# Voltage drop along an idealized resistance-free wire in a circuit?

If you connected the positive terminal of a battery to the negative terminal to a battery with a wire with (hypothetically) no resistance, and are asked to give the voltage drop of a segment of wire in the circuit, how would you determine this?

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Since you specified the wire, but not the battery, as being ideal, the result will be that there will be no voltage on the wire. Instead, the battery will be shorted out.

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Isn't it better to state that there is no voltage drop between a battery terminal and any point on the wire than "no voltage on the wire"? –  Bentley4 Aug 31 '13 at 23:15

If you are asking what happens when a constant-voltage source with infinite current capability is short-circuited by 0 resistance, it's not a well-defined question.

In practice, the source has an intrinsic resistance > 0 and the wire has a resistance > 0 and if that is your question, Ohms law (or a voltmeter) and basic circuit theory will give you answers but I don't know exactly what you are asking currently.

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If you have an ideal voltage source with 0 internal resistance and 0 external resistance (0 resistance of the wire) then the voltage drop is indeterminate since it is of the form $\infty \times o$. The question then is not well defined.

However in practice, we represent a practical voltage source by an ideal voltage source in series with a resistance $r$ (the internal resistance of the practical voltage source). This will be in series with an external practical wire of resistance $r\prime$.

Applying Kirchhoff's law, we have,

$I \times (r + r\prime) = V$

or, $$I\times r\prime = V - I\times r$$

$V$ is the voltage of the source, $I$ current through the wire.

Therefore the voltage drop is, $V - I \times r$

or $I \times r\prime$

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Let $V$ be the e.m.f of the battery/ voltage source

Let $i$ be the current through the wire of resistance $R$.

Actually $i= \frac{V}{R}$, as per the question, $R$ tends to ZERO, hence $i$ tends to a very very large number.

Let the drop across the wire be $V_1$ then $V_1=$(current through the wire)*(resistance of the wire) {as per Ohm's law}

so: $V_1= iR$

hence: $V_1=\frac{V}{R}R$

so: V_1=V.

Thus, voltage drop across ANY segment of the wire connected will be equal to the drop $V$ {e.m.f of the battery}

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Are you sure about this? If you divide the wire in two pieces you get $2V$ voltage drop, if you divide it in three $3V$? Can you clarify a bit? –  Bernhard Oct 20 '12 at 9:54