Question

04 (2) A rectangular pulses, E (t) is applied to the RC circuit as shown in Figure Q4(2). (1) Show that the circuit can be modelled as RC + V = (1) -4(-2) (ii) Calculate the response, or with C) = 0 ning Laplace transform. mark b) Consider the network circuit as shown in Figure Q4 (b) with initial current and their derivatives are zero fort = 0. 1) Show the loop current is can be formulated as dig dt +250i, = 20 (6 xaric) (ii) Compute the loop current ist) using Laplace transform 1-300 WW Figure Q4(a) 200 1-0100 OW 100 W 013 Figure Q4 (6)

Answer 1

Ect) can be written in terms of ult). 1) ELE) = uit)- ult-2) R=1062 + Elt) (=14F Vet) e. write kvt in hoop Ect) Ritu is - current through capacitor i=c du at dv dt :: Ect) = =RC + V

+V uct) - 417-2) = Rc du dt RC du +V uit)-417-23 dt 10 Laplace equivalent circuit 186 ES) + 10oss Ves) Es) 25 = 1/3 co-es2) VCS) - ES) x 10 s 108-10€ 106 E(S) s loostoof ی Nes) = 100 EU) 1601065) VCS) = EOS 11 clu Cice?) SI) ST)

-25 e VCS) S(SUI) (1+595 Vos) B D (+*+5) S S $+1) 1 $=4 $(5+1) Ito SED Best scatola (sty - 1 15=1 25 22(9) c = $225 Ot! Stres skstols.o olsa -26-1) e H-35 (145) s (Son Ś + (-) e (?) Sto is + + S+! St 2 vit) - e 1 St! sti 2 -t NCE) = e -t e ult) vet) Ncts ē' ult) (e-o) e ult) = (1-389-4) et ult)

V(t)=6.389 ē' ult) b ) at tão 20 4V O 1H V + > 1 in write kivh in hi 4 = 101, +10 (11-12) 4=2011-1012 - 2011 = 4+1012 i. 4+1012 20 i = 1 / + /- 2 12 write KUL in 22: 102-11) +2012V0

1(2-) +20 lg +V = 0 =0.1 dia V = L diz de dt 10i2-noii +2012+ ou die dt diz +30 2 -101, dt put oin on diz diz: +3873 -0 ( +' 12) = on dia +3012 -2-512 =0 dt +2512 diz dt diz dt 25 is - 2 = 2 diz +250 12 = 20 dt i) Laplace equivalent circuit TO 10 OS S ICS)

write kul in first loop :- 4 = (10+10) 1, () -101215) 놓 = 401, (5) --10 I205) 20 I, (S) *+101205) I, (S) IS) - 4 +18 S 26 2 15 - Firs) + I2CS) 55 2 write KvL in outer loop. - (10+20 ) 126) -101,15) +0.4512(S) = 0 30 12 (5) HISI25) = 10 Ii(5) (30+015) 1,5) =10 as +126) Flo 105 125) XS a (30+015) +599) 2 (30+0:15-5)1205) (95+045 ) 1,0) = I (S) s 25+0.15)

2 ७) sxon (5+25) XIO E(S) s(s+250) ICS) 20 S65+250) F215) h B + 5+250 A = 5 I (s) so &x20 $(57250 $=0 20 A = c+250 A= 25 B = (5+250) (s) 569) 15=-250 BE 20 - 2 25 -250 I2CS) 2 25 255 (5+2559) {a(t) = 2 2 5 Co-e250t) ult)