Problem 3 Inside a typical flash unit for a disposable camera, a capacitor is us

Question

Problem 3 Inside a typical flash unit for a disposable camera, a capacitor is used to store the charge that lights the flash bulb. The capacitor is charge by hooking it up to a 350 volt emf produced by a small trans former in the camera. The resistance ofthis part ofthe circuit is 30000 and the capacitor is 200pF. These may be treated as an RC circuit. (20 points) a) Once the capacitor reaches a voltage of 300V, it is ready to flash. How long does it take the capacitor to reach this voltage? b) When the capacitor is at 300V, what is the energy stored in the capacitor. If it then discharges in lms, what is the po wer dissipated by the flash?

Explanation / Answer

given data

Vmax = 350 volts

R = 30,000 ohms

C = 200 micro F

Time constant of the ckt, T = R*C

= 30000*200*10^-6

= 6 s

a) while charging a cpacitor, voltage across capcitor at time t after closing the switch,

V = Vmax*(1 - e^(-t/T))

let at time "t" Vltage across capacitor = 300 volts

Apply, V = Vmax*(1 - e^(-t/T))

300 = 350*(1 - e^(-t/T))

300/350 = 1 - e^(-t/T)

e^(-t/T) = 1 - 300/350

e^(-t/T) = 0.142857

t = -T*ln(0.142857)

= -6*ln(0.142857)

= 11.67 s <<<<<<<<<<<-------------Answer

b) Energy stored in the capacitor, U = 0.5*C*V^2

= 0.5*200*10^-6*300^2

= 9 J <<<<<<<<<<<-------------Answer

power dissipated by the flash = Energy dissipated/tike taken

= 9/(2*10^-3)

= 9000 Watts <<<<<<<<<<<-------------Answer

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