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What is the current through the wire?

  1. Feb 23, 2005 #1

    I'm pretty sure I did this right but it seemed abnormally easy so I wanted to check.

    A 10-m length of wire consists of 5.0 m copper followed by 5.0 aluminum, both of diameter 1.0 mm. A voltage difference of 85 mV is placed across the wire. What is the total resistnace (sum) of the two wires? What is the current through the wire? What are the voltages across the alumnium part and across the copper part?

    FYI Resistivity of copper = 1.68E-8, resistivity of aluminum = 2.45E-8

    So for the first question I calculated the two resistances and added them using R = p (L/A), where that p is really a rho (resistivity)

    I got Rcopper part = .11 = 1.68E-8 x 5.0 (pi x (5E-4)^2)
    and Raluminum = .17 (calculated the same way, except with the rho of aluminum
    Added together = .28

    Then, for current I used V=IR so I = .3

    And finally, I calculated I = .39 for copper and .25 for aluminum using V = IR again.

    Is this really this simple?
  2. jcsd
  3. Feb 23, 2005 #2
    nope, "And finally, I calculated I = .39 for copper and .25 for aluminum using V = IR again"

    Is wrong. Look into how current is effected in serial components.

    The rest looks good enough for government work.
  4. Feb 24, 2005 #3

    I looked in my book and it doesn't say how current is affected in serial components. Is it proportional to the individual resistivity?
  5. Feb 24, 2005 #4


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    Recall that electric current is the flow of electrons. An Amp is defined as a fixed number of electrons flowing per second. Now assume that you have a wire with x Amps of current flowing. If you were able to watch them in any given second you would see x electrons per second pass through a cross section of the wire. Now if you were to watch an adjacent cross section you must see the same number of electrons pass, since you neither lose nor gain electrons at any point. You can apply this logic to every point in the series circuit to conclude that the current must be constant at all points of a wire.
  6. Feb 24, 2005 #5
    Oh, wow, that makes total sense. Looking back at the question, I was supposed to do something with voltages. I am not sure I know whether the same would hold true for voltage that does for current?
  7. Feb 24, 2005 #6


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    You now know the current in the wire. You also know the resistance in each section of the wire, Apply Ohms Law to each type of wire to find the voltage drop. You should be able to opserve the relationship with the total voltage.
  8. Feb 24, 2005 #7
    Sorry Integral, I'm an idiot and figured that out about a nanosecond after I posted. Now my stupidity is out there for all the world to see.

    Thanks :)

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