6.1 Resistance, Resistivity & Conductor Dimensions

Resistance, Resistivity & Conductor Dimensions

In electrotechnical engineering, electrical resistance ($R$) is defined as the property of a material or circuit element that opposes the flow of electric charge (current). When a potential difference (voltage) is established across a conductive path, an electric field accelerates mobile charge carriers—free electrons—through the material. As these electrons drift through the crystalline lattice of the conductor, they repeatedly collide with vibrating positive metal ions.

Each collision transfers kinetic energy from the drifting electrons to the lattice ions, increasing thermal kinetic energy and causing the conductor to warm up. This energy transformation, known as Joule heating ($I^2R$), represents the physical manifestation of electrical resistance. Resistance is measured in the SI unit Ohm (symbol $\Omega$), named in honour of German physicist Georg Simon Ohm. By definition, a conductor exhibits a resistance of one Ohm ($1\ \Omega$) when a constant potential difference of one volt ($1\text{ V}$) across its terminals produces a current of one ampere ($1\text{ A}$).

---

The Four Factors Determining Conductor Resistance

The electrical resistance of any conductor depends on four distinct physical variables:

1. Length ($L$): Resistance is directly proportional to conductor length ($R \propto L$). As an electric charge travels through a longer path, it encounters a proportionally greater number of lattice ions, leading to more frequent collisions and higher cumulative opposition. Doubling the length doubles the electrical resistance.
2. Cross-Sectional Area ($A$): Resistance is inversely proportional to cross-sectional area ($R \propto \frac{1}{A}$). A larger cross-sectional area provides a broader conductive pathway with more parallel routes through which mobile electrons can drift simultaneously. Doubling the cross-sectional area halves the electrical resistance.
3. Material Type (Resistivity, $\rho$): Different elements and alloys possess differing atomic structures, valence electron densities, and lattice configurations. The intrinsic property quantifying how strongly a given material opposes charge movement is called its resistivity (symbolized by the Greek letter rho, $\rho$).
4. Temperature ($\theta$): The kinetic energy and vibration amplitude of the conductor's lattice ions vary directly with temperature. In metallic conductors, rising temperature causes the lattice ions to vibrate more violently across larger amplitudes, significantly increasing collision cross-sections and raising electrical resistance.

---

Material Resistivity ($\rho$)

Resistivity is the intrinsic electrical resistance of a uniform sample of a material having unit length and unit cross-sectional area—specifically, a solid cube measuring one metre on each side ($1\text{ m} \times 1\text{ m} \times 1\text{ m}$). Resistivity is expressed in Ohm-metres ($\Omega\cdot\text{m}$), or in micro-Ohm metres ($\mu\Omega\cdot\text{m}$).

$\rho = \frac{R \times A}{L} = \frac{\Omega \times \text{m}^2}{\text{m}} = \Omega\cdot\text{m}$

Materials with very low resistivity are excellent electrical conductors, whereas materials with very high resistivity serve as insulators. Below are standard resistivity values for common engineering materials evaluated at a reference ambient temperature of $20^\circ\text{C}$:

Copper versus Aluminium in Installation Practice

Comparing the resistivities of copper and aluminium reveals an important engineering relationship:

$\frac{\rho_{\text{Al}}}{\rho_{\text{Cu}}} = \frac{2.82 \times 10^{-8}}{1.72 \times 10^{-8}} \approx 1.64$

Because aluminium's resistivity is approximately 1.64 times that of copper, an aluminium cable requires roughly a 64% larger cross-sectional area than a copper cable of identical length to achieve the same electrical resistance. However, aluminium's mass density is only one-third that of copper. As a result, for equivalent current-carrying capacity, an aluminium conductor weighs roughly half as much as a copper conductor, making aluminium the economical choice for long-span high-voltage overhead grid lines.

---

The Conductor Resistance Formula: $R = \frac{\rho L}{A}$

Combining the physical relationships yields the fundamental resistance formula:

$R = \frac{\rho \times L}{A}$

Where:

• $R$ = Conductor resistance in Ohms ($\Omega$)
• $\rho$ = Material resistivity in Ohm-metres ($\Omega\cdot\text{m}$)
• $L$ = Conductor length in metres ($\text{m}$)
• $A$ = Conductor cross-sectional area in square metres ($\text{m}^2$)

Formula Transpositions

Depending on the unknown quantity required in examination questions, the formula transposes as follows:

$\rho = \frac{R \times A}{L}, \qquad L = \frac{R \times A}{\rho}, \qquad A = \frac{\rho \times L}{R}$

Converting Square Millimetres ($\text{mm}^2$) to Square Metres ($\text{m}^2$)

Because one linear metre contains $1,000$ millimetres ($10^3\text{ mm}$), one square metre consists of a square $1,000\text{ mm}$ wide by $1,000\text{ mm}$ high:

$1\text{ m}^2 = 1,000\text{ mm} \times 1,000\text{ mm} = 1,000,000\text{ mm}^2 = 10^6\text{ mm}^2$

To convert an area from $\text{mm}^2$ to $\text{m}^2$, you must divide by $1,000,000$ or multiply by $10^{-6}$:

$A(\text{m}^2) = A(\text{mm}^2) \times 10^{-6}$

Standard UK metric cable sizes convert to base units as follows:

• $1.0\text{ mm}^2 = 1.0 \times 10^{-6}\text{ m}^2$
• $1.5\text{ mm}^2 = 1.5 \times 10^{-6}\text{ m}^2$
• $2.5\text{ mm}^2 = 2.5 \times 10^{-6}\text{ m}^2$
• $4.0\text{ mm}^2 = 4.0 \times 10^{-6}\text{ m}^2$
• $6.0\text{ mm}^2 = 6.0 \times 10^{-6}\text{ m}^2$
• $10.0\text{ mm}^2 = 10.0 \times 10^{-6}\text{ m}^2 = 1.0 \times 10^{-5}\text{ m}^2$
• $16.0\text{ mm}^2 = 16.0 \times 10^{-6}\text{ m}^2 = 1.6 \times 10^{-5}\text{ m}^2$
• $25.0\text{ mm}^2 = 25.0 \times 10^{-6}\text{ m}^2 = 2.5 \times 10^{-5}\text{ m}^2$

Cross-Sectional Area of Circular Conductor Cores

Where an examination problem specifies the diameter ($d$) or radius ($r$) of a solid circular conductor core rather than stating the area directly, apply standard circular geometry:

$A = \pi r^2 = \frac{\pi d^2}{4}$

If measuring a conductor core with a digital micrometer reveals a diameter of $1.78\text{ mm}$: $r = \frac{1.78}{2} = 0.89\text{ mm}$ $A = \pi \times (0.89)^2 \approx 2.488\text{ mm}^2 \approx 2.5\text{ mm}^2$

---

Step-by-Step Worked Calculations

Worked Example 1: Resistance of a $2.5\text{ mm}^2$ Radial Run

Problem: Calculate the resistance of a single copper conductor $50\text{ metres}$ long having a cross-sectional area of $2.5\text{ mm}^2$ at an ambient temperature of $20^\circ\text{C}$. Take copper resistivity as $\rho = 1.72 \times 10^{-8}\ \Omega\cdot\text{m}$.

Solution:

1. State given values in base SI units:
   • $\rho = 1.72 \times 10^{-8}\ \Omega\cdot\text{m}$
   • $L = 50\text{ m}$
   • $A = 2.5\text{ mm}^2 = 2.5 \times 10^{-6}\text{ m}^2$
2. Apply the resistivity equation: $R = \frac{\rho L}{A} = \frac{1.72 \times 10^{-8} \times 50}{2.5 \times 10^{-6}}$
3. Calculate the numerator: $1.72 \times 10^{-8} \times 50 = 8.60 \times 10^{-7}$
4. Divide by the cross-sectional area: $R = \frac{8.60 \times 10^{-7}}{2.5 \times 10^{-6}} = 0.344\ \Omega$

Worked Example 2: Resistance of an Aluminium Sub-Main

Problem: A commercial sub-main feeder uses a $120\text{ metre}$ single-core aluminium conductor with a cross-sectional area of $16\text{ mm}^2$. Calculate its resistance at $20^\circ\text{C}$ given aluminium resistivity is $2.82 \times 10^{-8}\ \Omega\cdot\text{m}$.

Solution:

1. Identify quantities:
   • $\rho = 2.82 \times 10^{-8}\ \Omega\cdot\text{m}$
   • $L = 120\text{ m}$
   • $A = 16\text{ mm}^2 = 16 \times 10^{-6}\text{ m}^2$
2. Substitute into formula: $R = \frac{2.82 \times 10^{-8} \times 120}{16 \times 10^{-6}} = \frac{3.384 \times 10^{-6}}{16 \times 10^{-6}} = 0.2115\ \Omega$

Worked Example 3: Transposition to Determine Required Area

Problem: A lighting sub-circuit requires an $80\text{ metre}$ run of single-core copper conductor. Design criteria dictate that the conductor resistance must not exceed $0.55\ \Omega$ at $20^\circ\text{C}$. Calculate the minimum required cross-sectional area in $\text{mm}^2$ and select the appropriate standard metric size.

Solution:

1. Transpose the resistance formula for area $A$: $A = \frac{\rho L}{R}$
2. Substitute values ($\rho = 1.72 \times 10^{-8}\ \Omega\cdot\text{m}$): $A = \frac{1.72 \times 10^{-8} \times 80}{0.55} = \frac{1.376 \times 10^{-6}}{0.55} \approx 2.502 \times 10^{-6}\text{ m}^2$
3. Convert back to $\text{mm}^2$ by multiplying by $10^6$: $A = 2.502 \times 10^{-6} \times 10^6 = 2.502\text{ mm}^2$
4. Select the standard cable size: A $2.5\text{ mm}^2$ conductor would yield a resistance slightly over the $0.55\ \Omega$ limit ($R = 0.5504\ \Omega$). Therefore, to guarantee compliance, the next standard size of $4.0\text{ mm}^2$ must be selected.

---

Geometric Proportionality and Scaling Effects

Examiners regularly test candidate understanding of proportional relationships without providing full numerical values:

• Doubling Conductor Length ($L \rightarrow 2L$): $R_{\text{new}} = \frac{\rho (2L)}{A} = 2 \times \left(\frac{\rho L}{A}\right) = 2R$ Doubling length doubles resistance ($2\times$).
• Halving Conductor Length ($L \rightarrow 0.5L$): Halving length halves resistance ($0.5\times$).
• Doubling Conductor Cross-Sectional Area ($A \rightarrow 2A$): $R_{\text{new}} = \frac{\rho L}{2A} = \frac{1}{2} \times \left(\frac{\rho L}{A}\right) = 0.5R$ Doubling area halves resistance ($0.5\times$).
• Doubling Conductor Diameter ($d \rightarrow 2d$): Because area depends on the square of the diameter ($A = \frac{\pi d^2}{4}$): $A_{\text{new}} = \frac{\pi (2d)^2}{4} = \frac{\pi \times 4d^2}{4} = 4A$ Substituting $4A$ into the resistance equation: $R_{\text{new}} = \frac{\rho L}{4A} = \frac{1}{4} \times \left(\frac{\rho L}{A}\right) = 0.25R$ Doubling diameter quadruples area and reduces resistance to one-quarter ($0.25\times$).
• Doubling Length AND Doubling Diameter: $R_{\text{new}} = \frac{\rho (2L)}{4A} = \frac{2}{4} R = 0.5R$ Doubling both length and diameter halves the original resistance.