Question

by the line Two balanced Y-connected loads, one drawing 10 kW at 0.8 power factor lagging and the other 15 kW at 0.9 power factor leading, are connected in parallel and supplied by a balanced threc-phase Y-connected, 480-V source. (a) Determine the sourcec current. (b) If the load neutrals are connected to the source neutral by a zero-ohm neutral wire through an ammeter, what will the ammeter read? 2.45 Three identical impedances Z=30/30 are connected in A to a balanced thrce- phase 208-V source by three identical line conductors with impedance Z (0.8+/0.6) per line. (a) Calculate the line-to-line voltage at the load terminals (b) Repeat part (a) when a A-connected capacitor bank with reactance (j60) 2 per phase is connected in parallel with the load. 2.46 Two three-phase generators supply a three-phase load through separate three-phase lines. The load absorbs 30 kW at 0.8 power factor lagging. The line impedance is (14/16) per phase between generator GI and the load, and (0.8 + jl) Q per phase between generator G2 and the load. If generator GI supplics 15 kW at 0.8 power factor lagging, with a terminal voltage of 460 V line-to-line, determine (a) the voltage at the load terminals, (b) the voltage at the terminals of generator G2, and (c) the real and reactive power supplied by generator CG2. Assume balanced 2.47 operation

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Answer 1

For the Question 2.45

a) Source current = 30.08 A with 0.9999 pf lagging

b) Ammeter reading connected in neutral wire is = 0 A

© 2-45 By Given information about circuit Gannected in parallel loads are - - - - - - 1,Ibr // - sc saat Load 1 Loada as given loads are balanced so load I has impedance zi in each phase and load a has impedance Zg in each phase. let line to line voltage Vy= Vanlol = V₂ = kort a Nova I Neelal line carrents & Ia, Ib and Ic for Y load Iline a I phase so for load a line current I Lie I phase, Ian - for load a line current Ila - Iphase = Ian Active power in a circuit P = BU, IL Cosa for 3-6)

lire voltage of source = 480 line voltage of loads = 4804 c Parallel connection] Load I active lower Piz loko at of power factor lagging P2 a 15 kew at ang power factor leading led angle of voltage zo * Ve=48020 v For load, Jale Ia, L. where cost=0.8 = Jan Lbi [liven] for leada Ilga Jan L462 where cosba=0.9 - (Given ] in load It Cave sign because heating ] . lagging current ) Pia Bll. I la Costi 10,000 = B (489) () (0.8) 10000 Iaia Ta = 15.04A J3X480 X0.8 Jan = 15.04 36.87º A r dia Cos (0.8)] Pa = B Vez Ile corda 15000 = 3(480) [192] (0.9) 5000 Jag = (53) (480) 20.93 20.05 A Ia, = 20.05 125.840 A [Qa= Cog' (0.47]

Now apply RCL at node m. Ja = Jar tāds = 15.04 1-36.87° +20.05 125.84° Ila Ia - 30.08 (-0.540A So source carrent= I= 30.08 A C at 0.9999 lagging (by Now connect load neturals to source netural- 16 Jan Y IN C 12 apply KCl at Node o "INIAI Az = IN - let me dared current of load 1 and load a care I me and I ne respectively, RCL at Ne hart I bet Icha Ina

As given in question all the loads are balanced and for a balanced 3-6 network theme taral all the phare currents will adds upto zero so laut I be tal caso = INI Jan + ide + Ice to = IN2 IN : Initt N2 = 0+0= O A so Ammeter reading = oA