For Figure 2: The 3-phase load is connected to a balanced 2300V, abc source. In doing this, 3 conductors are used with the conductors having an impedance of Z₁ = 0.5 + j2 ohms each. After connection, the load receives 120 kVA at 0.6 power factor lagging. Determine the line currents. (Hint: Assume the load is Y-connected and use single-line representation.)

For Figure 2: The 3-phase load is connected to a balanced 2300V, abc source. In doing this, 3 conductors are used with the conductors having an impedance of Z₁ = 0.5 + j2 ohms each. After connection, the load receives 120 kVA at 0.6 power factor lagging. Determine the line currents. (Hint: Assume the load is Y-connected and use single-line representation.)

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

See similar textbooks

Similar questions

Give your background about the AC systems from single phase and three phase systems to their connections including their currents and voltages laws with the suitable diagrams.
In an installation there is a three-phase load of 300 kVA with a power factor of 0.8 behind, in parallel with a three-phase load of 150 kVA with a power factor of 0.6 ahead. A pure reactive element is connected in parallel, and a unity power factor is obtained at the power input. Indicate the reactant of the element, with a positive sign if it is a bank of reactors, or with a negative sign if it is a bank of capacitors. Just put the corresponding sign followed by the numerical value, without units. The numerical value must correspond to the three-phase kVARS. You can use integer values with no decimal places or up to two decimal places
The Figure shows a balanced Y-A 3Ø circuit. The phase voltages of the Y-connected source are V, = 11020°, V, = 1102 - 120° and V. = 1102120°, Volts. The line impdedances are Z,=10+j5 Q each. The impedances of the A-connected load are ZA = 75+j255 N each. Determine: Line Currents, Phase Currents (load), Power drawn by the load. Ve ZA Vb
Shown are two buses connected via a transmission line. Connected to Bus 1 is a generator(modelled as a voltage source V1), and to Bus 2 is a generator( with Voltage V2) and a load. The line impedance Z = R + jX is not known. Three-phase load is forecasted as 100MW + 60MVar. Generator at Bus 2 is scheduled to deliver 40MW and its excitation is adjusted for the machine to deliver 55 MVar. Voltage at Bus 1 is 241 KV angle 10 deg, while voltage at bus is 230 kV; use V2 as the reference for all angles. 1. What is the current (RMS) in amperes drawn by the load connected to Bus 2? 2. How much real power (in MW) is delivered by the generator connected to Bus 1? 3. What is the total real power loss along the line, in percent of the load? 4. What is the X/R Ratio of the line impedance?
In a three-phase system the per-unit phase voltages are 100° , Vôn = 0.8Z180° Van 0.920° , Von = 0.92 - а What are the per-unit symmetrical components V0) and V)? Express in polar notation.
(c) In the Y-A system shown in Figure Q2(c), the source is a positive sequence with VRN = 24020° V and phase impedance, Zp = (2 – j3) N. Determine; VRN R Zp VYN Y N Zp ZP VBN B Figure Q2(c) (i) All line voltages, VRY, VYB, and VBr-. (ii) All line currents, IR, Iy and Ig. (iii) Total real power absorbs by the loads.
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. ..
Using a test equipment, Vmax was found to be 5V while Vmin is 1V. The distance between voltage maximas is 14 cm and the first maximum was found to be 7 cm from the load. What is the load impedance if the characteristic impedance of the line is 75 ohms?
Overhead lines.
1. Given the circuit below, the phase voltages are given by Van = 20040° Vrms. The source impedance is 5 + j7 per phase and the wire impedance is 8 +j10 per phase. Let the load impedance be Zy = 50 ohms. Find: Van Vin Ven a. II, and Ic b. c. VAN, VBN, and VCN, and VAB, VBC, and VCA Z₂ Z₂ Z₂ Zw Zw Zw B C le N
Q1. For the system shown in Figure Q1, draw an impedance diagram in per unit, by choosing 100 kVA to be the base kVA and 2400 V as the base voltage for the generators. 10 KVA 2400 V G₁ Z=j0.2pu 20 KVA 2400 V Z=j0.2pu G₂ T₁ Transmission Line Z= (50+j200) 40 KVA 2400/9600 V Z=j0.1pu Fig Q1 T₂ 80 KVA 10/5 kV Z=j0.1pu M 25 KVA 4 kV