Question

Tom 6. Refer to figure 27-71 for this problem. Given the following information: Ep = 240 Vac, Np = 650 turns, Es1 = 460 V, Z1 = 800 12, Es2 = 208 V, Z2 = 48 , Es3 = 12.8V and Z3 = 5.2.2, find the following: currents in all winding (primary and secondaries), number turns in each secondary winding. ☺

Answer 1

Basics This a passive electrical device that converts electrical energy from one voltage to another voltage level. In a transformer, the primary power meeting is equal to secondary power rating as energy is conserved. If we step-up the voltage, the current gets step-down automatically to maintain constant power in: both primary & Secondary side of transforme. Let Up: supply voltage to primary side eide, Ip is primary side current, tp = primary power, Np = primary both pro J A T Iiiii 1 1 1 Let Us, Is, P. & Ns be secondary side voltage, current , power & turns respectively.

IP NP : مقنننناا WE > D Zs - Secondary side load impedance. - Secondary following relations hold true for ideal transforme - TO Ve = np i - Vs 0 No Ip = Ns 1 - © Is Np Let Zp = primary side reflected impedance. Zp I Np Zg I No These formulces listed above holds true for multiple secondary transformer.

Let us solve the ☺ Considering problem now. primary & secondary (1) of transforme -1 240 - 650 Ngu - 1245.833 - X 0.575 => Ns = 1 245,833 Turns. 1 2 6 Tarot , Iso = 0.875A Ipi= Reflected current in primary due to Iss in secondary @ PENS 5 Ip, = 12451833 x 0.575 Isi Np 650 Ip, = 1.10214 ® consider priserny & secundarzy (2) of Free formen fp = np => 240 - 650 Esar Nis2 208 Nisa =) Ns 2 = 563.33 Tums. 2 563 NS2 Turns.

a 208 = Iso = Esz CZ2 4.33 A 18 Ip = Reflected curent due to Isa in secondary Ip2 Msa Isa Mpy Jp = 4.33 X 3.755 5 6 3.33 650 A © Considor primary & secondary (3) of Transformer N2 =) 240 *650 Esz C 12.8 NS2 a) Ns3 = 34.67 turms 35 Tume Ist zz - ts3 Zz - 2.46 A 12.8 5.2

Ip acument in primary clue to Iss in secondan IP3 = Noa - Asz Isz Mp MP 34.67 62.46 7 73 650 650 -) IP2 = 0.1312 A Now The Potal primary cument is given by Sum of reflected cuments due to secondary ТР- Ігр +1, +Тоз 021+3.755 to 1312 = 4.9883 A B2-8002 oor Hence, Ip= 4.9883A. 1 & Esi= 4600 { Isi = 0.575A Nsi=1246 & Es2=208 Mp=650 & Isz = 4.33 N2 = 563 Fp240V Ip=4.9883 A o u82=22 EESZ 12.8 E Isg =2.46 E NS2=35 ♡ 23- 5-202