The figure (Figure 1) shows a simplified circuit for a photographic flash unit.

Question

The figure (Figure 1) shows a simplified circuit for a photographic flash unit. This circuit consists of a 9.0-V battery, a 50.0- k Ohm resistor, a 140 - mu F 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. How long does it take to charge the capacitor to 5.0V ? Express your answer using two significant figures.

Explanation / Answer

V = V0(1-e^-t/RC)

=> 5 = 9*(1-e^-t/(50000*140*10^-6))

=>t= -ln(1-(5/9))*(50000*140*10^-6) = 5.6765115135143016 sec

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