The conduction model, temperature dependence of resistance and electric power

Last week you defined resistance macroscopically: $R = \rho L/A$. This week you answer two questions: why does resistance arise in a metal, and where does the energy carried by the current go? The exam has three question types on this topic: drift velocity calculations, change of resistance with temperature, and power (bulb brightness, heater time, electricity bill). Power questions cost the most points, because it is easy to pick the wrong one of the three formulas.

Golden rules:

1. The drift velocity is very small: $v_d \sim 10^{-4}$ m/s. The lamp lights instantly because the field spreads through the wire at nearly the speed of light, not the electrons.
2. In a metal, resistance rises when heated: $R = R_0[1 + \alpha\,\Delta T]$, $\alpha > 0$.
3. For power, the three formulas say the same thing: $P = I\Delta V = I^2R = \Delta V^2/R$. Choose the formula that contains the quantity that is constant.
4. In series the current is common, in parallel the voltage is common. This is the key to brightness comparisons.

Drift velocity

When there is no field inside the wire, the free electrons run very fast ($\sim 10^6$ m/s) with thermal motion but in random directions; the net flow of charge is zero. When a field is applied, a small common velocity is added to this random motion: the drift velocity $v_d$.

In a wire of cross-section $A$, with $n$ charge carriers per unit volume (charge carrier density), the charges in a volume $A\,v_d\,\Delta t$ pass through a cross-section in time $\Delta t$:

$\Delta Q = n\,q\,(A\,v_d\,\Delta t) \quad\Rightarrow\quad I = n\,q\,v_d\,A, \qquad J = n\,q\,v_d$

Example. A copper wire with a cross-section of 2.5 mm² ($n = 8.5\times10^{28}$ m⁻³) carries 10 A. What is the drift velocity, and how long does an electron take to advance 1.0 m?

A  = 2.5 mm² = 2.5e-6 m²
vd = I/(n e A) = 10 / (8.5e28 × 1.60e-19 × 2.5e-6) = 2.9e-4 m/s
t  = 1.0 m / vd = 3.4e3 s ≈ 57 minutes

It takes an electron almost an hour to advance one meter. Yet when you close the switch the lamp lights immediately: all the electrons in the wire start drifting at the same time, because the electric field is set up along the wire very quickly.

PITFALL · $v_d$ versus thermal speed. $v_d$ is used in current calculations. In the calculation of the distance between collisions (the mean free path), the real average speed of the electron is used, that is, the thermal speed ($\sim 10^6$ m/s). Whoever mixes the two is off by a factor of $10^{10}$.

The microscopic model of conduction

In the simple model, the electron accelerates in the field with $a = eE/m_e$, collides with an ion, scatters in a random direction, and accelerates again. If the average time between collisions is $\tau$ (mean free time), the average extra speed the electron gains is:

$v_d = \frac{eE}{m_e}\,\tau$

Putting this into $J = n e v_d$:

$J = \frac{n e^2 \tau}{m_e}\,E \quad\Rightarrow\quad \sigma = \frac{n e^2 \tau}{m_e}, \qquad \rho = \frac{m_e}{n e^2 \tau}$

The model tells two things. First, $J \propto E$ comes out: Ohm’s law follows from here. Second, $\rho$ depends on the collision time: as the temperature rises the ions vibrate more, collisions become more frequent, $\tau$ becomes smaller, and $\rho$ becomes larger.

Mean free path: $\ell = v_{\text{avg}}\,\tau$. For copper, with $\tau \approx 2.5\times10^{-14}$ s and $v_{\text{avg}} \approx 1.6\times10^6$ m/s, $\ell \approx 40$ nm, about 150 times the distance between atoms.

Resistance and temperature

Over a limited temperature range, resistivity changes approximately linearly with temperature:

$\rho = \rho_0[1 + \alpha(T - T_0)], \qquad R = R_0[1 + \alpha(T - T_0)]$

$\alpha$ is the temperature coefficient of resistivity, with unit $(^\circ\text{C})^{-1}$. In metals it is positive (copper $3.9\times10^{-3}$, platinum $3.92\times10^{-3}$, tungsten $4.5\times10^{-3}$). In semiconductors (silicon, carbon) it is negative: as they heat up the number of carriers increases and the resistance falls.

Example: lamp filament. The resistance of a tungsten filament is 20 Ω at 20 °C. If its temperature in operation is 1500 °C, what is its approximate resistance?

ΔT = 1500 − 20 = 1480 °C
R  = 20 [1 + (4.5e-3)(1480)] = 20 × 7.66 = 1.5 × 10² Ω

For such a large $\Delta T$ the linear formula is only a rough estimate, but it gives the important result correctly: the resistance of a cold bulb is about one seventh that of a hot bulb. This is why bulbs usually fail at the moment they are switched on: the initial current is very large. A bulb is not an ohmic element.

PITFALL · the reverse question. In the question “The resistance rose from 200 Ω to 253.3 Ω, what is the temperature?”, the $\Delta T$ you find is added to the initial temperature. $\Delta T$ itself is also among the choices.

Superconductors

The resistance of some materials drops abruptly and exactly to zero below a temperature called the critical temperature $T_c$. These materials are called superconductors. For mercury $T_c = 4.2$ K; for some ceramic copper oxides $T_c$ is above 100 K, so they can be cooled with liquid nitrogen. Because the resistance is zero, a current started in a superconducting loop continues without a battery. The strong magnets of MRI machines work on this principle. On the exam it usually comes as a conceptual question: “decreases linearly to zero” and “becomes an insulator” are wrong choices.

Electric power

When a charge $\Delta Q$ passes through a potential difference $\Delta V$, it releases an energy $\Delta Q\,\Delta V$. Dividing by time gives the power:

$P = I\,\Delta V$

For a resistor, using $\Delta V = IR$ gives two more equivalent forms:

$P = I^2 R = \frac{(\Delta V)^2}{R}$

In a resistor this energy turns into heat (joule heating). Its unit is the watt: $1\ \text{W} = 1\ \text{J/s} = 1\ \text{V}\cdot\text{A}$. Electric companies measure energy in kilowatt-hours: $1\ \text{kWh} = (1000\ \text{W})(3600\ \text{s}) = 3.6\times10^{6}$ J.

Example: heater. A 1500 W heater connected to a 220 V outlet. What are its resistance and the current it draws?

R = ΔV² / P = 220² / 1500 = 32 Ω
I = P / ΔV  = 1500 / 220  = 6.8 A
Check: I² R = 6.82² × 32.3 ≈ 1500 W

Example: why transmit at high voltage?

A power plant sends 100 kW over a line with a total resistance of 2.0 Ω. What power goes to heat in the line when the transmission voltage is 10 kV and when it is 1.0 kV?

10 kV:   I = P/ΔV = 100e3 / 10e3 = 10 A    loss = I² R = 100 × 2.0 = 200 W    (0.2%)
1.0 kV:  I = 100e3 / 1.0e3 = 100 A          loss = 1.0e4 × 2.0 = 2.0e4 W       (20%)

Increasing the voltage 10 times reduced the loss 100 times. Using the formula $\Delta V^2/R$ with 10 kV here would be wrong: 10 kV is the output voltage of the plant, not the drop across the line. The real drop across the line is $IR = 20$ V, and $(20)^2/2.0 = 200$ W gives the same result.

Bulb brightness

The bulb’s label (“60 W, 120 V”) gives you its resistance: $R = \Delta V^2/P$. The power on the label holds only at the voltage on the label. How brightly the bulb glows in another circuit depends on the $I^2R$ power it dissipates in that circuit.

• Connected in parallel, both at 120 V: the 100 W one is brighter (the values on the labels).
• Connected in series, 120 V in total: the current is common, and $I^2R$ is larger for the larger resistance. The resistance of the 60 W bulb is larger (240 Ω against 144 Ω), so it glows brighter. Both dissipate far less power than on their labels.

TIP · first step in a power question. First find the resistance of each element, then compute the current or voltage in the circuit, and only then go to $P$. Don’t carry the power on the label directly into the circuit.

Common mistakes