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Light-bulb and capacitance

  1. Jan 26, 2008 #1

    A90

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    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=([tex]\Delta\phi[/tex])/(I)
    C=(Q)/([tex]\Delta\phi[/tex])

    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. jcsd
  3. Jan 26, 2008 #2

    hage567

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    Homework Helper

    You have Ohm's Law. You can use that to find the current (and therefore charge).
     
  4. Jan 26, 2008 #3

    A90

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    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
     
  5. Jan 26, 2008 #4

    hage567

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    OK, for a discharging capacitor

    [tex]I(t) = I_o e^{\frac{-t}{RC}} [/tex]

    Do you recognize this equation?
     
  6. Jan 26, 2008 #5

    A90

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    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).
     
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