Question

Problem 2 mA In the given RC circuit, a capacitor is connected to a resistor in series and is getting charged after closing the Switch. The time constant of the circuit is 10 (S). R = 109 (0) C =? (F) A- Calculate the capacitance of the capacitor. B- How much time does it take for the capacitor to become fully (about 99%) charged? C- If we close the Switch and the capacitor starts getting charged, what is the charge at the 5th second of Switch being closed? D- How much time does it take after closing the Switch for the current in the circuit become 0.5X108 (A)? Switch V = 10 (V) E- What is voltage across the capacitor at the 5th second of Switch being closed? F. What is the energy stored in capacitor at the 5th second of Switch being closed? -H

Answer 1

Date R-100) C= ?(F) - V-10(0) @ Given: Time Constant (%) - 108 We know that In Re Circuit we have Re car 108 10F Switch R 1012 Hence, capacitance of capacilos ie 10€ Fauad () [A-1085 Now, time required to felly charge capacitou (about 99%) is given by t= 55 (t. 5x time couslout] t = 5x los - 50s + lênce, 50s auve srequired to fully charge the capacitor B-50s Now, the switch is closed, We know that Olt) = 6(1-0 . 108€ lov (1-20) 10 tel1-0 0.5) 3.93×1086 Hence, change at 5th second of switch being closed és 3.93%20-80 Tc-3.93 x 1080 I Gt) * 0.5x108 A = lov ERC) Qt) & C.V (lee * te k/RC) 102

Dato 1/10 0.5 th ln (05) it 6.935 Hence time required és 6.938 for curent in the circuit to become 0:5*108 (A) 10 -6.938) We have alt) = 3.93 x 1080 when t=5 Hence, Vo = OCV) - 3.93x1080 Tölf (4) = 3.93V Therefore vollage across copacilor at som second és 3.93U TE-3-930 allow Energy cloud.co Capcilo 10 = 1/2 CU2.. U1*108* (3-43) U. 4.42x10 8 Joule Thereforo, erogy sloud is capaci loc of sth second es 7.92x1085 TF_7.72*1085 2