Question

An electronic flash attachment for a camera produces the flash by using the energy stored in a 750 capacitor. If a 9V battery is used and the circuit has a 4000 ohm resistance, (a) find the time constant (how long does it take the flash to charge to 63% of its final value), (b) what initial current flows when the battery is connected? (c) how long does it take to charge the capacitor if I is constant?

Answer 1

PS connected in series with Consider at time t, cunent Q. Given capacitor C = 750MF resistor R = 4000 ohm and charge on and voltage v=g Volt Batery R mu 114) 2(+ CKT is rot) capacitor is 26+) KUL at loop {v, zo с Apply + > sum

using ist) = des = V de Å (v- el R RC 52 In (ev-2) -I ein -t/RC -V + i +R + 2(+) =0 C PHIR 아을 steve) at RC de (CV-2) de d+ at RC CV-9 2 t de dt integrate Both sides CV-q RC os initially cabocitor is unchanged t=0 (-o) In CV- - RC curo RC -t/RC Cu-2 e 24t) cv(1- CV (a) let at t=2 charge on C is 2 = 63 % of Q. 2- 2 0.63% So +Z/Rc O. $ e e so t =RC The time constant of CKT is (4000) ( 750) x10 = 3X10 Exlossee 3 sec The time constant Ps <=3x108/10 sec 3 sec cupent in CKT is Plt) de tt d7 RC ict) V -T/RC Charge at time t , 241-6 (1-ety 0) = to é e/Rc T=Rc てこ (b) The Alpc e R R ro at t=0 initial cunent V ift-o ) R . -iſt-o) 2.25X10 eunent is - 2.25 mA ҶООО = 2.25 mA The initial

(6) The ек has the constant canent which zero the only when 014) = Pret (l- entre) Q(+) = Pency e-560 ） capacitor is fully charged using after t=50 (i- O(+) = Qack (1-0) N. ( - 0.00 %) = 0.993 Q max So After t=52 the charge on capacitor is 99.9% of maximum charge anex=ev .. which mean ofter t=52 Capacita is fully charged +=57=5RC = 5 x 13x106)S 15 x 10' sec xrof t = 15 sec The time taken by capacitor is 15 x10*sec x100 so The time taken by capacitor is 15 sec 6