Part A The figure (Figure 1) shows a simplified circuit for a photographic flash

Question

Part A 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-k12 resistor, a 140-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 6.0 V? Express your answer using two significant figures Submit My Answers Give Up Figure 1 9.0 V 50.0 k 140F

Explanation / Answer

During charging of a capacitor, the voltage equation across a capacitor is given by:

V = Vo*[1 - e^-(t/RC)] 6/9

= [1 - e^-(t/RC)] 3/9

= e^-(t/RC) ln (1/3)

= -t/RC t = RC* ln 3

= 50*10^3*140*10^-6 * ln 3

= 7.7 s answer

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