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My textbook says,

"The resistance of wires is usually so low compared with other devices in a circuit that you can ignore wire resistances when measuring or calculating the total resistance. The exception is when there are large currents. If the current is large, the resistance of wires may be important."

Why is this true? If the current is large, then shouldn't the resistance decrease for a constant voltage (V=IR)?

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  • $\begingroup$ Could be due to the temperature rising, since larger currents cause larger heat dissipation, $P=VI$. Changing temperatures may have significant effects on the resistance. $\endgroup$ – Steeven Mar 28 '17 at 15:14
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Resistance of a material is $$R=\rho\frac {l}{a} $$ where $\rho $ is the resistivity of the material, $l $ is the length of the conductor, and $a $ is the cross sectional area.

For a given conductor, all these are constant. So resistance of a given conductor does not depend on current or voltage.

It however, depends on temperature: $$R=R_0 (1+\alpha×t) $$ where $R $ is the new resistance, $R_0$ is the initial resistance, $t $ is the change in temperature, and $\alpha $ is the coefficient for increase in resistance per unit rise in temperature.

For copper, resistivity is very low, so resistance is also low.

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If the current is large, then shouldn't the resistance decrease for a constant voltage (V=IR)?

You have got that backwards. The resistance of a metal wire is constant. It only depends on the length and diameter of the wire, and the material it is made from. (OK, there is a small change in resistance if the temperature changes, but that is not usually very important).

When the current is large, the voltage between the ends of the wire (given by $V = IR$) becomes large, and that may affect the voltages in the rest of the circuit.

For example, consider a car starter motor. That is powered from the car's $12\text{V}$ battery, and in cold weather when the engine is hard to turn over, the current required may be as high as $200\text{A}$. If the wires between the battery and the motor have a resistance of only $0.1\Omega$ (which would normally be considered "small"), the maximum current the battery could send though those wires is $12\text{V}/0.1\Omega = 120\text{A}$, Since the resistance of the motor is not zero, the actual current would be less than $120\text{A}$, and the motor probably wouldn't turn at all, or turn too slowly to start the car.

If you look at a car, you will find the wires between the starter and the battery are very thick compared with the rest of the electrical wiring, to make their resistance very small.

Another issue is that passing a large current through a resistance generates heat, and the heat (in watts) is given by $I^2R$. For a high powered household electrical device (for example a washing machine) the current may be as high as $10\text{A}$. If the resistance of the electrical wiring in house wiring to the washer was $0.1\Omega$, that means $10\text{W}$ of heat are being generated in the wires, all the time the washer is operating. The build up of heat over a long time may be enough to start a fire, if there is nowhere for it to "escape" to - for example if the wires are attached to the wood frame of the house and there is no air circulation to cool them down!

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For conductors, the resistance increases with increase on temperature. Though the intensity of that increase depends on the material of the conductor.

There exists an expression which related the resistivity of the conductor, the mass of the charge carriers (mass of an electron, $m_e$ as we are dealing with metallic conductors), the charge on a single carrier ($e$ for electrons, since we are dealing with metallic conductors) and their number density ($n$).

The expression is given by:

$\rho$ = $m_e$/$n$ $e^2$ $\tau$

The symbol $\tau$ represents something called 'relaxation time', which is the average interval between two successive collisions of electrons in the conductor.

The picture is something like this: As the current increases, the conductor starts to heat up (due to joule heating) and this heating results in an increase in the kinetic energy of the electrons of the conductor. As the kinetic energy of the electrons increases, they collide more frequently resulting in a decrease in $\tau$, the average interval between two collisions of electrons.
Mathematically, decreasing $\tau$ will increase $\rho$ which will ultimately increase $R$, the resistance of the wire under consideration (as $R$ = $\rho$ $l$/$A$, where $l$ and $A$ are the length and area of cross section of the conductor, which, in your case, are fixed)

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That's the trouble with textbooks like yours,they make generalizations which are silly. It's lazy (or ignorant) writers really. Electrical circuits span a huge field from tiny micro electronic devices through to electric arc smelters (and loads of other stuff). Any text book writer OUGHT to know that and be more careful.

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