Question

Exercise 26.51 5 of Part A In the cat shown in the figure out -5.90 E 28.0 V, and he was recently the capacitor is uncharged and the weich Sie in position. The swich then moved to position 2, so that the begin och be the charge on the actor a long time whethewiichis moved to position 27 C Submi Best - Part 6 Alter the switch has been a ponton 2 for 200 me, the charge on the capacitor la pressured to be 110 pc. What is the value of the mentace RT VON R. gure Switch S Switch S in position 1 in position 2 Part How long after the switch is moved to position will the charge on the capacitor bequal to 100% of the final al und in part! WA ? R Submit Request Ans

Answer 1

E=28 v c=5.90 MF I R Part A long time capacitor will to the e. mif After a have a voltage equal of the eincuit. Charge voltage & Capacitance So, 8 vx e -6 0 28 x 5.90x10 Q = 1.652 x 10-4 C Part B the relation between charge and Circuit is we know time of charging in the CS Scanned with CamScanner

9 = $(1-é/RC) 11ouc 92 so, at t = 3x103, 3x103 So, 110x10 6 - 1.652x104 (i e RC 3x10 10/Rc 7 0.666 - 1 - - e 3x10 o = 0.334 -3 - 3x10 -) RC 1) en (0.334) Ra 3x 10-3 en (0-334) 5.9816-6 Ra 463.67 2 Answer CS Scanned with CamScanner

of o. 1/60) So, Q (1 Consider that theat at time. 463-67 x 5.9x10 6 In lo Pante let us ts. Charge will be 907 e 908 100 t enool. RC t +/Rc 0.1 e RC enlo + - 3 t 6-299 x 10 S CS Scanned with CamScanner