Lesson 3, Topic 1
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Factors Affecting Electrical Resistance of a Conductor

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The electrical resistance of a conductor is affected by the following factors:

• The temperature of the conductor.
• The length of the conductor.
• The cross-sectional area of the conductor.
• The nature of the material of which the conductor is made.

1. Temperature: The resistance of a conductor increases as the temperature of the conductor increases, i.e.

â‡’ $$\scriptsize R \propto T$$

2. Length: The resistance of a conductor is directly proportional to the length; as the length of a conductor increases, the resistance also increases, i.e.

â‡’ $$\scriptsize R \propto L$$

3. Cross-sectional Area: The resistance of a conductor increases as the area of the conductor decreases, which reveals that the resistance of a wire or a conductor is inversely proportional to the area, i.e.

â‡’ $$\scriptsize R \propto \normalsize \frac{1}{A}$$

4. The Nature: The nature of material of the conductor also affects the resistance of a conductor.

Combining all factors shows that resistance is directly proportional to the length of a conductor and inversely proportional to the area.

â‡’ $$\scriptsize R \propto \normalsize \frac{L}{A}$$

âˆ´ $$\scriptsize R = \normalsize \frac{\rho \:\times\: L}{A}$$

$$\scriptsize \rho$$ = a constant called resistivity of a material

Resistivity is the resistance, per unit length, of a conductor of unit cross sectional area.

âˆ´ $$\scriptsize \rho = \normalsize \frac{R\:\times\:A}{L}$$

Where R is in ohms, A in m2 and l in m

The unit for $$\scriptsize \rho$$ is ohm-metre (Î©-m)

Example 1:

Calculate the resistivity of a wire of length 5m of cross-sectional area 1.4×10-6m2, its resistance is 3.0Î©

Solution:

$$\scriptsize R = \normalsize \frac{\rho L}{A}$$

$$\scriptsize \rho = \normalsize \frac{RA}{L}$$

= $$\frac{3.0 \; \times \; 1.4 \times 10^{-6}}{5}$$

= $$\frac{0.0000042}{5}$$

= 0.00000084

= 8.4 X 10-7Î©m

Example 2:

Find the resistance of a wire of lenth 0.85m, radius 0.25mm and resistivity 4.5 x 10-6Î©m

Solution:

Data given in the question:

l = 0.85m

$$\scriptsize \rho = 4.5 \: \times \: 10^{-6} \: \Omega m$$

The cross-sectional area of a wire is directly proportional to the square of its radius

A = $$\scriptsize \pi r^2$$

$$\scriptsize R = \normalsize \frac{\rho \:\times\: L}{A}$$

$$\scriptsize R = \normalsize \frac{4.4 \:\times \: 10^{-6} \:\times\: 0.85}{\pi (2.5 \: \times \: 10^{-4})^2}$$

$$\scriptsize R = \normalsize \frac{3.74 \: \times \: 10^{-6}}{1.963 \: \times \: 10^{-7}}$$

= 19.081 ohms

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