Resistivity & Conductivity: Definition, Causes, Formula & Units (w/ Chart)

Resistivity and conductivity are two sides of the same coin, but both are crucial concepts to grasp when you’re learning about electronics. They’re essentially two different ways of describing the same fundamental physical property: how well electric current flows through a material.

Electrical resistivity is a property of a material that tells you how much it resists the flow of electric current, while conductivity quantifies how easily current flows. They’re very closely related, with electrical conductivity being the inverse of resistivity, but understanding both in detail is important for tackling problems in the physics of electronics.

Electrical Resistivity

The resistivity of a material is a key factor in determining the electrical resistance of a conductor, and it is the part of the equation for resistance that takes into account the differing characteristics of different materials.

Electrical resistance itself can be understood through a simple analogy. Imagine that the flow of electrons (the carriers of electric current) through a wire is represented by marbles flowing down a ramp: You would get resistance if you placed obstructions in the path of the ramp. As marbles bumped into the barriers, they would lose some of their energy to the obstructions, and the overall flow of marbles down the ramp would slow down.

Another analogy that can help you understand how current flow is affected by resistance is the effect that passing through a paddle wheel has on the speed of a current of water. Again, energy transfers to the paddle wheel, and the water moves more slowly as a result.

The reality for current flow through a conductor is closer to the marble example because the electrons flow through the material, but the lattice-like structure of the nuclei of the atoms are obstructions to this flow, which slows the electrons down.

The electrical resistance of a conductor is defined as:

R = \frac{ρL}{A}

Where ρ (rho) is the resistivity of the material (which depends on its composition), length L is how long the conductor is and A is the cross-sectional area of the material (in square meters). The equation shows that a longer conductor has higher electrical resistance, and one with a bigger cross-sectional area has a lower resistance.

The SI unit of resistance is the ohm (Ω), where 1 Ω = 1 kg m2 s−3 A−2, and the SI unit of resistivity is the ohm-meter (Ω m). Different materials have different resistivities, and you can look up the values for the resistivity of the material you’re using in a calculation in a table (see Resources).

Electrical Conductivity

Electrical conductivity is simply defined as the inverse of resistivity, so a high resistivity means a low conductivity, and a low resistivity means a high conductivity. Mathematically, the conductivity of a material is represented by:

σ = \frac{1}{ρ}

Where σ is the conductivity and ρ is the resistivity, as before. Of course, you can re-arrange the equation for resistance in the previous section to express this in terms of the resistance, R, cross-sectional area A of the conductor and the length L, depending on what the problem you’re tackling calls for.

The SI units for conductivity are the inverse of the resistivity units, which makes them Ω−1 m−1; however, it is usually quoted as siemens/meter (S/m), where 1 S = 1 Ω−1.

Calculating Resistivity and Conductivity

With the definitions of electrical resistivity and conductivity in mind, seeing an example calculation will help to cement the ideas introduced so far. For a length of copper wire, with a length L = 0.1 m and a cross-sectional area A = 5.31 × 10−6 m2 and a resistance of R = 3.16 × 10−4 Ω, what is the resistivity ρ of copper? First, you need to re-arrange the equation for resistance to get an expression for resistivity ρ, as follows:

R = \frac{ρL}{A}
ρ = \frac{RA}{L}

Now you can insert values to find the result:

\begin{aligned} ρ &= \frac{3.16 × 10^{−4} \text{ Ω} × 5.31 × 10^{−6}\text{ m}^2}{0.1 \text{ m}} \\ &=1.68 × 10^{−8} \text{ Ω m} \end{aligned}

From this, what is the electrical conductivity of the copper wire? Of course, this is quite straightforward to work out on the basis of what you’ve just found, because conductivity (σ) is just the inverse of resistivity. So conductivity is:

\begin{aligned} σ &= \frac{1}{ρ} \\ &= \frac{1}{1.68 × 10^{−8}\text{ Ω m}} \\ &= 5.95 × 10^7 \text{ s/m} \end{aligned}

The very low resistivity and high conductivity explain why a copper wire just like this is probably what’s used in your home to deliver electricity.

Temperature Dependence

The values you’ll find in a table for the resistivity of different materials will all be values at a specific temperature (generally chosen to be room temperature), because resistivity increases with increasing temperature for most materials.

Although for some materials (like semiconductors such as silicon), resistivity decreases with increasing temperature, an increase with temperature is the general rule. This is easy to understand if you go back to the marble analogy: With the barriers vibrating around (as a result of the increased temperature and therefore the internal energy), they’re more likely to block the marbles than if they were completely stationary throughout.

The resistivity at temperature T is given by the relationship:

ρ (T) = ρ_0(1 + α(T – T_0))

Where alpha (α) is the temperature coefficient of resistivity, T is the temperature you’re calculating the resistivity at, T0 is a reference temperature (usually taken as 293 K, roughly room temperature) and ρ0 is the resistivity at the reference temperature. All temperatures in this equation are in kelvins (K), and the SI unit for the temperature coefficient is 1/K. The temperature coefficient of resistivity generally has the same value of the temperature coefficient of resistance, and tends to be of the order of 10−3 or lower.

If you need to calculate the temperature dependence for different materials, you simply need to look up the value of the appropriate temperature coefficient and work through the equation with the reference temperature T0 = 293 K (as long as it matches the temperature used for the reference value for resistivity).

You can see from the form of the equation that this will always be a resistivity increase for increases in temperature. The following table contains some key data for the electrical resistivity, conductivity and temperature coefficients for various materials:

\def\arraystretch{1.5} \begin{array}{c:c:c:c} \text{Material} & \text{Resistivity, }ρ \text{ (at 293 K) / Ω m} & \text{Conductivity, } σ \text{ (at 293 K) / S/m} & \text{Temperature Coefficient,} α \text{/ K}^{−1} \\ \hline \text{Silver} & 1.59 × 10^{−8} & 6.30 × 10^7 & 0.0038\\ \hdashline \text{Copper} & 1.68 × 10^{−8} & 5.96 × 10^7 & 0.00386\\ \hdashline \text{Zinc} & 5.90 × 10^{−8} &1.69 × 10^7 & 0.0037\\ \hdashline \text{Nickel} &6.99 × 10^{−8} & 1.43 × 10^7 & 0.006\\ \hdashline \text{Iron} & 1.00 × 10^{−7} & 1.00 × 10^7 & 0.00651\\ \hdashline \text{Stainless Steel} & 6.9 × 10^{−7} & 1.45 × 10^6 & 0.00094\\ \hdashline \text{Mercury} & 9.8 × 10^{−7} & 1.02 × 10^6 & 0.0009\\ \hdashline \text{Nichrome} & 1.10 × 10^{−6} & 9.09 × 10^5 & 0.0004\\ \hdashline \text{Drinking water} & 2 × 10^1 \text{to} 2 × 10^3 & 5 × 10^{−4} \text{to} 5 × 10^{−2} & \\ \hdashline \text{Glass} & 10^{11} \text{to} 10^{15} & 10^{-11} \text{to} 10^{-15} & \\ \hdashline \text{Rubber} & 10^{13} & 10^{-13} & \\ \hdashline \text{Wood} & 10^{14} \text{to} 10^{16} & 10^{-16} \text{to} 10^{-14} & \\ \hdashline \text{Teflon} & 10^{23} \text{to} 10^{25} & 10^{-25} \text{to} 10^{-23} & \\ \hdashline \end{array}

Note that the insulators in the list don’t have established values for their temperature coefficients, but they are included to show the full range of values of resistivity and conductivity.

Calculating Resistivity at Different Temperatures

Although the theory that resistivity increases when temperature increases makes sense, it’s worth looking at a calculation to underscore the impact that an increase in temperature can have on the conductivity and resistivity of a material. For the example calculation, consider what happens to the resistivity and conductivity of nickel when heated from 293 K to 343 K. Looking at the equation again:

ρ (T) = ρ_0(1 + α(T – T_0))

You can see that the values you need to calculate the new resistivity are in the table above, where the resistivity ρ0 = 6.99 × 10−8 Ω m, and the temperature coefficient α = 0.006. Inserting these values into the above equation allows the new resistivity to be easily calculated:

\begin{aligned} ρ (T) &= 6.99 × 10^{−8} \text{ Ω m} (1 + 0.006 \text{ K}^{−1} × (343 \text{ K}- 293 \text{ K})) \\ &= 6.99 × 10^{−8}\text{ Ω m} (1 + 0.006 \text{ K}^{−1} × (50 \text{ K)}) \\ &= 6.99 × 10^{−8}\text{ Ω m} × 1.3 \\ &= 9.09 × 10^{−8}\text{ Ω m} \end{aligned}

The calculation shows that a fairly substantial increase in temperature of 50 K only leads to a 30 percent increase in the value of the resistivity, and thereby a 30 percent increase in the resistance of a given amount of material. Of course, you could then go on and calculate the new value for conductivity on the basis of this result.

The impact of an increase in temperature on the resistivity and conductivity is determined by the size of the temperature coefficient, with higher values meaning more of a change with temperature and lower values meaning less of a change.

Superconductors

The Dutch physicist Heike Kamerlingh Onnes was investigating the properties of different materials at very low temperatures in 1911 and discovered that below 4.2 K (i.e., −268.95 °C), mercury completely loses its resistance to the flow of electric current, so it’s resistivity becomes zero.

As a result of this (and the relationship between resistivity and conductivity), their conductivity becomes infinite, and they can carry a current indefinitely, without any loss of energy. Scientists later discovered that many more elements exhibit this behavior when cooled to below a certain “critical temperature,” and are called “superconductors.”

For a long time, physics offered no real explanation of superconductors, but in 1957, John Bardeen, Leon Cooper and John Schrieffer developed the “BCS” theory of superconductivity. This posits that the electrons in the material group into “Cooper pairs” as a result of interactions with the positive ions making up lattice structure of the material, and these pairs can move through the material without any impediment.

As an electron moves through the cooled material, the positive ions forming the lattice are attracted to them and slightly change their position. However, this motion creates a positively charged region in the material, which attracts another electron and the process begins again.

Superconductors owe many potential and already realized uses to their ability to carry currents without resistance. One of the most common uses, and the one you’re most likely to be familiar with, is magnetic resonance imaging (MRI) in medical settings.

However, superconductivity is also used for things like Maglev trains – which work through magnetic levitation and aim to remove the friction between the train and the track – and particle accelerators like the Large Hadron Collider at CERN, where the superconducting magnets are used to accelerate particles at speeds approaching the speed of light. In the future, superconductors may be used to improve the efficiency of electricity generation and improve the speed of computers.

References

Resources

About the Author

Lee Johnson is a freelance writer and science enthusiast, with a passion for distilling complex concepts into simple, digestible language. He's written about science for several websites including eHow UK and WiseGeek, mainly covering physics and astronomy. He was also a science blogger for Elements Behavioral Health's blog network for five years. He studied physics at the Open University and graduated in 2018.