Question

Problem 2 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 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?

Answer 1

Problem-2 Vo R = 109 (2) C= P(f) HH AI Switch. V= 1010) Time constant = 10(s) (A) capacitance of the capacitor - Twine constant for RC circuit = RC RC = 10 (given) 109x6 =10 C = 10X 10-9 C = 108€f C = lox 10-9 (C= lonel Destration of voltage assess capacitor vasiatim cieth time Apply KVL in the circuit -V+ IR+ 2 = 0 or V = Vet VR

We know IR = Ic = c dve dt So, Via Vc t RC duc dt. V = Vct Rc d Vc at (VR= IR) (Vi = v) V-Vc = Ro dve → dvc SV-Vo dt. dt RC Ve - ln (V-Ve) = t it O RC 0 luf V - Vc) -t RC V - VG V etRc 2- t/RC I 1-ve Vo = V(1 - et RC This shows the variation et Voltage assess capacitor with time.

I Cave at Ic = c d f V (1-e-HRc)] dt CV e ARC Utc) V e-t RC Variation of cussent flowing through the Ic X R We know Ic = dac dt SIodt t V R Se are at o t А. V etRC To -cv [e-the-l] (Qc = cv [ 1 - e-tke Variation of charge flowing through capacitor with time the

(6) Tinie for the capacitos to the fully charged. (9%) Usung eh Vc = V(l-e-HRc) for fully charge Vc = V (voltage assess capacitor = supect voltage) - etre = 1 e t/RC =0 at to This is possible So let us see for capacitor to bel9% charged time Vc = 0.99V Put in 0.99 V = V(l- & TRG) 0.99=1-e-t/RC e-t/RC = ooL lo o This gives it u 5RC So, at time to GRC), the capacitor is fully charged

(C) Charge at the 5th second of switch being closed. Using equation (C) le = CV 1- e-tire] Time constant RC = 10 R = lot 2 t = 5 C = 108 F and v=lov. ple = 109x10[1-6-510] 10-7 [1-4:12] 10-4 (1-0,6065) Qc = 0.39 35x107c Qc = 3.94x108c- Charge at t = 5th sec. (8) Time for current to become 0.5X108 A using egh I= Ve HRC R 0,5X 10-8 = 10 e-t/10 109 0.5=et/10

ln (0.5) = -t 10 0.6931 10 t = 6.931 seconds of switch habe se) Voltage at 5th second Msing egn (4) closed. being Ve - Vil-e-th) Vc = 10 [1-6 5110] Nes 10 (1-0.6065) as t=5 RC=10 V = 10 Vo lox 0.3935 I Vc = 3.935 Volts If Energy stored at 5th second. Energy Love? Ix 10-8 X (3.935) 7.742x 10-8 I Vc= 3,935 V C = 108 F 2 energy