# 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).