Question

The RC charging circuit in a camera flash unit has a voltage source 275 V and a capacitance of 125 F.

(a) Find its resistance R if the capacitor charges up to 90.0% of it's final value in 15.0 s.

(b)Find the average current delivered to the flash bulb if the capacitor discharges 90.0% of its full charge in 1.00 ms

Answer 1

Q = Qmax *( 1- e^-t/RC )

Q = 0.90* Qmax , so

0.90* Qmax = Qmax *( 1- e^-t/RC )

0.90 = 1- e^-t/RC  

e^-t/RC = 1-0.90 = 0.10

-t/RC = ln(0.1)

***check the data is C = 125 F or 125 μF , I took 125 F as in question ***

R = -t/ ( ln(0.1) *C) = -15 / [ ln(0.1) *125]

R =  0.0521153 ohm answer

(b)

I = ΔQ/Δt = (Qf- Qi ) / Δt =

full charge = C*V = 125*275 = 34375 C

90 % = 0.90*34375 =30937.5

i = 30937.5/ (1*10^-3 ) =3.09375e7 A

i thnik your data may be wrong plz let me know in that case