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Change in potential energy across wires differing composition

  1. Feb 23, 2012 #1
    1. The problem statement, all variables and given/known data

    9 V battery is connected to wire made up of three segments of different materials connected in a series: 10 cm copper, then 12 cm iron, then 18 cm tungsten. All wires are 0.26 mm diameter. Find potential difference across each wire.

    p Copper= 1.7 x 10-8
    p Iron = 9.7 x 10 -8
    p Tungsten = 5.6 x 10-8

    26 mm diameter = 0.13 mm radius or 1.3 x 10-4 m

    2. Relevant equations


    area of wire (1.3 x 10-4)(1.3 x 10-4) x pi = 5.31 x 10-8 meters squared

    3. The attempt at a solution

    Copper Resistance
    R= (1.7 x 10-8 ohm m)(.1m)/5.31 x 10-8 meters squared
    R=0.032 ohm

    Iron Resistance
    R=(9.7 x 10-8 ohm m)(.12m)/5.31 x 10-8 meters squared
    R=0.22 ohm

    Tungsten Resistance
    R=(5.6 x 10-8 ohm m)(0.18 m)/5.31 x 10-8 meters squared

    R total series = r1 + r2 + r3
    = .032 + .22 + .19
    = 0.442 ohm

    V copper = vtotal * Rcopper/R total
    = 9 V * .032/.442
    = 0.65 V

    V silver = vtotal *Rsilver/R total
    =9 V * .22/.442
    =4.48 V

    V Tungsten = vtotal * Rtungsten/R total
    =9V *.387/.442
    = 3.87 V

    Volts all add up to 9 so 2nd Law is good but my answers are wrong somewhere...

  2. jcsd
  3. Feb 24, 2012 #2


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    Homework Helper

    I like your reasoning - and your answers - so check that you have transcribed the figures correctly, and looked at the correct answers in you book.
  4. Feb 24, 2012 #3
    Thanks Peter. The prof had changed the numbers on the website from those provided in the book. I had overlooked one of them.
  5. Feb 24, 2012 #4


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    Homework Helper

    Note: when I worked this out I ignored the diameter - since they were all the same, as well as the 10-8 on each of the resistivities, [they all had the same factor] since I knew at the end I was only going use the resistances in the form of fractions - where those values would cancel anyway.
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