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Homework Help: Potential Difference and Power Dissipation Problem

  1. Mar 12, 2010 #1
    A potential difference V is applied to a wire of cross-section area of 1 unit, length 1 unit,
    and conductivity σ. You want to change the applied potential difference and draw out the
    wire so the power dissipated is increased by a factor of 30 and the current is increased by a
    factor of 4 .

    a) What should be the new values of the length?

    b) What is the new cross sectional area?

    What i've done so far is:
    Linitial = 1 unit, Ainitial = 1 unit
    Initial power Pi = I2 * R, R = ρ * Linitial / Ainitial = 1/σ (as ρ = 1/σ, )
    Pi = I2 / σ
    Pf = 30 * Pi
    If2 * Rf = 30 * I2 / σ

    Substituting If = 4 * I
    16 * I2 * L / σ * A = 30 * I2 / σ
    16 * L / A = 30
    16 * L = 30 * A

    L * A = volume of wire = Linitial * Ainitial (as volume does not change)
    => L * A = 1 * 1
    A = 1 / L
    Substituting
    16 * L = 30 * (1/L)
    L2 = 30 / 16
    final length L = √1.875 = 1.369 units

    But it's wrong. (Its electronic and I only have 5 guess left (I've added 1 to it too for 2.369, didn't work either))
     
  2. jcsd
  3. Mar 13, 2010 #2

    collinsmark

    User Avatar
    Homework Helper
    Gold Member

    Well, if it's any consolation, I got the same answer you did. 1.369 units. :confused:
     
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