Question

For the electrical circuit below, let R1=30 Ohms, R2=15 Ohms, R3=10 Ohms, R4=20 Ohms, C=0.1 F and Vin=10V and complete the following.

For the electrical circuit below, let R1 = 30 12, R2 = 152, R3 = 102, R1 = 20 2, C =0.1 F and Vin = 10V and complete the following. a) Use circuit analysis to derive the differential equation for the capacitor voltage, V.(t). b) Find the solution for V«(t) and sketch it for t=0, t and . c) What is the time constant, t, for this circuit? How long does it take V (t) to reach its final value? d) Find the capacitor current, ic(t) and sketch it. e) Find the designated output voltage, Vout (t), between terminals a and b, then sketch it. f) How long does it take Vout (t) to reach its final value? R1 R3 a + ic(t) R2 Vin R4 Vout + Vc(t) с b

Answer 1

Given, R,= 3on Ra=15n Rz=lon c=o.lt Vin lov R4 =2on. Circuit Rz-lon 0 R,=3an R2 = 150 + 3 Rudon vout =lov recht Apply Nodal analysis, •VCt) - 10 + ic (t) vct) -0 =0 30 + 30 ic (t) = c dve (t) i CCL VCt-vect c duc Ct) Vct-vect) dt dvect) to dt 15 VCt vect) 153 - 3 dvect)

act) 3 dvect) -+ avect vCt) - 10 30 tie Ct) + vet) 30 vct tic Ct 3 cdy(t) dt (c=01F) te dvect) اليا I dvc (t) = 1 dt =) VCA dt z dvect) tave 15 2 7 dorect) dt 15 1 +1 ve (t) + I dvect) 3 PO 30 a dve (t) + vect) 3 15 .ct) 3 I dvc (t) t vel 5 at 15 dvect) +5 yect) 5 at dvect) at thvelt = 5 J differential equation of capacitor vo Hage.

Toansto tor & Apply Apply lap ducat? - Sve(s) 5 35 VCCs) (for Resocuits duelt) + 1 vect) = 5 - Sve(s) + 1 ve(s) = 1/2 ve(s) €5+1 35(S+ / St - Apply innedse Laplace toans foom, ve(t) = 5 [i-ett @ Time constant for given circuit, Shoot ciscuit --> vo Hoge source, 315 zon - 15 0.1 s 3on lor ->) 30x30 602 T = Reac 1503 315 Col Boxol T=3=time, constant

we got Vc Ct) L = 3.16 3 [i- ést at to, vect) = 5(1-1)=0 tar, vect) = 5(1-03 time constant = 5 [l-el too, Vect) = 5 [1-6] = 5(1- = 5(1 - 1 =5[1-o rect) 5. 3.16 t-T (t=59)?t

[=0.If Capacitor Curdent, ic Ct)= c dvect) dt we got welt) = 5 (-est ic Ct) = 1 x5 dci-e-Ist] 10 at It 11 2 iCt)= icct)

- 10 - loe & t. C = -56-1 ezt time taken to deach its final value is, (v (t)) T= time constant = 5x3 = 3sec 15 sec from circuit, voy(t) = 20 Ct) 30 out = 2x15 3 dvcGt) tave 3 a dt ducct) dt ECA) = 5 [i-e dt uct) vCt) are Ct) ext] } =>d (3-52-54) +23[1-e ] +10 - 10 e lot 3 - to a - 10-25 e-/st 33 Coult) = 10-25 est 3

At t=0, vout (t) = 10-25 At t-T, vout (A) - le 50 10-5e' .6.9343 At tno vout Ct) Esteady state rout(t) value) 6.343 con اد T-3 (t=5T P time taken to reach its final valee of Crout (t) is 57 =) 5x3 #15sec