitted, but several of the currents are known. Specifically, Iad = 1 A. (a) Compute Iab , Icd , Ide , Ife , and Ibe . (b) If Vba = 125 V , determine the value of the resistor linking nodes a and b . Spring 2014, Homework #4, Problem #3 Answers: (a) Iab = –9 A , Icd = 8 A , Ide = 9 A , Ife = –10 A , Ibe = 1 A , (b) 13.9 9 Assume the system shown below is balanced, Rw = 0 , Van = 208 V , and a positive phase sequence applies. Calculate all phase and line currents, and all phase and line voltages, if Zp is equal to (a) 1 k , (b) 100 + j48 , (c) 100 – j48 . Spring 2014, Homework #4, Problem #5 Van 2080 V Van 2080 V Van 2080 V Vbn 208 120 V Vbn 208 120 V Vbn 208 120 V Vcn 208 240 V Vcn 208 240 V Vcn 208 240 V Answers: (a) Vab 36030 V , (b) Vab 36030 V , (c) Vab 36030 V Vbc 360 90 V Vbc 360 90 V Vbc 360 90 V Vca 360 210 V Vca 360 210 V Vca 360 210 V I aA 2080 mA I aA 1.87 25.8 A I aA 1.8725.8 A I bB 208 120 mA I bB 1.87 145.8 A I bB 1.87 94.2 A I cC 208 240 mA I cC 1.87 265.8 A I cC 1.87145.8 A 10 For the balanced three-phase system shown below, it is determined that 100 W is lost in each wire. If the phase voltage of the source is 400 V , and the load draws 12 kW at a lagging power factor of 0.83, determine the wire resistance Rw . Spring 2014, Homework #4, Problem #6 Answer: 225 m 11 The balanced circuit shown below has Vab = 380 VRMS . Determine the line and phase currents in the load when Z = 9 + j12 . Spring 2015, Homework #4, Problem #2 I aA 14.67 83.1 A RMS Answers: I bB 14.67 203.1 A RMS I cC 14.67 36.9 A RMS 12 A three-phase balanced load is fed by a balanced Y-connected source with a line-to-line voltage of 220 VRMS . It absorbs 1500 W at 0.8 power factor lagging. Calculate the phase impedance if it is (a) connected , and (b) Y connected . Spring 2015, Homework #4, Problem #3 Answers: (a) 61.9 + j46.5 , (b) 20.6 + j15.5 13 Determine the power delivered to the load in the circuit below. Spring 2015, Homework #4, Problem #4 Answer: 265 W 14 In the circuit below, each impedance Zp is a parallel combination of a 1-mF capacitance, a 100mH inductance, and a 10- resistance. One line voltage is Vab 2080 V . The sources have a positive phase sequence and operate at 50 Hz. Each wire resistance is Rw = 1 . Determine the total (real) power absorbed by the load. Spring 2016, Homework #4, Problem #3 Answer: 1.68 kW 15 In Circuit #4, determine the line current IcC . Spring 2016, Homework #4, Problem #4 Answer: 1 24 A 16 In the circuit below, the balanced Delta-connected load draws a total apparent power of 15 kVA at a lagging power factor of 0.87 . The source is balanced and follows a positive phase sequence. One phase voltage is Van 1700 V . The line impedances are negligible. Determine the RMS amplitude of the line current IbB . Spring 2016, Exam #2, Problem #3 Answer: 41.6 ARMS 17 A balanced three-phase, negative-sequence, Wye-connected source with Vcn 2400 VRMS is connected to a three-phase Wye-connected load. Each phase of the load has an impedance of 10 + j15 Ω . The impedances in the lines are ZaA = 1 + j3 Ω , ZbB = 2 + j2 Ω , ZcC = 1 + j2 Ω . Determine the RMS amplitude of the line current into phase C of the load. Spring 2016, Exam #2, Problem #4 Answer: 115 ARMS 18 A balanced, positive-sequence, Y-connected, three-phase source is connected to a balanced Y-connected three-phase load. One phase of the source is known to be Vbn 480 120 VRMS . Each phase impedance in the load is 15 + j10 . Line impedances are negligible. (a) Determine the magnitude of the line current IaA . (b) Compute the total (real) power absorbed by the load. (c) Calculate the power factor at which the source operates. Spring 2016, Final exam, Problem #5 Answers: (a) 26.6 ARMS , (b) 31.9 kW , (c) 0.83 19