Question

Question 1 1 pts Consider the RC circuit below with a 9V emf battery. At first the switch is in the position shown, touching neither electrode, and the capacitor is discharged. The switch is then pushed to left position. 4 ΚΩ w 8k2 w 90022 E 33uF PWW 2 k2 What current flows immediately after the switch is closed to the left position in mA? Question 2 1 pts With the switch remaining in the left position, how much time does it take for the capacitor to reach 90% of its maximum charge ins? ("Maximum charge" means the charge the capacitor would have if the switch were left in the left position forever). Question 3 1 pts The switch is in the left position for a very long time, so that the capacitor reaches its maximum charge. It is then switched to the right position to discharge. How much faster did the capacitor lose 90% of its maximum charge relative to how long it took to build it up (in the previous question)? Answer as a ratio of time to discharge divided by the time to charge. The answer therefore has no units.

Answer 1

Question 1:- As per the theory of charging of capacitor Current at time t is given by Ilt) 0 E et/RC E R Here E = emf of battery, R= resistance of resistor C = Capacitance of Capacitor Given E = 9V, R = 4K 12 = 4x103.1 , C = 334F = 33x1067 t = 0 Put these values in o Core=1) E e/Rc E I(0) Tolm R 9 v 4 x1032 -3 2.25 x 10 A I (o)= 2:25 mA Ilo) =

Question (2) As per the theory of charging of capacitor charge of capacitor at time t is given by q = EC ['- et/RC] q=% [l-etik ]-©; Here qo= EC= maximum charge Given, q= 0.9% => 9/%=0.9, R=48R, C=33lic from @ ét/Rc et/Rc a ya => e = 1 -e slovo =) HRC - (290) *t=Rclu [7-(413] += (4x18 2) ( 33x10 f) lu o.g 10.304 s t-(0.1325) (2:30) series, their Question 3 :- 8K-2 and zkr are in resultant resistance - SKR+Zk2= 10k_2 This lokar is in parallel with goo 250gK2). Their resultant resistance will be 1 ㅗ ㅊ > R=0.826 kr loka • Кл. R ~ 826 2 R

this 826 r resistor. Capacitor will dish charge through As per the theory of dishcharging of capacitor, charge is given by at time t HRC = q étre qo q q=% e tIRC No += RC lu (wo) t=48262)(33x1 quo = 0.0029 5 gare) 33 x106f) lu Oig vo -3 = 9.54 x 10 0.0029 S 0:304 s time to dishcharge time to charge time to discharge (9.54x163 ) ( time to charges