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Current homework problem

  1. Dec 3, 2009 #1
    1. The problem statement, all variables and given/known data
    A light bulb has a straight metal wire filament of 50 micro metres diameter and 10mm long and when powered with a 5V constant voltage supply it draws 0.5A of current. What happens to the current if the filament is replaced with one of three times the diameter?


    2. Relevant equations
    I=nAvQ ??



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 3, 2009 #2
    Re: Current

    Look at the formula for the resistance of a resistor in terms of its physical dimensions.
     
  4. Dec 3, 2009 #3
    Re: Current

    Thanks

    So
    R=l * rho /A
    where l is the length of the conductor
    rho is the electrical resistivity of the material
    A is the cross-sectional area of the conductor

    So V=IR
    I(1) = V/R(1) and I(2)=V/R(2) where I(1) is the inital current
    taking a ratio
    I(1)/I(2) = [V/R(1)]/[V/R(2)] = R(2)/R(1) = [l(2)*rho*A(1)] /[l(1) * rho* A(2)]
    rho is the same for each and l is the same for each so
    I(1)/I(2) = A(1)/A(2)
    = (pi * r(1) ^2) / (pi* r(2) ^2)
    =r(1)^2 / r(2) ^2 where r(1) and r(2) are the corresponding radii of the filaments
    =(25*10^-6)^2 / (75*10^-6)^2


    so I2=4.5A

    Is that right?
     
  5. Dec 5, 2009 #4
    Re: Current

    I agree with your answer.

    My own method would be:

    Step 1. begin by transcribing the phrase "the filament is replaced with one of three times the diameter" into the phrase "the filament is replaced with one with nine times the cross sectional area."

    Step 2. Nine times larger area means nine times smaller resistance.

    Step 3. Look at I=V/R. If we make the denominator of any fraction nine times smaller, the fraction become nine times larger.
     
    Last edited: Dec 5, 2009
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