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 m^{2} 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^{-6}m^{2}, its resistance is 3.0Î©

**Solution**:

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

= \( \frac{0.0000042}{5} \)

= 0.00000084

= 8.4 X 10^{-7}Î©m

### Example 2:

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

**Solution:**

**Data given in the question:**

l = 0.85m

\( \scriptsize \rho = 4.5 \: \times \: 10^{-6} \: \Omega m \)radius = 0.25mm

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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