128, An equipment has an impedance 0.9 p.u. to a base of 20 MVA, 33 kV. To the base of 50 MVA, 11 kV, the p.u. impedance will be (a) 4.7 (c) 09 (b) 20.25 (d) 6.75
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- The 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......Use the information below to answer questions 24 und 25. The voltage across and the current through a load are (10+ j5Y and (2-jiArespecively. 24. What is the reactive power in VArconsumed? D. 0 A. 15 C. 20 D. 25 25. What is the pawer factor of the load? A. 0.8 leading B. 0.8 Jagging. -D. 0.6 ligging C 0.6 leading6. What is the effect of load inductance on the ripple current? 7. i effect of chonnine Bock and boost 8. Wh 9 What chopping 10. At what duty evcle does the load current rinnle current he
- 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 components2. The pu impedance of a circuit element is 0.15. If the base kV and base MVA are halved. Then the new value of pu impedance of the circuit element is A. 1.2 C. 1.1 3. The pu impedan B. 1.0 B. ambient tempera- ture D. 0.8 E. nonePlease show you detailed solution Number 10 A voltage of 250 volts at 50 Hz is impressed on a circuit having a 60-ohm resistor, 35-?F capacitor, and a 0.25-henry inductor in series. Determine power factor. A. 0.979 C. 0.924 B. 0.862 D. 0.707
- what will be the total and unit for impedance if the given is 8mA and 9mV?A parallel connection of a RL branch with a C branch is connected across a 100V AC mains. At first R = 10 ohms, L = 20mH and frequency of 1000rad / s, the current measured is 2.2361A at 89.44 leading power factor. A) Determine the initial capacitance. B) However, a fault occurs on the capacitor branch making its capacitance 20% lower and a resistance of 5 ohms is detected. If this faulty circuit is rerun but at a frequency of 500 rad / s, determine the new current that will flow through the circuitAssum rms for the phase current
- Power in a three phase delta system with balanced load is equal to (Sqrt (3)) (VL) (IL) (p.f.) (Sqrt (3)) (Vph) (Iph) (p.f.) (Sqrt (3)) (VL) (Iph) (p.f.) O (Sqrt (3)) (Vph) (IL) (p.f.)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 SoThe concentration ratio of parabolic trough power plant is typically O a. from 30 to 40 O b. more than 1500 O c. less than 10 O d. from 10 to 20