Question

On an RC circuit

4.- En un circuito RC donde C=1 200 F, R=500 12, E(t) = 2 si 0 <t < 5, -2 si 5 <t < 10, 0 si 10 <t las unidades de E(t) son Voltios. Encuentre la carga en Colulombs en el Condensador en el tiempo t=15 segundos. Tome q(O)=0. (20 puntos).

Find the charge in Colulombs on the Capacitor at time t = 15 seconds. Take q (0) = 0.

Answer 1

Given in that question Re cirmir and R = 50012 f 200 and When o etxs EH): when -25tt 10 O when tout charge H4) and 40=0 11 alt=1s Stepat applying KVL in this cirmir R 614) = Rilts + 2 sidt + 2 Elt) 2 с NOW 114) da of 1) Where, Now, م charge puting rawe of it) in equation ☺ and R de 2 + 1 - 9= get Elt or, de dit + 200 500 Plt) 500 or, de dt + EN) Re 9 Eles OY, d9 dt +0.49a NOW. applying capace transform Laplace transform in equation w 30, 3 $415) 40) + 0.4 445) [ 12. -SH e 500 att J-rest dt $ SH Hoe 10 •Now constant laplace transforma -St ae dt

Now. given 940) 0 5-st 16 So, 2103 [s00-1] = 30 е dt s ! -st dh 그 $ w * 90(0047 * [*] - no [a] saputo *** 2013 (104) - Sa Latest 1 t FIOS 2855 OY, 45) 11 250 s (5+0.4) OS 07. 915) 250 2.5 Sto.4 2.5 s 2.5 Ston e ston NOW. taking invera cap cau lapcau trans form of both side - M+) = 1 250 fos - 2 2.5€ 04t) + 0-417-10)1 2.5- 2.se U (ATO) 10) Juct ult-s) -09(+-5) 2-5- 2.5e Now, calculating charge in cololombs on the Capacitor at t= 15 sec. H16) = to [2.s-xsēʻx 248–250 *= fs+252 9 |

or, 9115) = 2.5 -0.006 0.006 - 0.34 + 0.05 250 | 35 - 1990 ori 1415) = 0.008799 e .: ch charge in columbs on the capacitor at t is see is 0.008799 Columbs.