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Problem with Capacitance

  • #1

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!!
 

Answers and Replies

  • #2
berkeman
Mentor
56,875
6,850

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 [tex]I = C \frac{dV}{dt}[/tex]
 
  • #3
34,065
9,940
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.
 
  • #4
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?
 
  • #5
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).
 
  • #6
34,065
9,940
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.
 

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