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In circuit analysis, is current through wires defined?

  1. Jan 29, 2013 #1
    1. The problem statement, all variables and given/known data
    This is not a homework problem, but related to my circuits class.

    In the typical electrical engineering simplification, the resistance of wires in the circuit is taken to be 0 Ω. Also, the potential difference between two points along the same wire is 0 Volts.

    Does this mean that current through wires is undefined? Would engineers consider the current zero or infinite?

    2. Relevant equations

    voltage = current * resistance

    3. The attempt at a solution
    Maybe Ohm's Law does not apply here, and there is another way of looking at the situation.
     
  2. jcsd
  3. Jan 29, 2013 #2

    phinds

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    wires in circuits generally have such a tiny resistance relative to that of the other objects in the circuit that their effect is negligible. Why would you expect otherwise? circuit analysis becomes unnecessarily complicated if you add in the tiny amounts of loss due to wires.
     
  4. Jan 29, 2013 #3

    rude man

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    The wire can be considered to have zero resistance and the current still finite because V = IR but if R → 0 then V = 0 for any finite I. I is determined by other impedances in the circuit containing the zero-R wire.

    In reality there is always some finite R for any wire (and inductance too, and some capacitance to nearby nodes ...) and for a stretch of such a wire V = IR still holds.
     
  5. Jan 31, 2013 #4
    In circuit analysis (at least when Faraday's law doesn't apply), you may assume that the wires are inexistent and that all your components connect directly where they need to.
     
  6. Jan 31, 2013 #5

    CWatters

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    Neither. You would need to consider the circuit as a whole...

    In this case the resistance of the wire (0 Ohms) is going to be much less than other circuit elements that are in series with the wire. Those other circuit elements will dominate the overall resistance and the current flowing. The voltage drop down the wire is not the same as voltage source.

    Imagine you have a real world battery. These behave like an ideal voltage source (Videal) in series with a small resistor representing the internal resistance of the battery (Rint). lets say that's 0.1 Ohms.

    If you short out the battery using your ideal wire the current will be....

    Ishort = Videal/(Rint + Rwire)

    Rwire = 0 so
    Ishort = Videal/Rint

    The voltage drop down the ideal wire is 0V. The resistance of the wire is 0V but the current is Videal/Rint.
     
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