Question

1) A 220 V/60 Hz three-phase source supplies two loads. The first load is A-connected presenting P = 5 kW and Q=-3 kVAr. The second is Y-connected presenting a resistance of 8 Q and an inductance of 16 mH. Compute: a. The magnitude of the current passing in each phase/element of load 1 (A-connected.) b. The (three-phase) active power consumed by the second load (Y-connected). c. The (three-phase) apparent power supplied by the source. (15 marks)

Please show steps to solving.

Answer 1

0 Given 308 Source is arou/Getty » a 220v. frequency of bottz Voltage First : 8-Comeeted load y p=5kw Q=-3 kVAr seconds To connected load Grasr Lolem -@) from the data of do Conected lood, S=SP2+02 Jawi 63k]2 25.620950 KVA - For a connected lood, line voltage is equal to phore voltage as youpha 220v . and I = 3x Iph - s;= 3x Uphtspn 55.63095x10^= 3 x 220 x Spins of Don = 5.83095 x103 ; 8.8347A ph 3x220 2. Magnitude of current passing on each phane / element of lood I. (A-connected] Pas fp = 8.82.00

06) Y- Comested load: Raar 22 16mko JewL . Impedance due to inductance a ) 1991) La jxQtTxbox 16m Rt IXL =;6.031552 .: Srrechчсе рот рал. , ,,= 220v = 8+$ 6.03185) in Y- Comection 220 Ph Ve=3x von Tgn = Upb bis 22000 . 22.00 -- 1699-37.0152° A 2/ph ft 36.03265 & active power (3-0) confermed by the second load for connected Pa 3 Yoh Iph Coso = 3x 220 x 12.677. Cos (0-632.0152) 5 3.857 kW Orvar ex - Ps= 3.857&w ] © Since P2= 3.87 kW Cosp= 0.79844 Qe = P2 Tcunos = a bo e^6. 7 2.900 kVAr. qands B PE 702 B, Tamp

Apparent power of loadel (&-connected → 5,= P +10= 643) KVA Apparent power of Lada 2 Cer connected 7822 Patha 28.857 +12.908) KVA & 3-6 Apparent power supplied by the Source , sasitse → S = 6-3) + (3.87+52.905) KVA ... 18.887.10.092] KVA is = [6-654–30.092] kva