# Problem with Capacitance

## Homework Statement

Imagine that you have a light-bulb that has a resistance of about 10 ohms and that can tolerate a maximum voltage of 3 volts. Imagine that you want to connect this to a charged capacitor large enough to keep the bulb glowing reasonably brightly for more than 10 seconds. Roughly what should the capacitor's capacitance be?

## Homework Equations

VC(t) = V0e(-t/RC)

## The Attempt at a Solution

I feel as if I need another equation for this problem, because I don't have 4 variables to plug in! The starting potential of the capacitor should be 3 volts, right?

So

VC(t) = (3 volts)e(-10 seconds/(10 ohms)*C)

So now I have unknowns of C and VC, and I need to solve for C. Clearly I'm missing something, but I'm not sure what... Can somebody point me in the approximately right direction? Thank you!!

berkeman
Mentor

## Homework Statement

Imagine that you have a light-bulb that has a resistance of about 10 ohms and that can tolerate a maximum voltage of 3 volts. Imagine that you want to connect this to a charged capacitor large enough to keep the bulb glowing reasonably brightly for more than 10 seconds. Roughly what should the capacitor's capacitance be?

## Homework Equations

VC(t) = V0e(-t/RC)

## The Attempt at a Solution

I feel as if I need another equation for this problem, because I don't have 4 variables to plug in! The starting potential of the capacitor should be 3 volts, right?

So

VC(t) = (3 volts)e(-10 seconds/(10 ohms)*C)

So now I have unknowns of C and VC, and I need to solve for C. Clearly I'm missing something, but I'm not sure what... Can somebody point me in the approximately right direction? Thank you!!

Good start. :-)

I would probably start with the equation $$I = C \frac{dV}{dt}$$

mfb
Mentor
The light-bulb will get dimmer as the voltage goes down, but it should not get too dim. You can define what exactly that means by fixing Vc(10s) to some reasonable value.

So I can arbitrarily say that Vc(10 s) is, say, 1.5 volts? And then solve for C, which would give 1.4 farads. Isn't that considered a really really high capacitance?

Would dV/dt just be the derivative of the right side of my equation above? That is, dV/dt = -(V0/RC)*e-t/RC?

Then, since I = C*dV/dt,

I = -3 volts/10 ohms * e(10 s)/((10 ohms)*C)

I feel like that doesn't help, because then I just have another unknown (I).

mfb
Mentor
So I can arbitrarily say that Vc(10 s) is, say, 1.5 volts?
Right.
And then solve for C, which would give 1.4 farads. Isn't that considered a really really high capacitance?
Maybe 20 years ago, but now those are cheap elements available everywhere.