Calculating Capacitance for Optimal Light-Bulb Performance

  • Thread starter Thread starter EzequielSeattle
  • Start date Start date
  • Tags Tags
    Capacitance
AI Thread Summary
To keep a 10-ohm light bulb glowing brightly for over 10 seconds with a maximum voltage of 3 volts, the required capacitor capacitance can be estimated using the equation VC(t) = V0e(-t/RC). By assuming a final voltage of 1.5 volts after 10 seconds, calculations suggest a capacitance of approximately 1.4 farads. This value, while high, is now considered feasible due to the availability of large capacitors. The discussion emphasizes the need for defining acceptable voltage levels over time to determine capacitance accurately. Understanding the relationship between current, voltage, and capacitance is crucial for solving the problem effectively.
EzequielSeattle
Messages
25
Reaction score
0

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!
 
Physics news on Phys.org
EzequielSeattle said:

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}
 
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).
 
EzequielSeattle said:
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top