Concept explainers
A balanced three-phase load is connected to a
(a) Line currents for phases A, B, and C.
(b) Line-to-neutral voltages for all three phases at the load.
(c) Apparent. active, and reactive power dissipated per phase, and for all three phases in the load.
(d) Active power losses per phase and for all three phases in the phase conductors.
Trending nowThis is a popular solution!
Chapter 2 Solutions
Power System Analysis and Design (MindTap Course List)
- The three-phase source line-to-neutral voltages are given by Ean=100,Ebh=10+240, and Ecn=10240volts. Is the source balanced? (a) Yes (b) Noarrow_forwardA three-phase line, which has an impedance of (2+j4) per phase, feeds two balanced three-phase loads that are connected in parallel. One of the loads is Y-connected with an impedance of (30+j40) per phase, and the other is -connected with an impedance of (60j45) per phase. The line is energized at the sending end from a 60-Hz, three-phase, balanced voltage source of 1203V (rms. line-to-line). Determine (a) the current, real power. and reactive power delivered by the sending-end source: (b) the line-to-line voltage at the load: (C) the current per phase in each load: and (d) the total three-phase real and reactive powers absorbed by each load and by the line. Check that the total three- phase complex power delivered by the source equals the total three-phase power absorbed by the line and loads.arrow_forwardFigure 2.33 gives the general -Y transformation. (a) Show that the general transformation reduces to that given in Figure 2.16 for a balanced three-phase load. (b) Determine the impedances of the equivalent Y for the following impedances: ZAB=j10,ZBC=j20, and ZCA=j25. ZAB=ZAZB+ZBAC+ZCZAZCZA=ZABZCAZAB+ZBC+ZCAZBC=ZAZB+ZBAC+ZCZAZAZB=ZABZBCZAB+ZBC+ZCAZCA=ZAZB+ZBAC+ZCZAZBZA=ZCAZBCZAB+ZBC+ZCAarrow_forward
- In a balanced three-phase Y-connected system with a positive-sequence source, the line-to-line voltages are 3 times the line-to-neutral voltages and lend by 30. (a) True (b) Falsearrow_forwardA three-phase line with an impedance of (0.2+j1.0)/ phase feeds three balanced three-phase loads connected in parallel. Load 1: Absorbs a total of 150 kW and 120 kvar. Load 2: Delta connected with an impedance of (150j48)/phase. Load 3: 120 kVA at 0.6 PF leading. If the line-to-neutral voltage at the load end of the line is 2000 v (rms), determine the magnitude of the line-to-line voltage at the source end of the line.arrow_forwardLet a series RLC network be connected to a source voltage V, drawing a current I. (a) In terms of the load impedance Z=ZZ, find expressions for P and Q, from complex power considerations. (b) Express p(t) in terms of P and Q, by choosing i(t)=2Icost. (c) For the case of Z=R+jL+1/jC, interpret the result of part (b) in terms of P,QL, and Qc. In particular, if 2LC=1, when the inductive and capacitive reactances cancel, comment on what happens.arrow_forward
- Two balanced three-phase loads that are connected in parallel are fed by a three-phase line having a series impedance of (0.4j2.7) per phase. One of the loads absorbs 560 kVA at 0.707 power factor lagging, and the other 132 kW at unity power factor. The line-to-line voltage at the load end of the line is 2203V. Compute (a) the line-to-line voltage at the source end of the line. (b) the total real and reactive power losses in the three-phase line, and (c) the total three-phase real and reactive power supplied at the sending end of the line. Check that the total three-phase complex power delivered by the source equals the total three-phase comp lex power absorbed by the line and loads.arrow_forwardFor a balanced- load supplied by a balanced positive-sequence source. the line currents into the load are 3 times the -Ioad currents and lag by 30. (a) True (b) Falsearrow_forwardA single-phase source is applied to a two-terminal, passive circuit with equivalent impedance Z=3.045, measured from the terminals. The source current is i(t)=22cos(t)kA. Determine the (a) instantaneous power, (b) real power, (c) reactive power delivered by the source, and (d) source power factor.arrow_forward
- It is stated that (i) balanced three-phase circuits can be solved in per unit on a per-phase basis after converting - load impedances to equivalent Y impedances. (ii) Base values can be selected either on a per-phase basis or on a three-phase basis. (a) Both statements are true. (b) Neither is true. (c) Only one of the above is true.arrow_forwardFigure 3.32 shows the oneline diagram of a three-phase power system. By selecting a common base of 100 MVA and 22 kV on the generator side, draw an impedance diagram showing all impedances including the load impedance in per-unit. The data are given a follows: G:90MVA22kVx=0.18perunitT1:50MVA22/220kVx=0.10perunitT2:40MVA220/11kVx=0.06perunitT3:40MVA22/110kVx=0.064perunitT4:40MVA110/11kVx=0.08perunitM:66.5MVA10.45kVx=0.185perunit Lines I and 2 have series reactances of 48.4 and 65.43, respectively. At bus 4, the three-phase load absorbs 57 MVA at 10.45 kV and 0.6 power factor lagging.arrow_forwardConsider a three-phase Y-connected source feeding a balanced- load. The phasor sum of the line currents as well as the neutral current are always zero. (a) True (b) Falsearrow_forward
- Power System Analysis and Design (MindTap Course ...Electrical EngineeringISBN:9781305632134Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. SarmaPublisher:Cengage Learning