Question

In the druit of the figure below, the switch has been open for a long time. It is then suddenly closed. Take 8 - 10.0 V. R - 40.0 KN, A) - 195 kn, and C=145 w w R (a) Determine the time constant before the switch is closed 3 4075 5 (1) Determine the time constant after the switch is closed. (c) Let the switch be closed to. Determine the current in the switch as a function of time. (Assume is in A and is in s. Do not enter units in your expression. Use the following as necessary. t.) 5.128E-5 Need Help? Additional Materials book

Answer 1

tirse constant Ofte switch is closed is 2.82755 0. E- lov, Ris 40k 6, R2- 19565, (= 14.5'uef a Befpuie switch is closed concent is docus ces, RI + E C C R2 Time constant, t= t. Req Reg. c . To find Real cohere Rea is thevinin's sexistance across c we replace call sources by their internal resistances and find equivalent across co R de Req M R2 Req= Re R2 6 After switch is closed Ri j T= (Ritra) C =4 t- (195+40) * 10 10²0 14.5*10-6S T. 235x14.5*10-35 IT 3.40755 :. time constant before switch is closed is 3.40755 cercent is drawn as to determine t tz C: Req. € R2 to find Req, we shost embie and find Req across cu Rea = R2 Cap R is shorted ) ..T= Rzxc (195*10*x 14.54106) s : 2.87755 E Rea WRr T-

Befrore switch is Josed ca t=0, the capautes la serata connected for a long time and hence t gets feally charged and be comes open ciiceited capacitor is open, worent in loop, IOA &ori =Y E- Vero-) ISOR Vc 10-) – OR2=0 y Uc (0-) = Jou Rr When Seeltch is closed col t=0; capacitet doepoot allow sudden change of voltage, RI - Capacito's voltage at of, Vcrot) - VC10-) 2. Uccotl=1ou At tro, circuit is drawn as capacito is connected across R2, estisse energy be dissipated across R2, ICHI Is Il u stored will Welt) E R2 is given as, D) Uc (8) ou thence capacitor's voltage VciH) = Velosse [vero)- Ucico)] ett 0+(10-0)e-tit where t= 2.8275s (determined in & Velto lo e -tie Capaciteer's worient, IcH4) Colucit) -C.d. [doe-ti] & lo entit RC - to re-tyt x 10 Jc449= 51.28e-68715) HA dt 3A 195 In closed loop .

E = I Ri a4 I = 10 x 103 A 40 I, 250 x 10-6 A | 1. - 250 hen applying KCl ad node It Ict)= Is ::I5-(250 - 51.28 e-t58225 Then Is-(250+51-28 e-Oº353 +) 40 : woent through switch as a function of time is Is- (250+51.28e-0.363t) x 10-6A Gr Is - (250 + 51.280 28245 ) x 10-6 A