the figure shows a simplified circuit for a photograph flash unit. this circuit

Question

the figure shows a simplified circuit for a photograph flash unit. this circuit consists of a 9.0-V battery, a 50.0-kOhm, a 140-microF capacitor, a flashbulb, and two switches. Initially, the capacitor is uncharged and the two switches are open. To charge the unit, switch S1 is closed; to fire the flash, switch S2 (which is connected to the camera's shutter) is closed.

http://session.masteringphysics.com/problemAsset/1122589/2/Walker.21.66.jpg

How long does it take to charge the capacitor to 4.0V?

Explanation / Answer

v = V(1-exp(-t/RC)) ==> 4 = 9(1-exp(-t/7) ==> 1-4/9 = exp(-t/7) ==>ln(9/5) = t/7==> t = 7*ln(9/5) = 4.114 s

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