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- 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.)Please help with this balanced three-phase circuit problem.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......
- 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) 0.9 (b) 20.25 (d) 6.75What is load curve4. An Open-Wye/Open-Delta transformer, as shown in Figure, is formed by two identical single-phase transformers. The primary line-to-line voltage is 12.47 kV, and the secondary line-to-line voltage is 208 V. Take2) The one-line diagram of a three-phase power system is as shown in Figure. Impedance are marked in per-unit on a 100-MVA, 400-kV base. The load at bus 2 is S2=15.93 MW - j33.4 Mvar, and at bus 3 is S3-77 MW + j14 Mvar. It is required to hold the voltage at bus 3 at ...L0 kV. Working in per-unit, determine the voltage at buses 2 and 1. (Please determine the voltage magnitude within the specified limits: 410-450 V. By considering the voltage magnitude value you determined yourself, solve the question.) V₁ j0.5 pu V₂ j0.4 pu S₂ Figure V3 S3What 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 generatorA 50 Hz, Scott connected transformer supplies an unbalanced two phase load at ( 400 V) per phase. For the leading phase the load has a resistance of (13.3 ohm) and capacitor of (318) micro farad in series. For the other phase the load consists of resistance (10) ohm and an inductance of (42.3) mH. Calculate both mathematically and graphically the lines currents on 3 phase side. The main transformer primary to secondary turns ratio is (12/1).6.5% ans ex A so kvA. 2o00/20 V, So He single-phase transformar has impedance drof of8% and resistance drop of 4%. Caleulate. The Requlation of the Thams former at full- Load o8 P.s Legging. also find The Power factor at which Voltage requlation will Be Zero. Sols 0.5 leading. s alculate The % Noltege Requlation af atransformar in w hich the Percantage resistance drof is 1% and Percentage dre is 6% when The Power fador is 0.8 lagging ) > unity aMS 38 % ans 1 % (シ→ 0.8 leading . amS - 2.2%in distribution transformers; What is 30 degrees phase shift in delta-star connection? What are the disadvantages? (can you explain with drawing?)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 circuitSEE MORE QUESTIONS