4.) An unbalanced four-wire Y-Y circuit is shown in Figure 3. (a) Find the line currents ia, ib, ic, an in. (b) Find the complex power absorbed by each phase of the load. +- 30 cos(10) V A 2012 зн m-m 402 8H m 30cos (10-120°) V ic 63 9H - с (30 cos (10x + 1200) V N
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- X₂=0.1 1 XL +03 Soto X₁ = 0.2 X = 0.1 X = 0.1 Above is the one-line diagram of a simple power system. Each generator is represented by an emf behind the subtransient reactance. All impedances are expressed in per unit on a common base. All resistances and shunt capacitance are neglected. The generators are operating on no load at their rated voltage with their emfs in phase. A symmetrical fault occurs at bus 1 through a fault impedance Z, =j0.08 per unit. a) Using Thevenin's Theorem, determine the impedance to the point of fault and the fault current in per unit. b) Find the bus voltages and line currents during the fault. 2 O otoDraw Wye- Delta three phase transformers connection ( no-load and full-load) in Lt-spiceThe following single-phase loads are connected to a 415V three-phase supply. Z, = (25 - j15)0, Z, = (45+ j10)2 and Z, = (5+ j30)2. Answer questions 37 and 38. -. What is the current in amperes in the red phase? A. 6.65233.69 D. 5.60233.69 B. 8.23/30.96 C. 14.23430.96 Find the magnitude of the current in the neutral conductor in amperes. A. 18.16 с. 21.33 B. 14.23 D. 12.54 E. 15.68 . A three-phase supply energizes three resistive single-phase loads which draw the following currents: 1, =84, 1, = 6A and I, = 24. What is the magnitude of the neutral current? A. 8 В. 16 С. 5.29 D. 12.49 E. 5.61
- Given the following phase currents: Ia= 3.5+j3.5; Ib=-j3; and Ic+-2.6+j1.5. Determine the negative sequence current for phase A. A. 0.313+j0.678 B. -0.313 -j0.678 C. -0.313+j0.678 D. 0.313-j0.678A 400 MVA synchronous machine has H, = 4.6 MJ/MVA and a 1200 MVA machine has H₂=3.0 MJ/MVA. The two machines operate in parallel in a power plant. Find out H relative to a 100 MVA base.A three-phase alternator generating unbalanced voltages is connected to an unbalanced load through a 3-phase transmission line as shown in figure. The neutral of the alternator and the star point of the load are solidly grounded. The phase voltages of the alternator are Ea = 1020º, Eb = 102-90°, Ec = 10/120⁰ E E E The positive sequence component of the load current is (A) 1.3102-107⁰ A (B) 0.3322-120° A (C) 0.9962-120º A (D) 3.5104-81⁰ A +) + j1.002 0000 j1.002 0000 j1.0Ω vovo j1.0Ω 0000 j2.002 voor j3.002 oooo
- In a balanced system, the phasor sum of the line-to-line voltages and the phasor sum of the line-to-neutral voltages are always equal to zero. (a) False (b) TrueThree single-phase transformers, each rated 10MVA,66.4/12.5kV,60Hz, with an equivalent series reactance of 0.1 per unit divided equally between primary and secondary, are connected in a three-phase bank. The high-voltage windings are V-connected and their terminals are directly connected to a 115-kV three-phase bus. The secondary terminals are all shorted together. Find the currents entering the high-voltage terminals and leaving the low-voltage terminals if the low-voltage windings are (a) Y-connected and (b) - connected.A single-phase, 120V(rms),60Hz source supplies power to a series R-L circuit consisting of R=10 and L=40mH. (a) Determine the power factor of the circuit and state whether it is lagging or leading. (b) Determine the real and reactive power absorbed by the load. (c) Calculate the peak magnetic energy Wint stored in the inductor by using the expression Wint=L(Irms)2 and check whether the reactive power Q=Wint is satisfied. (Note: The instantaneous magnetic energy storage fluctuates between zero and the peak energy. This energy must be sent twice each cycle to the load from the source by means of reactive power flows.)
- Four conductors meet at a junction. the current are i1= 5sinwt, i2= 3sin(wt + 90) and i3= 2sin(wt-90). calculate the rms value of i4.For this kind a problem, how do you know how the voltage source is configured ( wye or delta). The problem start getting the phase current for phase 1 and 2 , but do we have to pretend that voltage source to delta load is connected like y-delta to convert the voltage to delta source? Or how does it work?R 20 B Y R 20 400V, 60Hz B IR ZB = 55 < 0° 0 Figure 4: Star Unbalanced Load IR ZR IB ZY ZR = 30 < 0° 0 Fault Zy17 <0°02 = A balanced load with ZR = ZB Zy = ....<00 Q was supplied a three phase 50 Hz, 100 V between line (balanced source). A fault on phase 'B' causes an open circuit, which unbalanced the load as shown in Figure 4. Similarly, a fault on the load of phase 'Y' cause a short-circuit, which unbalanced the load as shown in Figure 5. Using complex notation determine and analyse the individual phase currents and the neutral current in both figure.