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- The instantaneous power absorbed by the load in a single-phase ac circuit, for a general R LC load under sinusoidal-steady-state excitation. is (a) Nonzero constant (b) Zero (c) Containing double-frequency componentsConsider the three single-phase two-winding transformers shown in Figure 3.37. The high-voltage windings are connected in Y. (a) For the low-voltage side, connect the windings in , place the polarity marks, and label the terminals a, b, and c in accordance with the American standard. (b) Relabel the terminals a, b, and c such that VAN is 90 out of phase with Va for positive sequence.Problem 1 - Series en Parallel AC networks [19] Look at the circuit in Figure 1 and determine the following: (a) Total Admittance. (b) Total Impedance. (c) Total Current (l:). (d) Current (I1) through impedance Z2. (e) Current (12) through impedance Z3. (f) Current (I3) through impedance Z4. (g) Is this an inductive or capacitive circuit? A. B Zs 220V;50HZ Figure 1 (h) Voltage across Z1. (i) Voltage across A and B. G) Voltage across Zs. Z1 = 3 + j5 ohm Z2 = 10 + jo ohm Z3 = 5 + j15 ohm Z4 = 10 – j30 ohm Zs = 20 – j30 ohm Admittance and Impedance in rectangular notation. All currents and voltage in polar notation. Take voltage as reference.
- The reactance of a generator is given as 0.1 pu based on the generator of 17 KV, 300 MVA. Determine the pu reactance on a base of 17 KV, 250 MVA. New pu reactance is.......In the system shown in Figure 1, the transformers are connected star-star with both star points grounded and the generators are connected in star with thier star points grounded. The system base is 15 MVA. The transformers all have reactances of 0.04 p.u. on this 15 MVA base. The reactances of all other elements are given in Table 1 (in 2) and the voltage levels are given in Table 2. p.u. G1 p.u. T1 jö Per-Unit Convert all values to p.u. on a 15 MVA base. Xa= p.u. Xc₂= XL = V BABE G1 2 X 9 T3 Figure 1: A section of the distribution system T1 L Table 1: Sequence reactances (2) 3 G1 L G2 0.3 0.59 0.01 4 L 9/10 10 Fault Voltage What is the voltage at bus 3 (in Volts) after the fault has occurred? Vp= V T2 5 T2 34 10/4 Table 2: Voltage bases (kV) G2 4 T3 10/9 | G2 Fault Current A three-phase fault with a fault reactance of 0.01 p.u. occurs at bus 3. Calculate the fault current flowing at the fault point in KA. Ip=-j KA SoQ2 (a) Three-phase generator consist of three set of single-phase generators, with voltages equal in magnitude but differing in phase angle by 120° apart. (i) Sketch phasor diagram of the three-phase voltages VA, VB and Vc (ii) Plot the three-phase voltages waveform in Q2(a)(i) in single plot.
- Power factor correction is usually done by adding _________ to the load circuit.Select the correct response:a. switchb. capacitorc. inductive loadsX₂=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 otoThe p.u. impedance value of an alternator corresponding to base values of 13.2 kV and 30 MVA is 0.2 p.u. Then the p.u. impedance value of an alternator for the new base values of 13.8 kV and 50 MVA is......
- What is the main direct cause of reactive power in AC system?A. Resistance of transmission linesB. Inductance and capacitance in the loadsC. Ideal transformer connected in the systemD. Power produced by generatorQ4/For the three-phase power network shown in Figure. the various components are: GI: 100 MVA, 0.30 pu reactance. G2: 60 MVA, 0.18 pu reactance. Transformers (cach): 50 MVA, 0.10 pu reactance. Inductive reactor X: 0.20 pu on a base of 100 MVA. Lines (each): 80 ohms (reactive); neglect resistance. with the network initially unloaded and a line voltage of 110 kV, a symmetrical short circuit occurs at midpoint E of line 2. Calculate the short circuit MVA to be interrupted by the circuit breakers A and B at the ends of the line. T3 38 L1 L2 G2 Bas 12 T4 BusQ2) A 13.2-kV single-phase generator supplies power to a load through a transmission line. The load's impedance is Ztoad 500 236.87° ohm , and the transmission line's impedance is Zine = 60 253.1° ohm. To reduce transmission line losses to 0.0103 of its losses without using the transformers design and use two transformers T1 between the generator and the transmission line and T2 between the transmission line and the load.