Question

pls answer 7&8

10 ks2 10 kg vo Problem 7) The switch in the circuit has been open for a long time. Find the time constant and the voltage on the capacitor as t oo and sketch the voltage vc(t) on the capacitor for t > 0. 10v 1 mA 20 k223 40 uF - Velt t = 012 2012 We Problem 8) The switch in the circuit has been closed for a long time before it opens at t = 0. Determine the current through the inductor i(t) for time t > 0, including the time constant 1, the initial current at t = 0 and the final current as t oo. 12 V 3300 310 mH L( Last update: 4/9/2020 747pm Page 2

Answer 1

Solution:

Problem (7)

solution solution ③ smce the circuit was open befose to TV (0) - Ovolt Alao since the capacitor volt age cannot Change in lontone ously = 00(07) = 0,0)= Ovolle --0 After switch is closed at tzo once again at tro, the circuit reaches Steady State Steady state > Capacitor open che The resultant Circuit is - lok v lox man lov @ Olma $10%, 2% (20) KCL at node v gives CRT at ito) (Steady State) or 110/tv zo 7-10 IMA tv 30k =0 TOK

10K+201 on v[ to 757-2 or r= 15 volts Now Buy veltage Division 076692 V204 - 15x3 - lovolto (2006)= lovelt) ---0 Now, the genezial equation of voltage across capacitor for txo is given by 104) = 360)+(00-(m) evolte -- T= ReqC Reg -= equivalent resis tince Alen bey Capacity after twitching lok lok * Reg= 20k//202 = lok T= Reg C = OK X 40MF= 0-4 seconds 7

Thes 77 = 0-4 seconds |-- ► Substituting all the values, eq ® geves r(t) = 10 + (0-10) 804 volts Juckr 10[1- e 25t5 volte poca) 100k --- 6.3210 - -.- t=220•4Sec

Problem (8)

Problem ® • Befose switching at tso ie at t=ó o 6<o), the cicuit B in Steady - State Steady state → Inductor is Whost ckt The resultant circuit i 122 122 I 2002 I humanung 12V @ 30ng (0) Req , ckt at (t=0) steady State pea In the above ckt Rega 24 +(30/20) = 362 Now I 12 - 11/12 = 1 ampere Now Pay cessent Division Rule 4(0)- 2,90 3 * 15 mmpere or iz 607= 0-2 Ampere 30720 2. Ampere

Vow, since the inductor current cannot change instantaneously C07)= < (= 0-2 ampere Ans Now aflex switching the 12 V sousce is dis compree- ted from the inductor I Ans + the current ik preobrnit in the inductor at t=0, decay and the energy c di ssepabad in the resistors. Hence at t70 / (%) = 0 Ampere/ Now, the general equation of current through an inductor for txo is given by {@= ï() +(46-469)e Memperes. -- © az Y Rea Reg-7 Equivalent resistance sllen by inductor after switching

- 12-2 2052 83or < Rear EE Reg= 30+30= 502 7 te Whey = lon1/50 = 0.2 msec IT = 0.2msec Substituting all the values, eq o gives {(A) = 0+(02-0) e Poemsec amperes = -5000t 0.2 e amperes 1 - 56004 (t) = 0.2e Amperes A it) 40. 2A 0.074 k OV (20.2msec