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Homework Help: Change in power dissipation

  1. Oct 6, 2008 #1
    1. The problem statement, all variables and given/known data

    a wire of resistance r is drawn to double its length. assume a constant voltage and a fixed volume, by how much does the power dissipation change?

    2. Relevant equations


    3. The attempt at a solution
    so when i plug in 2L for L, it just ends up canceling anyway. the answer in the back of the book says it should decrease by 3/4, so i don't understand how that would happen
  2. jcsd
  3. Oct 6, 2008 #2
    How about the relation

    [tex] P = \frac{V^{2}}{R} = \frac{\left(EL\right)^{2}}{R} [/tex]

    How does the voltage change when the length is doubled? Does this help?
  4. Oct 6, 2008 #3
    thanks. ok, so then i plug in 2L and I get 4E2L2/R, then i plug in 2LE/I for R and it becomes 2LIE, but this still would not give me the right answer when subtracting from the Power with L
  5. Oct 6, 2008 #4
    For the initial condition

    [tex] E_{initial} = \frac{V}{L} [/tex]

    Now, double the length. So,

    [tex] E_{final} = \frac{V}{2L} \rightarrow E_{final} = \frac{1}{2}E_{initial} [/tex]

    Remember, the voltage is "constant." Does this make sense yet?

    EDIT: Whoops. Made a mistake...fixed it. Sorry, my mind isn't working right. Now plug it into the previous relation that I provided for the power and you will see the 3/4 difference.
    Last edited: Oct 6, 2008
  6. Oct 7, 2008 #5


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    You are both thinking too 1-dimensionally about this problem; let's call the volume of the wire [itex]\tau[/itex]. Assume that it is a cylindrical shape, initially with radius [itex]R_i[/itex] and length [itex]L_i[/itex], so that [itex]\tau=\pi R_i^2L_i[/itex]. You are told that the volume of the wire remains constant, so what happen to the radius R as the wire is stretched to a length [itex]L_f=2L_i[/itex]?
    What does that do the cross-sectional area [itex]A=\pi R^2[/itex] of the wire?

    The resistance [itex]r[/itex] will change as a result of the changing cross-section. Remember, for a cylindrical wire of length [itex]L[/itex] and cross-sectional area [itex]A[/itex] with constant resistivity [itex]\rho[/itex], the resistance is given by [itex]r=\rho \frac{L}{A}[/itex].

    So, its the change in the resistance due to the changing cross-sectional area of the wire that accounts for the different power dissipations.

    Also, you are told to treat V as a constant, so why would you write V=EL and calculate a change in V?

    Just use [itex]P_i=\frac{V^2}{r_i}[/itex] and [itex]P_f=\frac{V^2}{r_f}[/itex] to calculate the change in power dissipation [itex]P_f-P_i[/itex].
  7. Oct 7, 2008 #6
    I liked your approach. I didn't consider it since the student had already supplied relevant equations that would get the job done. I don't understand this comment though. The voltage V never changes, since it is constant. If you double the length, the electric field E must be halved to keep V constant. So, the result is the same, but I would say that your approach is better since it takes into account the fixed volume of the wire.
  8. Oct 7, 2008 #7


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    How is the result the same? If you plug in a constant V and r into P=V^2/r you get the same power dissipation. Even if you use your expression involving E and L you get


    Clearly, any change in the dissipated power must come from the changing resistance of the wire.
  9. Oct 8, 2008 #8
    Whoa! You are right. Usually, I don't make such simple mistakes. Sorry for leading you astray Briteliner. I should have kept my mouth shut on this one. =)
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