Question

The RC charging circuit in a camera flash unit has a voltage source of 305 V and a capacitance of 121 µF.

(a)Find its resistance R (in ohms) if the capacitor charges to 90.0% of its final value in 12.2 s.

(b)Find the average current (in A) delivered to the flash bulb if the capacitor discharges 90.0% of its full charge in 1.04 ms.

Answer 1

Given
RC circuit with voltage source V = 305 V

Resistor , R = ? , capacitor with capacitance C = 121*10^-6 F

Let the maximum charge of the capacitor is, Q

we have the charging of a capacitor relation is

q(t) = Q(1-e^(-t/T))

a)

given t = 12.2 s

q(t)= 90*Q/100 = 0.9*Q

and T is time constant of RC circuit, T = R*C

substituting the values

0.9*Q = Q(1-e^(-(12.2/(R*121*10^-6))))

solving for R, R = 43788.37 ohm

b) here we should calculate the maximum charge on the capacitor from the relation
   Q = C*V
   Q = 121*10^-6*305 C

   Q = 0.036905 C

now 90% of Q is , q = Q*90/100 = 0.036905*90/100 C = 0.0332145 C

time duration is Dt = 1.04*10^-3 s
q = I*Dt

I = q/Dt

I = 0.0332145/(1.04*10^-3) A

I = 31.94 A
the average current is I = 31.94 A