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 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? Hi

Answer 1

Q. Give Re CKT When At t=0 the capacitor starts Switch is closed the Charging through Resistor R= 10% consider at time + Rolor ww CKT has ist and ht 2 (+) cunent change on capacitor is 214) Apply KVL at loop (+) switch z Vi i -V+PHIR + v=1OVOH 온 R i(t) = < (v-ų) z prtl=de d+ de dt (CV-2) А. de dt RC integrate Both sides CV-q dg + dt = os too copacitor is Cu-q Uncharged so q=0 2(t-o)=0 2 in ev--2) t-o RC Inlev-q ev- -t/PC ev-q e 2 (H = ev (1-2 )

time t , 21+) = aline) Chooge in capacitor at with time constant T-RC OCYOSS so The cunent through CKT at time t Plt) - de -The dt RC i(t) = ve -the A The voltage the capacitor YEM Oct) V 11) ver-ete) AI time constart a=RC L'T = 1 Ofer 10x10 The capacitance of capacitor is Caloo 6 To beome fully charge Q(H = 99% pax Q1+1 = 0.99 Qnox OS -9 C=2 = 108 109 041 - Chart Cr-ette) othe = Inloo 0.99 Onor = Opor (l- -th = 1-0.99-0-01 0.99 I-e the the 100 e ine = Inloo 흗 = 4.605 so t=4.62 t=4.6) x10 SPC The time to charge 991. is 7=846.0 sec (( Charge at t= 5 sec -5/10 2(+-5) evinné) 46 sec = cv loé e 8/0)

Amox Qmax 7 noximum charge on capacitors -7 CV 108x10 10 Colomb 104 11-02) = 0.8646 X10 = 8.65X10 at t=5sec is 8.65x10 Coorb Olt = 55) -8 The charge on copaci tor at time t ift) = 0.5X10 PA (0) ter -te V -te е 10 => 0.5X108 .8the 10 The 109 -the ette 0.5=e 0.5 the Both sides 2=0 take lograthmic Elna loln2 = 6.93 sec + at in 2 so =t T vo-eko) The time taken by CKT t = 6.93 sec to become the cament 0.5X108 A. (€) The voltage 050 cis t = 5 sec V(+-5°C) 10 lloj) = 8.65 Volt The voltage is 8.65 volt. Le V(+)? The energy stored in C ot time t is U- at t=5666 U-¢ ¢{U[t=5))" (8.65)- 37.38 x103 stored in capacitor of time t=5sec is U = 37. 38 x16° J 2 8 X 10 8 The energy