Question

CO TI, t=0 2H 4 1) 2A 12

Determine the value of the current in the inductor for the time = 200ms

Answer 1

9.18 o circuit at t=ó steedy state inductor act as a short circuit and it will not allow suddon change in Gissent in it. through . IL (ot) just ILCO-) = Il co) q BA 5 q Prior At instant After the oporsting sie oportet pirot t=0 6-2 then Il co-) Inductor 402 2A $1212 shoof'ct 0 chron't a to 6. Wh t=0 Ilco) 41112 Porsellel Combination 41112 12A -32 4812 1274 4812 16 3-20 4 from Gosand division sale 6 3X2 3 610 % ILCO) 3 X2 g 6+3 Il Co-) / Amp.

20 un o Giderit @ t=0 and and too MA to L260) 342 V ZA & 2* tige do circuit @ to and too of * At to switch is open and semain open till too Honce 2A Active source is disconnected A from "H" inductor the arront Howing through this time. t=0 * 7 t=(oo) IL COO) = 0 Inductor has two type of Response initial and final and ac to thoses Response formula inductor is through Ict) = (coo) + [icot) -ico)] ette inductor during according an of total Cursent :. Ta t=4Rea We have value. Time onstant value Lp Reqsec 4/Reg see. Req= Equivalent Resistance seen from the inductor's terminals. 62 wut = 102 4-- Req= L = 2H Req

B Time Gnstant c= -gec Y Reg 2% てっ 0.2 sec. 0.2 sec from standard formula of inductor Gosent ict) [ I.co) * I.com] the e Amp. Il(t) = Icroo) + [Inco) IL(0) = 2 Amp . ILCoo) o Amp. 0.6667 Amp T = 6.2 gec e -4.2 IL (+) = 0+ [8.6667-0] Ilct) = et/012 0.6667 e Amp. # ILCt) Current @ t= 200ms put t= 0.2 in equation Z= 0.2 sec. IL (0-2) = 0.6667 e 0.2 e 0.2 = 0.667 é? IL (0.2) O. 24 52 Amp. Ans.