metallic conduction and resistance-temperature dependence

In a metal some valence electrons are shared by the whole crystal: they form a free-electron gas that can move through the sample. Under an electric field $E$ the electrons acquire a drift velocity $v_d$, and the current is $I = n e A v_d$. Resistance appears because electrons collide with the vibrating lattice ions, and each collision disrupts the directed motion. When the metal is heated, the amplitude of the thermal vibrations grows, collisions become more frequent, the mean free time $\tau$ decreases, and the resistivity $\rho$ increases. Over a moderate temperature range this dependence is nearly linear: $\rho = \rho_0(1+\alpha\,\Delta T), \qquad \Delta T = T - T_0.$ Because $R = \rho L / A$ for a conductor of constant cross-section, the resistance obeys the same law: $R = R_0(1+\alpha\,\Delta T).$ Here $\rho_0$ and $R_0$ refer to the initial temperature $T_0$, and $\alpha$ is the temperature coefficient of resistance; for metals $\alpha > 0$ and is of order $10^{-3}\ \text{K}^{-1}$. The formula is used together with Ohm's law for a section of a circuit, which links the measured $I$ and $U$ to the resistance through $R = U/I$.

A copper wire has resistance $R_0 = 10.0\ \Omega$ at $T_0 = 20\ ^\circ\text{C}$. The temperature coefficient of copper is $\alpha = 3.9\times10^{-3}\ \text{K}^{-1}$. What is the resistance of the wire at $80\ ^\circ\text{C}$?

The filament of a lamp is made of tungsten ($\alpha \approx 4.5\times10^{-3}\ \text{K}^{-1}$). At room temperature $20\ ^\circ\text{C}$ its resistance is $R_0 = 20\ \Omega$, and in operating mode $R = 240\ \Omega$. Estimate the temperature of the filament.

A student measures the resistance of a metal resistor: at $0\ ^\circ\text{C}$ it is $R_1 = 4.00\ \Omega$, and at $100\ ^\circ\text{C}$ it is $R_2 = 5.56\ \Omega$. (a) Find $\alpha$ from these two points. (b) Predict $R$ at $400\ ^\circ\text{C}$ within the linear model. (c) Give one reason why the prediction may be inaccurate.

Self-check in a simulation (safe). Use a virtual laboratory — for example, PhET "Resistance in a Wire" or a circuit simulator with a thermistor.

1. Set the length and cross-sectional area of the wire and switch on the $\rho(T)$ model.
2. Record the resistance $R$ at five temperatures, including $T_0 = 20\ ^\circ\text{C}$.
3. Plot $R(T)$ and check that the points fall on a straight line; find $\alpha$ from the slope: $\alpha = \dfrac{R-R_0}{R_0\,\Delta T}$.
4. Compare the value of $\alpha$ you obtained with the tabulated value for the chosen metal and explain the difference.

Safety rules. Never test the temperature dependence by connecting an incandescent lamp to a 220 V mains supply. For a home measurement use battery power and a multimeter only; measure the cold filament resistance when the lamp is switched off and completely cooled.

Show that the resistance of a metal grows almost linearly with temperature, and link the slope of the graph to the temperature coefficient $\alpha$ and to the mechanism of electron collisions with lattice ions.

A graph of $R(T)$: a straight line passing through the point $(T_0,R_0)$ with slope $\alpha R_0$; the same figure marks the slope $\dfrac{\Delta R}{\Delta T}$. Beside it, a sketch shows ions at the lattice sites and free electrons; arrows indicate the electron drift, and two panels with small and large amplitudes of ion vibration illustrate why heating makes collisions more frequent.

A $60\ \text{W}$ lamp is connected to a $230\ \text{V}$ mains supply. In steady operation the tungsten filament ($\alpha \approx 4.5\cdot10^{-3}\ \text{K}^{-1}$) is at about $2700\ ^\circ\text{C}$, its resistance is $R_\text{hot} = U^2/P = (230\ \text{V})^2/60\ \text{W} \approx 882\ \Omega$, and the current is $I = P/U \approx 0.26\ \text{A}$. At $20\ ^\circ\text{C}$ the temperature rise is $\Delta T \approx 2680\ \text{K}$, so $R_\text{cold} = R_\text{hot}/(1+\alpha\Delta T) \approx 882/13 \approx 68\ \Omega$, and at switch-on the current reaches $230\ \text{V}/68\ \Omega \approx 3.4\ \text{A}$, about 13 times the steady value.

The filament is a metal: the current is carried by free electrons accelerated by the field, which collide with the vibrating lattice ions. On heating, the ions vibrate with larger amplitude, scattering becomes stronger, and the resistivity grows as $\rho = \rho_0(1+\alpha \Delta T)$. For a fixed geometry $R=\rho L/A$ rises by the same factor — this is what links the measured cold and hot currents to the temperature coefficient $\alpha$.

A $12\ \text{V}$ battery feeds a starter through a copper cable of length $1.2\ \text{m}$ and cross-section $25\ \text{mm}^2$. At $20\ ^\circ\text{C}$ the resistivity of copper is $\rho_0 = 1.7\cdot10^{-8}\ \Omega\cdot\text{m}$, so $R = \rho_0 L/A \approx 1.7\cdot10^{-8}\cdot1.2/(25\cdot10^{-6}) \approx 8.2\cdot10^{-4}\ \Omega \approx 0.8\ \text{m}\Omega$. At a current of $150\ \text{A}$ the voltage drop along the cable is $IR \approx 0.12\ \text{V}$ and the heating power is $I^2R \approx 18\ \text{W}$. Within a few seconds the copper warms to $60\ ^\circ\text{C}$ ($\Delta T = 40\ \text{K}$, $\alpha = 3.9\cdot10^{-3}\ \text{K}^{-1}$): the resistance rises by about $16\%$ to $0.95\ \text{m}\Omega$, and the voltage drop becomes $0.14\ \text{V}$.

The cable is a metallic part of the circuit, so its resistance enters Ohm's law for a complete circuit: the current is set by the sum of the resistances. The resistivity of copper depends on temperature as $\rho = \rho_0(1+\alpha\Delta T)$, and $R=\rho L/A$; therefore as it warms the resistance grows and the current through the starter decreases. At the same time the power $I^2R$ is released, which heats the cable further — a feedback loop.

An $800\ \text{W}$ toaster runs on $230\ \text{V}$; in steady operation at about $500\ ^\circ\text{C}$ its resistance is $R_\text{hot} = U^2/P = (230\ \text{V})^2/800\ \text{W} \approx 66\ \Omega$, and the current is $I = P/U \approx 3.5\ \text{A}$. For nichrome $\alpha \approx 4\cdot10^{-4}\ \text{K}^{-1}$, so with $\Delta T = 480\ \text{K}$ the factor $1+\alpha\Delta T \approx 1.19$. Hence at $20\ ^\circ\text{C}$ the resistance is $R_\text{cold} \approx 66/1.19 \approx 55\ \Omega$, and the switch-on current $230\ \text{V}/55\ \Omega \approx 4.1\ \text{A}$ is only $19\%$ above the steady value.

The mechanism is the same: conduction electrons collide with lattice ions, and a higher temperature increases scattering. But in the disordered alloy nichrome the extra scattering from thermal vibrations is only a small fraction, so $\alpha$ is small and the resistance changes little with temperature. Through Ohm's law for a complete circuit this means an almost constant current $I = \varepsilon/(R+r)$ and power $P = I^2R$, which is what a heating appliance needs.