4000 = Blue phase current = 50 A p. f. = 0.7 lagging Calculate the current in the neutral wire. Take V RN as reference, 1 Y.

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1 EEE405 EECC EGEEG CES - EXECSE -phae, 4-wire, 80, 50Hz pwer upply with the phae equence i cnnected t a balanced tar lad. Each phae f lad cnit f a capacitr f 0µF in erie with a 00Ω reitr. Determine: (a the phae vltage; (b the impedance at each phae; and (c the line current. 80 (a 9.4 (b 06. Ω 6 (50(0x0 00 j Ω 9.4 (c w-wattmeter methd i ued t meaure the ttal active pwer taken by a -phae mtr. he line vltage i 80 and the tw wattmeter reading are 5.5kW and.5kw repectively. Determine: (a the ttal active pwer; (b the pwer factr; and (c the line current. 5.5 (.5 (a tal active pwer (.5 4kW (b tan φ { }. 075; φ (.5 pwer factr cφ c ( (c c φ, c φ (80(0.4 -phae 4-wire 80, 50Hz upply feed pwer t an unbalanced tar lad ha the fllwing meaured data at each phae. ed phae current 45 p. f. 0.8 lagging ellw phae current 5 p. f. 0.9 lagging lue phae current 50 p. f. 0.7 lagging Calculate the current in the neutral wire. ake a reference, ; ; (0 (c ( 0 (c (0 (c ( n unbalanced three-phae delta lad i cnnected t a 45,, -phae upply a hwn in fig. 4. Calculate the ED line current j0 Ω 0 + j Ω 5 j5 Ω Fig Current thrugh, j Current thrugh, j5 ( phae mtr f 80% efficiency i cnnected t a 80, 50Hz -phae upply a hwn in fig.. f the mtr winding i delta-cnnected and each phae ha a reitance f 8Ω and an inductance f 9mH, determine: (i the current m in the mtr winding and the current fed frm the -phae upply; (ii the utput pwer f the mtr; and (iii the pwer factr f the mtr. 5 (i 80, X πf π (50 (9x Ω. + jx 8 + j5.97 Ω Ω Fr the delta-cnnected mtr, --

2 EEE405 EECC EGEEG CES - EXECSE ( the current in mtring winding 8. the current frm -phae upply (ii cθ where θ 6.7 (80(65.99 c kW the utput pwer x efficiency (4.8 x 0.8 kw 7.86 kw (iii wer factr c θ c hree identical cil cnnected in tar are nw cnnected in parallel with the mtr t the upply a in part (a. he ttal pwer drawn by the cil i fund t be 4kW and the verall pwer factr 80, 50 Hz becme 0.9. Calculate: (i the pwer factr f -phae upply tar-cnnected lad; (ii the line current taken by the tar-cnnected lad; (iii the reitance and inductance f the cil. 6 fig. (i ttal active pwer + ( kw 76.8 kw ttal reactive pwer Q Q + Q where Q tanθ 4.8 x tan kar Since pwer factr f the circuit 0.9, θ c Q (Q + Q ( Q tan θ tan Q 0.9 Q (0.48( kar 0.9 kr Q tan θ pwer factr f the tar cnnected lad c [tan - 0.6] c (ii cθ cθ (80(0.97 (iii Fr a tar-cnnected lad, 80. Ω c θ. c4.57.ω X in θ. in Ω 7 a tar-cnnected unbalanced -phae lad i fed by a -phae 4-wire ytem. Explain what will happen if the neutral wire i being pen-circuited. b -phae, -wire, 80, equence pwer ytem i cnnected t an unbalanced tar lad a hwn in fig.. ake a reference and by uing the illman therem methd, determine: (i the diplacement neutral vltage O ; (ii the phae vltage at the tar lad; (iii the three upply line current; and (iv the ttal active pwer delivered t the lad. 7 X mh πf 4.6 O 0 + j8ω fig. m tr 0 + j5ω O 5 j0ω. When the neutral wire i dicnnected, the tar-pint vltage f the three-phae lad will n lnger be zer.. ptential will appear at the tar pint. t value will depend n the actual unbalanced lading cnditin. hu different value f phae vltage will be reulted at each phae f --

3 EEE405 EECC EGEEG CES - EXECSE the three- phae lad.. Over-vltage r under-vltage in either ne r tw phae may appear. (i ake a reference, the phae vltage phar are: y illman herm O j j j (0 0( (0 0( (0 0( O ( j (ii ( j9.48 O O ( 4. j99.04 O O O + O ( 4. j8 (iii O ( ( O ( ( O ( ( (iv tal active pwer, (7.55 (0 + (4.8 (0 + (5.8 (5 W.5 kw phae, 4-wire, 50Hz pwer upply f phae equence i Cil cnnected t a balanced tar lad cnit r 5Ω; 0.H f three cil a hwn in fig.. Each cil ha an internal reitance f 5Ω and an inductance f 0.H. Determine: (a the upply phae vltage; (b the pwer factr f the lad; and (c the line current fig. 80 a 9.4 b X π(50(0.. 4 Ω ; p.f c ( he pwer input t a k, 50Hz, -phae delta-cnnected mtr i meaured by tw-wattmeter methd. f the tw wattmeter reading are 50kW and 00kW repectively, determine: (a the ttal input pwer; and (b the pwer factr. 9 a W 50 kw, W 00 kw, he ttal input pwer W +W 550kW [ (W W ] [ (50 00 ] 59.8 φ 5. b tan φ W + W p.f. c phae 4-wire 80, 50Hz upply feed pwer t an unbalanced tar lad ha the fllwing meaured data at each phae.

4 0 EEE405 EECC EGEEG CES - EXECSE ed phae current 6 p. f. 0.8 lagging ellw phae current 45 p. f. 0.9 lagging lue phae current 60 p. f leading, Find the current flwing in the neutral wire. ake a reference, i.e. 0 ; 0 ; 0 6 (0 (c ( 0 (c (0 + (c ( delta cnnected reitive lad i fed by a 80, -phae upply a hwn in fig. 5. he pwer delivered t each lad are : ad 0kW 80 ad 50kW ad 80kW Calculate the current flwing in the ed line. ake a reference, cφ cφ W W , 50 Hz, three-phae -wire pwer upply upplie pwer t tw balanced three-phae lad cnnected in parallel. hey have the fllwing data. ad : ighting lad f 0kW at 0.5 p.f. lagging ad : Cntinuu mtr lad f 40k at 0.7 p.f. lagging. Determine: (a the line current taken frm the three phae upply; (b the reading f the tw wattmeter if the ttal active pwer upplied t the tw lad i meaured by tw-wattmeter methd; and (c the value f capacitance per phae f a delta-cnnected capacitr bank required t imprve the verall pwer factr t ake a reference, Fr lad : c φ ; φ c - ( c φ (45(0.5 fig. 5 Fr lad : S ; φ c - ( S (45 tal line current frm the -phae upply + ( he magnitude f line current, 8.07 and the pwer angle f the lad, φ 54. herefre the reading hwn n the tw wattmeter are: W c( 0 + φ (45(8.07 c( W W c( 0 φ (45(8.07 c( W tal active pwer, 0 kw + (40(0.7 kw 58 kw wer angle φ befre crrectin 54., wer angle θ after crrectin, c

5 EEE405 EECC EGEEG CES - EXECSE Capacitive reactive pwer required fr crrectin i:.(tan φ tan θ (tan 54. tan8. ar ar eactive pwer rating fr each capacitr per phae, ar ar Fr delta cnnectin, vltage acr each capacitr i 45. the capacitance per phae i: QC C 78.0 µ F/ phae ω (45 (π( W 8. S Q Q a tar-cnnected unbalanced -phae lad Fig. fue i fed by a -phae 4-wire 80, equence pwer ytem a hwn in fig.. Determine the current in the neutral 80 wire. (Ue a reference link b Explain what will happen if the link O cnnected t the neutral wire i being pencircuited. fue c With the link pen-circuited, determine: (i the vltage O at the tar pint O; and fue (ii the three upply line current. a reference, the phae vltage phar are: j0 j j j0 Ω j Ω 6 + j8 Ω.0.6 When the neutral wire i dicnnected, the tar-pint vltage f the three-phae lad will n lnger be zer. Different value f phae vltage will be reulted at each phae f the threephae lad. + + y illman herem O , 5 + j j, 6 + j8 (9.4 0( (9.4 0( (9.4 0(0. 5. O (0.8 j O O (ii j0 O O j O O j8-5-

ELG3311: Tutorial 2 R C = 75 Ω X M = 20 Ω R T = Ω X T = Ω. Solution: The following equivalent circuit is referred to the primary.

ELG3311: Tutorial 2 R C = 75 Ω X M = 20 Ω R T = Ω X T = Ω. Solution: The following equivalent circuit is referred to the primary. ELG3311: Tutrial rblem -1: The secndary winding f a transfrmer has a terminal vltage f v s (t) 8.8 sin 377t. The turns rati f the transfrmer is 50:00 (a 0.5). f the secndary current f the transfrmer is