Question

The circuit shown ammeters A₁ (left) and A₂ (right) mark 95.7 A and 30 A effective respectively, current Ī2 is 48° behind the generator voltage and current Ī1 is 15 ° ahead of Ī. If the generator (source) delivers a voltage of 380√2 sen (377t - 58°) volts 18.19 ° ahead of the current it supplies, determine:

a) The impedances Ż₁, Ż₂ and Ż₃ in modulus and angle

b) The sinusoidal current consumed by the impedance Ż₂     

                     

Z3|20.5692 + -1 za 98 I I I I 1 2 3

Answer 1

380 12 Sin ( 337 +- 589 Given Vg :: Vycoms) 380 (- 58° 377 > and vg is 18.19° ahead of current it Supplies. noe current Ammeter Ai ( 95-7 (-58°-18-190 = 95-7 <-76-19° and Ammeter Az current 30 A Із current I is 15 ahead of I (A, current) Toe I, 95.7 <- 76.19 +15° 95.7 <- 61.19° current is 489 behind the Va 380 <-58*- 480 Ip Z 2 380 <- 106 Zz 95.7 <-76-1977 A 23 - Aj Zz < 20.56 ga 1 Vg = 380 <-58° J. Iz LI, 1 Az o Toopt 3000 łoop

() Apply KUL in Loop '1' Vg zi I, zi Vg Il 380 <- 58 95.7 7-76.19° 3.9.7 <18:19° Apply Kol at Node 'A' I+ I+ Iz H 95.7 (-61. 19° +380 (+106° 1380 - 2020 → 95.7 <-76-196 = 22 95.7.8-76-199 95.7 (-61.19* + 2010° 380 <- 1060 22 9.65 <- 109.83 380 <-106° z z Zz 380-106° 9.65 (-109.839 39.38 <3.839 Apply KVL in Loop 'z'. I Zt 30<09 [23 (20.56 + 9] = > (95.7 461.190)(3.37 (18-19) + 3070*39 + 3023 (20.56 = 0 379.93 <-43' + 270 00 + 30 (20-560 X 23 606.05 (-35.31 30 <20.56 (23) 606-65 (- ? 5. 31° 20.20 <- 4-759 <- P0.56° IN 23 30 (6) from equation ©; IQ- 380 <-106 39.38 <3.839 9.65 <-109.83° US