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Resistivity what a concept

  1. Nov 26, 2003 #1
    So here is the problem

    Two cylindrical rods, one copper and the other iron, are identical in lengths and cross-sectional areas. They are joined, end-to-end, to form one long rod. A 12-V battery is connected across the free ends of the copper-iron rod. What is the voltage between the ends of the copper rod?

    I know that R = resistivity(L/A)

    so what I did was created a ratio to get the resistance of the new rod

    Riron/Rcopper = (resistivty of iron * L/A)/(resistivty of copper * L/A) = (resistivty of iron)/(resistivty of copper) = 9.7*10^-8/1.72*10^-8

    = a resistance of 5.64

    I believe that is the resistance of the copper and iron rod connected to each other however I do not know how it helps if it does at all
  2. jcsd
  3. Nov 26, 2003 #2
    It's a start.

    After all that ********, all you did was find the ratio of the resistance of the iron part to the resistance of the copper part.

    So now imagine that you have a circuit consisting of a battery and two resistors in series, connected by wire with 0 resistance. One of the resistors is the copper rod, the other is the iron rod.

    Remembering that for any resistor, ΔV = IR, can you write an equation that describes the changes in potential at each step going around the entire circuit?

    hint: let Rc = resistance of the copper rod. What is the resistance of the iron rod, in terms of Rc ?
  4. Nov 27, 2003 #3
    Not sure I follow

    I will have to look at this a little more closley becuase I am not sure I know where to start.
  5. Nov 27, 2003 #4


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    The problem as stated doesn't makes sense. If you connect a conductor to the two poles of a 12 volt battery, then the voltage (drop) between the two end IS 12 volts. You could then use "V= IR" to determine the current in the two parts, but "12 volts is 12 volts"!
  6. Nov 27, 2003 #5

    Of course it makes sense. There is a certain drop across the copper part, and an additional drop across the iron part. The combined drop is 12 volts. The question asks for the drop across the copper part alone.

    What's wrong with that?
  7. Nov 28, 2003 #6
    Is this correct

    I came up with the following

    Rcopper/Riron = (resistivty of copper * L/A)/(resistivty of iron * L/A) = (resistivty of copper)/(resistivty of iron) = 9.7*10^-8/1.72*10^-8

    = a resistance of .18

    Icopper = Iiron
    V= IR


    then we know that

    Vcopper = .18*Viron



    we also know that

    Vcopper + Viron= 25

    so with a little subsititution

    Vcopper + Vcopper/.18 =25


    Vcopper = 1.83

    This is driving me outta my head please let me know If i am correct
  8. Nov 29, 2003 #7
    Wait till you get to magnetism.

    You mixed up the resistivities, and you typed 25 instead of 12 for the voltage, but amazingly you ended up very close to my answer.

    I'll take your numbers for the resistivities; my book gives slightly different figures. But it's obvious that the 1.72 must be the copper, and the 9.7 is the iron.

    So here's how I set it up:
    Let R = resistANCE of the copper rod
    therefore IR = voltage drop across the copper rod
    since the two rods are identical in length & diameter:
    (9.7/1.72)R = 5.64R = resistANCE of the iron rod.
    I*5.64R = voltage drop across the iron rod
    The sum of the voltage drops across the two rods must equal the total voltage supplied, so
    I*R + I*5.64R = 12
    IR(1 + 5.64) = 12
    IR = 12/6.64 = 1.81 V

    as a check:
    voltage drop for the iron rod = 5.64*IR = 10.19 V
    1.81 + 10.19 = 12
  9. Nov 30, 2003 #8
    Thanks again

    I am glad that I finally got to the bottom of it.

    Luckily this is only a one semester class and as we are drawing to the end quickly I do not think we will get to magnetism. So I will have to conquer that another day
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