- #1

Lucille

- 31

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## Homework Statement

Consider the following circuit in a camera flash where a light bulb is in series with a charged capacitor:

a) Derive, but do not solve, the governing differential equation for the circuit. (Hint: treat the resistor and the light bulb as a single resistance.)

b) Assume Rlight = 5 ohms If at time t = 0 the switch is closed, the solution to the differential equation from part a) states that a current will develop in the circuit which will light the bulb according to: i(t) = Q/RC*e^(-t/RC) where Q is the initial charge present on the capacitor. We want the capacitor to discharge quickly to produce a short flash for photography. If we want i(t) to fall to half its maximum value in 0.01 s and the capacitor has a value of 160 muF, what should R be?

c) Calculate the value of the ratio of the instantaneous power dissipated in the resistor to the instantaneous power dissipated in the light bulb, i.e. Powerresistor/Powerlight.**Image can be found at http://www.chegg.com/homework-help/questions-and-answers/consider-following-circuit-camera-flash-light-bulb-series-charged-capacitor-derive-solve-g-q3553019

## Homework Equations

v=iR

v=1/C integral idt

P=i^2*R

## The Attempt at a Solution

For a)

I got 0 = R*di/dt + i(t)/C

I found an expression for total voltage and differentiated it w.r.t. time

b) I plugged in numbers and got that R = 85.2 ohms (but this seems large?)

so:

1/2Q/RC=Q/RC*e^(-t/RC)

1/2 = e^(-t/RC) -- R=Rt-Rl = 85.2 ohms

for c)

Presistor = i^2*Rr

Plight = i^2*Rl

Dividing, i^2 cancel so: Rr/Rl = 17.04?