Missing part in the text of the problem:
"a flash unit for a camera has a capacitance of1200μF."
Solution:
In a RC circuit, the charge of the capacitor at time t follows the relationship:
[tex]Q(t) = Q_0 (1-e^{- \frac{t}{\tau} })[/tex]
where [tex]Q_0 [/tex] is the full charge, and [tex]\tau = RC[/tex] is the time constant of the circuit.
We can isolate [tex]\tau[/tex] from the previous equation:
[tex] \frac{Q(t)}{Q_0} = 1-e^{ \frac{t}{\tau} } [/tex]
[tex] \frac{t}{\tau} = -ln(1- \frac{Q(t)}{Q_0}) [/tex]
[tex]\tau = - \frac{t}{ln(1- \frac{Q(t)}{Q_0} )} [/tex]
We can now using the data of the problem. We know that after a time t=22.0s, the capacitor is at 90% of tis charge, therefore [tex] \frac{Q(t)}{Q_0} = 0.9[/tex]. So we find
[tex]\tau = - \frac{22}{ln(1-0.9)}=9.55 s [/tex]
And from this value we can find the value of the resistance R, since we know that [tex]\tau = RC[/tex]. Given [tex]C=1200 \mu F = 1200 \cdot 10^{-6} F[/tex], we have
[tex]R= \frac{\tau}{C}= \frac{9.55s}{1200 \cdot 10^{-6}F}=7958 \Omega = 7.96 k \Omega [/tex]
