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In regards to the Sequence Networks of Impedance Loads, If the the positive sequence V1 = 1.03 angle(-8), the negative sequence voltage V2 = 0.03 angle(-34), and the zero sequence voltage Vo = 0, what is the a - phase voltage Va?
a. 1.05 angle(-8.7)
b. 1.00 angle(-7.2)
c. 1.01 angle (-9.0)
c.
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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.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.
- 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) 6.75The 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......A 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).
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- Problem 4- A large 3-phase industrial load draws 3.6 MVA at 0.85 power factor lagging from through a three- phase 12.47/7.2 kV 3-wire overhead distribution line. The phase conductor is #1 ACSR “Robin". Taking 3.6 MVA and 12.47 kV as base values, determine (a) The base impedance in (N) (b) The reactive power drawn by the load (p.u.) (c) The total current (p.u.) drawn from the power supplyQuestion 3. A three-phase impedance load balanced D a connected load in parallel with a balanced ? connected load made of bonding. D each arm for the load ?D = 6 + 6? and ? each arm for the connected load It is given as ?? = 2 + 2?. ? the connected load ?? = 1? is earthed through the neutral impedance. To this impedance load, the symmetrical component voltages ?0 = 10∠60°, ?1 = 100∠0° and ?2 = Unbalanced phase-to-earth voltages ??, ?? and ?? of 15∠200° volts were applied. a) Draw the zero, positive and negative sequence circuits. b) Find the complex powers imparted to each component circuit.2) 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 S3