Question

HE 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 0.5X10-(A)? Switch E- What is voltage across the capacitor at the 5th second V = 10 (V) of Switch being closed? F- What is the energy stored in capacitor at the 5th second of Switch being closed? [ Hi

please do all part a,b,c,d,e,and f asap

Answer 1

Page No. 1 R = 109(2) C=?(F) Switch V=10 (V) Given: Time constant (Y) = 103 We know that, In Re circuit we have y = 70 с toler los = 108 F R 109r Hence, capacitance of capacitor is 158 farad (f) A - 108 F Now, time required to fully charge capacitar (about 99%) is given by t =by t= 5x time constant ).

Page No.2 - ) 5x 103 508 c+ are required to fully charge Hence, 50 s the capacitor B - 503 Now, the switch is closed, We know that Qlt) =) olt) Q (1-et/R) C.v (1-étly) = 10*5.40v (1-é?) 16*c (1-2) 3.93 x 10 8 C c c Hence, charge at 5th second of switch closed is 3,93 x1080 beling C - 3.93 x 10°C

Page No.3 Also, I (+) (e-tlpc) R polo 0.5X10% Х n 10 n 109r .t los = 0,5 -t In (o.s) U los =) -0,693 -t los => t 6.935 time Hence, required is 6.93s for become Current In the circuit to OS X 108 (A) D-6.933

Page No.4 when We have QL+) 3.93 X10-8c t=5 Hence, Vet) = Q It) 3.93x1080 108 F =) Vit) = 3.93 V hobe There fane voltage across capacitor 5th second is 3.93 v at E - 3.93 Il Now, Energy stored in Capacitat (0) Ź CV 2 2 -> U X 10 x (3.93) ll 2 8 U 7.72 x 10 Joule Therefore, energy stored in capacitor at 5th second second is F - 7.7281085 7.72 x108J