Question

1SL 1. (30%). A single-phase source is applied to a two terminal passive circuit with equivalent impedance Z = 12/6: measured from the terminals. The reactive power delivered by the source is Q=-32/v2 Mvar and the power factor is p.f.=1/V2 leading, the source current is i(t) = 4v2 cos (120-t) kA. (a) Determine the complex power and the apparent power delivered by the source (b) Draw the power triangle. A (e) Determine the load impedance Z = 2/6, bel/bz 20

Answer 1

Leading p.f. means , current leads voltage=>capacitive load.

Solution - Given. i (4)=452 los (1205t) kA 11121282 1. Reactive power , Q = - 32 MVAR = - 3 2 2 x 10° VAR delivered by source 2..Power factor = (leading) 0-; <0 = current leads voltage a capacitive hood. 3. Source Current ilt) = 413 cos (1202 t) KA - 413 cos ( 1200 t)x10 A Since. S. Apparent Power Q = + Reartive power) _ (positive diretion) P.(Real Power) O For Logging Power Factor

® For leading power factor. P.(Real Power) To Sfabparent 10--ve (Reactive power) negative direet ion) Power) w Power factor = bef: = Cose = cos(@v=0;) Cost = 1 h O= cost ) = 45° tan4s = 121 Power Triangle 245 18=32 XROVAR 45° P 1 = 374 X105 Tnegative direction) p=*Exit W] (Real power) Appavant posers/p2 +Q - term worth of the mogle

- 5 = x10 s = 32 xlOG VA Apparant power, IS = 32 MVA Complex power, = P +jQ = 2 xid®+j ( a vos) va = (-i )ok va complex power = tz (1-j) mra

© Since, 름시에 에서 | A - - 3는 VAR I , B 4s Power S = 32MVA Triangle P름에 No=4s 1 185-3름 MAR S = 320VA 45 (1207t)kA | = 4VE xio³ lo (120nt) - 46 xrio' ( bkotr forn ) .. 1 - 4x10 , 9 = 0 -

P = f'm Im CoolOrO;) Cos or the leading) 32 x 100 = 12 in 1452 X103) cost-45) { Cosforoi) /h 8,-0;=-45 lean na gamine (o (453xi8) e nel 32 x 16 = 2x100 m fost 468): Cass=/h) 32x10" 8/2 x103 VLX2X13 .: 21 in proi 8/2 Xpot 2-450 452 X103 2=22-45 7=28=45