Question

1) For the circuit below, a) Find i(t) fort > 0 b) Find vu(t) fort > 0 [you can do this by using your answer to part a) and the relationship between voltage across an inductor and current through an inductor] c) Plot your answers to parts a) and b) on separate plots. I lists to V. (4) 2H 2) For the circuit below, if the capacitor is fully discharged for t < 0, a) Find i(t) fort 0 you can do this by solving part b) first and then using the relationship between voltage across a capacitor and current through a capacitor] b) Find ve(t) fort > 0 c) Plot your answers to parts a) and b) on separate plots. 20k. I ilt) to Vechi - 25 uf 3) For the circuit below, a) Find i(t) fort > 0 b) Find velt) fort > 0 c) Plot your answers to parts a) and b) on separate plots. w 24v 2004+ 36kr

Answer 1

This question is based on the network element properties.

ict) dt Switch is closed at to at tot, inductor acts like open circuit in ict) = 0 A at t ot @ For t o Applying Kirchoff's voltage law in the circuit in 6 = 10 it a di ww 1on15 di & 5i - 3 64 T s5.dto 5t e Integrating faclor; If - es. = e Multiply the above equation with "est" on both sides, 5t &st. di +50 st i = 3 e at d esti) = z best Integrating on bolksteles; est i = 3. est at + Integration I constant I 5 esti - zest + c. at t=0; i=0 in. 0-3 + c = c = -2 iese - 365t - 3 ( Substituting C value in egn. O = ilt)- (-e-5t) An, txo - 0.601-25t) A 6 voltage across, inductor; q (t) = r. d o yle) - da g e5t) - 2:1-3)(-5) 655 6e viteze © Pets Caries .. - 6v 1 3 L- (Amp (t) (0.6)-3 → (sec) t (see) )

20k tot tot (OV - (a ) capacitor; ict) = cdo 20k A (25 pes de + = 20 k. 2 At initial stage ie at t=0t; capacitor acts like short circuit ie icel = 0 = 0.5 mA Vct) = ob @ At to switch is closed. Apply Kul dan soko ww 10 = (20k) it I lidt () (254) vlt) since this Equation is having an integral term , and voltage is the estate variable for Ve-10) wy Vc - 10 S for simplificath purpose, = are Stat Substitute valves at later stage) Integrating factor; J.F. - eSkodt etc Multiply with If on both sides, due. e RC cument ict) = care RC at e Dutegrateng on both sides, T RO A (sa jetre ve = Se ke ta p etke e Rover et ke Itci T = 0.5 e 2t man v I substitute a RC LIRC e 2 des tutit ARC RC id i lt) on notoR at t=0 ; Ne = 0 ma) . 0 = RC +Glost Ost SaS . c ekonto. GE-AVN ontre av ( 0) Vects = u(1-ethe) Vo, txo | velt = 10 vu, R c = (20 k) (25 M) = 0.5 sean lov Vect) = 10 (1 - e 2 a to t (sec) O.

For te 12KR bt Vc Š 6 kot = Y For t o the circuit is. Assuming that the Circuit is in steady stale 24 at t= ě, where capacitor acts like open circuit. By voltage division rule: co-) = 24.6K = 8 Vu (6K+12 K) Since capacitor doesn't allow sudden changes in voltages, Y co- – Veco)= Vecot) = 8 V. For to switch position has changed. - Tilt L. Here c= 200 MF :: 12kn vnct) & 6kN & - ER R = (12k || 6k) (t) l (12K) (GK) (12kt6k) Apply KCL ; - 4 kr c. alve & C = 0 20044 Now; (t) = c. du = cq (8 e RC) į (t) = 4.8.1 e RC to volts) a = -2 e RC mani = -2 é %0.8 mA to 8V * characteristic equation; - D+ c = 0 D= RC :. The solution to above differential Equation is; VCA = Ke RC at t=0; (0) = 8 = Kie = 80k=8 Vect) = 8 e tee for tzo R.C = (4K) (200M) = 0.8 see VCt) = 8 e ons Un tyo. tcsec) it mo t (sec)

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