# RC circuit in a camera flash

## 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/...ies-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???

ehild
Homework Helper
It is correct. That circuit is not efficient!

Thank you! Would the power ratio also be correct? I wasn't 100% that the i^2 would just cancel..

ehild
Homework Helper
It is correct. The same current flows both through the resistor and bulb, so their square is also the same, and cancel :)

Perfect! Thank you so much! :)