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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 componentsA synchronous generator has been synchronized to an infinite grid of 13.8 kV and 60 Hz. The generator prime mover C/C is such that the no-load frequency is 62.5 Hz and the power slope is 1 MW/Hz. The AVR C/C (The reactive power against the terminal voltage C/C) is such that the zero reactive power voltage is 14.6 kV, while the slope is 0.4 MVAR/kV. Calculate: (a) the kVA loading of the generator and its power factor. (b) the settings of the prime mover and AVR such that the generator delivers 3 MW at 0.85 lagging power factor.5.5 kW, 1450 rpm. L-type equivalent circuit and parameters for a continuous sinusoidal phase of a star connected 3-phase ASM with line voltage 220 V and frequency 50 Hz are given as follows, ignoring core (iron) losses: R1=0.21 Ohm, R2=-0.18 Ohm, X1=X2-0.66 Ohm and Xm = 9.9 Ohm How many times the inrush current will be the rated current when the ASM is started directly? Answer options group 8.0815 2.6455 4.3763 7.1534 1.2558 6.3414 5.2280 3.7473
- 1980628cmid 11616 For reactive load, the power factor is equal to Select one: O a. 1 O b. 90 O c. 2 O d.0 If the voltage V, = 800 mV, then the current flowing in the primary winding Ip is: j2on Select one: O a. None of these O b. l, =40<0° mA O c.l, = 0.320<90° A O d. Ip =320<-90° mA if a Load absorbs an average power of 8 kW at a lagging power factor of 0.5 Select one: O a. -8000 + j13856.4 VA O b. None of these O c. 8000 - j13856.4 VA O d. 8000 - j13856.4 VA estion-299141-3 freactive power isWhile the instantaneous electric power delivered by a single-phase generator under balanced steady-state conditions is a function of time havi ng two components of a constant and a double-frequency sinusoid, the total instantaneous electric power delivered by a three-phase generator under balanced steady-state conditions is a constant. (a) True (b) FalseA 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.)
- 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-j0.6782. An alternating current varying sinusoidally with a frequency of 50 Hz has an rms value of40 A. Find:a. The instantaneous value of 0.0025 seconds after passing through maximumpositive value,b. The time measured from the maximum value when the instantaneous current is14.14 A.There is an installation with two equal triphasic loads of 300 kVA each and a power factor of 0.9 lagging. When connecting a pure reactive element in parallel, the reactive power at the entrance of the installation is 150 kVARS, with a lagging power factor. We can say that: a) A bank of reactors was connected and the fp deteriorated. b) A capacitor bank was connected and the pf deteriorated. c) A bank of reactors was connected and the fp improved. d) A capacitor bank was connected and the pf improved. e) None of the above
- The single-line diagram of a subtransmision system is presented in the figure below. At point B a three- phase short circuit has occurred. Calculate the voltage at points A, B, C and D. Three phase short circuit Infinite bus D +j0.2 j0.67 +10.70 +j1.05 V-1.00 puA 3 phase 200 MVA capacitor bank is to be switched off (disconnected) the grid that operates at 13.8kV (3-phase) at 60 Hz. The capacitor is connected in a WYE configuration, therefore the voltage across each phase is VLN=7.97KV. A restrike occurs at the zero crossing of the current. Determine the magnitude of voltage across the capacitor after the restrike. b. 0.05 mSec after the first restrike the arc extinguishes, which leaves a 1st charge trapped. Another restrike occurs again 3/8 of a cycle after the first restrike and is quickly extinguished. Identify the magnitude of voltage trapped across the capacitor after the а. second restrike.Consider a 10 kVA, 200/400 V, 50 HZ: single-phase tranformer has the folloving test results:0.C. test: 200 V, 0.6A, 63 W (L.V.side)S.C. test : 20 V, 25A, 85 W (H.V. side)Deternine:i- The eficiengy at 75 % offiull-load at 0.9 leading power factor.ii -The secondary terminal voltage on fill-load at tunity power facton