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.
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))
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 =
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
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