A Y-A system has source voltages of 415Z-105° V on its a-phase and 415Z135° V on its c-phase.Determine the type of sequence of this system.If this Y-A system has source, line and load impedances of 6/45°2, 15 - j25 Q and 102230° Qrespectively, obtain(a) The line currents(b) The load currents(c) The load voltages(d) The complex powers supplied by the sources

In this question, A balance star source to supplier to balance Delta load. Find the line current,…

Answered: A Y-A system has source voltages of…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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consider the following diagram of impedances in per unit of a common base a) get the YBUS admittance array and the ZBuz impedance array 213-j2 pu zt2-j0.15 zg2-j1.85 791-j0.9 211-0.1 1 212-j0.5pu 223-j0.5pu Va 1 p.u. Jv1-1p.u. 22-10 pu
18. A 415 V, 50 Hz, 3-phase supply of phase sequence RYB is connected to a delta connected load in which branch RY consists of R, = 100. branch YB conists of R₂ = 20 2 in series with X₂ = 60 Q and branch BR consists of a capacitor C= 30 uF. Take Vy as the reference and calculate the line currents. Draw the complete phasor diagrams. (Elect. Machines, A.M.I.E. Sec. B. 1989) 14.54, 1, 10.66 2172,92, 14.46 2-471 (1 = 7.78
Draw a 3-phase wye-connected generator with transmission lines connected to a 3-phase delta-connected load. Assume the lines connecting them have an inductance of j5 per phase. The generated voltages are 120 Vrms with phase 'a' having a phase angle of 0°. The loads are 10 Ohm resistors per phase. When drawing, make sure to label all 3 phases with voltage phasors, line inductances, and load resistances.
Question 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.
a system that has all zeros and poles in the left half of s-plane represents minimum phase systems ? T F -----------------
A three-phase alternator generating unbalanced voltages is connected to an unbalanced load through a 3-phase transmission line as shown in figure. The neutral of the alternator and the star point of the load are solidly grounded. The phase voltages of the alternator are Ea = 1020º, Eb = 102-90°, Ec = 10/120⁰ E E E The positive sequence component of the load current is (A) 1.3102-107⁰ A (B) 0.3322-120° A (C) 0.9962-120º A (D) 3.5104-81⁰ A +) + j1.002 0000 j1.002 0000 j1.0Ω vovo j1.0Ω 0000 j2.002 voor j3.002 oooo
A three-phase system has balanced conditions so that the per-phase circuit representation can be used as shown in Figure 1 Select the turns ratio of the step-up and step-down transformers that the system operates with an efficiency greater than 99 percent. Moreover, find the complex power (received or given) of all components in the circuit and the V1 and V2 voltages. The load voltage is specified as 4 kV rms, and the load impedance is 4/3 ohm. ..
In the figure, bus 1 is the reference bus, V1 = 1.06 p.u and the angle is 0.. A load with active power= 100 MW and reactive power = 50 Mvar is connected to bus 2. The line impedance is z12 = 0.12 + j0.16 p.u on a base of 100 MVA. Start with an initial estimate of |V2|0) = 1.0 p.u and 820) = 0•. Find |V2| () (one iteration). 100 MW 212 = 0.12 + j0.16 ++ 50 Mvar Vị = 1.0/0° О а. |V2| (1) = 0.4 p.u. Ob. |V2| (1) = 0.5 p.u. Oc. |V2| (1) = 0.7 p.u. d. (1) = 0.8 p.u. е. |V2| (1) = 0.6 p.u.
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20) Two AC generators A and B are in Parallel. Each generator has an emf of 1000V per phase and in phase, the phase impedance are ZA = 0.1+j2 ohms, ZB = 0.2+j3.2 ohms respectively. If the common load has an impedance of (2+j1) ohms per phase, solve the voltage per phase at the load
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