Question

Question: A three-phase line has an impedance of 1 + j2 Ω per phase. The line feeds two balanced three-phase loads that are connected in parallel. The first load is Y-connected and has an impedance of 20+j40 Ω per phase. The second load is ∆-connected and has an impedance of 30-j60 Ω per phase. The line-to-line voltage at the load end of the line is 415V. Taking the phase voltage Va as reference, determine:

a) The total current per phase from the source end (4 marks)

b) The magnitude of the line voltage at the source end of the line (2 marks)

c) The total 3-phase real and reactive powers absorbed by each load (4 marks)

d) The total 3-phase real and reactive power losses in the line (3 marks) e) The real power and reactive power supplied at the source end of the line (3 marks) f) If the second load is now connected in Y-connection with the same impedance of 30-j60 Ω per phase, what would be the current per phase supplied by the source? (4 marks)

Answer 1

3 V3 1tj2 Zline vs. Zload-1 = 20+140 = 10-71163.43°A. Vspha vzpnt. Iph (Zline) = 239.0 L0+(864(337) (1+j2) Given Zline= (1+j2) elph load- Zy= 20+140 load-2 Za= 30-360 ZA = 10-320 Zy Cewivalent) Vell-2)= 4150 Velph) - 4150 - 239.6609 Iph Ve= 239.610 to I Zload 2 10-320 I =239.6 5-35463.4'*. 201540 Iph = It Iz Iz = 239-6 5-35663.4+10.714634 10-520 2pn = 8.64|33.71 Yource Current = 8.6444 = 237.97 14.62. Vs(L-1) = V3X237.97=412-18V © for load-1 P= 1721.9w =3239-6)( 5235163-4) = (1721.9+j 3438.54) 9= 3438.54 VAR (lagsing) for load 2 $= 36V 5, ) 3( 239-6)(10.71 (-6343) = 3449.34 -36885-3 p=3443-4W Q= -6885-3VAR Q-0885.3 VAR Creading) @ line losses lloss = 3 Ipn?R = 3(8-64)X1 Qlass= 3 (Ipnt) xXline 3 (8.643² x 2 VA = 0.224 kW = 22 yow -0.447 KVAR. = 447VAR. e Vseph)- 237.97 24-62 3s(ph) = 8.64133.7. (s) = 3( vpn) (Jpnt)= 312379724-62)(864) (L-337) = 5-39-2.997 kw Ps= 5-39ky = 2.997 kvarlleadin) 0 = -2.997 KVAR CSS Scanned with CamScanner