Question

Q4: The switch in the has been open a long time before closing at t=0. Find iz(t) for t>0. 2.5 ΚΩ IL 15V 1 362.5H 2.5 UF 9 ma

Answer 1

Solution:

In the beginning of the problem we took the conditions for t= 0- steady state and calculated all the initial current and voltage.

Then at t > 0 we applied the Laplace transform and finally calculated the current in the inductor in mA.

Thank you..

1. 
   44. At is t= Switch in steady state is open cirinit t=0 open 2.5kn ДАА Velo) ) ISV IMA illo IN the steady state Inductol Capacitor e short & open carinit (ILO) circuit Velol 2. Гk M ILIO short Circuit V-ov + Velo) + IV Illo) above circuit Velöl=0 volt from the 0-15-2.5K X FLlo)=0 - 2.5k IL 10 = 15
   
2. 
   IL (5) = - 15 2.5k ILlol=-6mA In steady state at t= Illo l = -6mA Velo o volt At toot ON to, o I to switch is closed 205kn M Velo) = =o volt GMA 62.5H f Illot-tma eoma 2. Slut Velol=0volt short short circuit Ilo) = -6mA T 2.5 L110) = 62.5% 6.375 gma 2.5 ut 62.5H 15V
   
3. 
   205 kr +0.375 P IMA SX25M S Ela 62.5 xs Figure: Taking the Laplace transform of the Previow circuit diagraly 118) 118) 2.5 KN o mu II I IV 1 Il + 160377 gma axtos P +1 S 45 62.58 KCL at node v gm ILkt) + Iz to, S gm 5 TiH) + VIS) + Visztis Is S. unior 2.5k S for simplicity V18) =V
   
4. 
   дw) v-0.375 t SVXOS t S SV + 15 2. ГКXP 62.58 u 9m $ 0.016 V-6 Mp4 - 3 0.258 V X10 (0= US V+6) X10 S + Р -) -5 -} 0.016 V-6X10 +0.258 V X10 + 0.us vtt) X10 9m S S gms v(0.016 +0.258*X105 + 0.us X163 ) + 6810² 6810 -] - 3 gm = S v [ 258°x10 ++ us xo't 16 x 10 98153 = 25X107/ p2 + 0.16 Xio's + 0.6uxio "]v 9x104 ve s+ 1008 + 6100) 25 3600 = V[8² + 160stonoo 3600 = V(8+)? as
   
5. 
   36.00 V($) Volt 2 IL18) V(8) - 0,375 62.53 Illsl = VC) 0.37 62.53 A 62.58 Ilis? Ix 3600 1 62.58 (8 +80)2 0.375 62.55 Tc13) - 57.6 818+8012 ISE 0.31 62.58 und Partial fraction method of st part 576 B + A S + 818+802 18+80) (3480) A18+80) + B 318+80)+cxs 57.6 2 s (st80) 3 1540)
   
6. 
7. 
   A = 0.009 A=IM 05 160A +8OB to Equating 8 coefficient on both 0 A+B to - A B = -0.009 is 0.009 O 160 A +80B + C 0 1601 (+0.009) +801 (-0.009) +C 0 = 6.72 +1 c = 0-0.72 (s 0.72 (= -72om A = gm gm Ca-720m
   
8. 
   On substituting in ILIS) ILIS = Im_gmt gm +1-720)M 3+80 (st sol 6 m S ? IL(S) = 3 9 120 MA S S180 ist for Note i - ㅗ -at ILT e ult) sta -at ILT te ult) Sta)? Inveye la place la place transform of Ills sot lot 3 U1t) 9e UTH -720 te Ulu ma ! Taking Allt = -sot - sot 3-ge - 720 te Pilt uith MA