Light-bulb and capacitance

1. Jan 26, 2008

A90

1. The problem statement, all variables and given/known data
Imagine that you have a light-bulb that has a resistance of about 10 Ω
and that can tolerate a maximum voltage of 3 V. Imagine that you want to connect this to a charged capacitor large enough to keep the bulb glowing reasonably brightly for more than 10s. Roughly what should the capacitor's capacitance be?

2. Relevant equations
I=(dQ)/(dt)
R=($$\Delta\phi$$)/(I)
C=(Q)/($$\Delta\phi$$)

3. The attempt at a solution
It looks like I need to know something about the charge to solve this, or I need to cancel it out, but I can't seem to make any headway. How can I solve for the capacitance without Q? Am I missing something obvious?

2. Jan 26, 2008

hage567

You have Ohm's Law. You can use that to find the current (and therefore charge).

3. Jan 26, 2008

A90

Is that assuming dQ=Q and dt=10? I get:
I=(Δϕ)/R=dQ/dt
then, dQ=I*dt=(Δϕ*dt)/(R*Δϕ)=dt/R=1

The answer is supposed to be 1.4 F

4. Jan 26, 2008

hage567

OK, for a discharging capacitor

$$I(t) = I_o e^{\frac{-t}{RC}}$$

Do you recognize this equation?

5. Jan 26, 2008

A90

We have learned it with electric potential, rather than electric current, but I can see how one implies the other. With the data given I found I_0 (sorry, subscript keeps coming out as superscript- that is 'I naught') to be .3, but I'm not sure about I(t).