A Copper (Cu) rod of length 25 cm and cross-sectional area 3 mm$$^2$$ is joined with a similar Aluminium (Al) rod as shown in figure. Find the resistance of the combination between the ends A and B.
(Take resistivity of Copper = 1.7 $$\times 10^{-8}$$ $$\Omega$$m, Resistivity of aluminium = 2.6 $$\times 10^{-8}$$ $$\Omega$$m)

We need to find the total equivalent resistance of a parallel combination of a Copper ($$\text{Cu}$$) rod and an Aluminium ($$\text{Al}$$) rod between points $$A$$ and $$B$$.

1. Extract Given Values and Convert to SI Units

• Length of each rod ($$L$$): $$25\text{ cm} = 0.25\text{ m}$$
• Cross-sectional area ($$A_{\text{area}}$$): $$3\text{ mm}^2 = 3 \times 10^{-6}\text{ m}^2$$
• Resistivity of Copper ($$\rho_{\text{Cu}}$$): $$1.7 \times 10^{-8}\ \Omega\cdot\text{m}$$
• Resistivity of Aluminium ($$\rho_{\text{Al}}$$): $$2.6 \times 10^{-8}\ \Omega\cdot\text{m}$$

2. Calculate the Resistance of Each Individual Rod

The resistance of a conductor is given by the formula:

$$R = \rho \frac{L}{A_{\text{area}}}$$

For the Copper Rod ($$R_{\text{Cu}}$$):

$$R_{\text{Cu}} = (1.7 \times 10^{-8}) \times \frac{0.25}{3 \times 10^{-6}}$$

$$R_{\text{Cu}} = \frac{0.425 \times 10^{-2}}{3} = \frac{4.25}{3} \times 10^{-3}\ \Omega \approx 1.4167\text{ m}\Omega$$

For the Aluminium Rod ($$R_{\text{Al}}$$):

$$R_{\text{Al}} = (2.6 \times 10^{-8}) \times \frac{0.25}{3 \times 10^{-6}}$$

$$R_{\text{Al}} = \frac{0.65 \times 10^{-2}}{3} = \frac{6.5}{3} \times 10^{-3}\ \Omega \approx 2.1667\text{ m}\Omega$$

3. Calculate the Equivalent Resistance ($$R_{AB}$$) in Parallel

Since the two rods are connected side-by-side in a parallel configuration across terminals $$A$$ and $$B$$, the equivalent resistance is calculated as:

$$R_{AB} = \frac{R_{\text{Cu}} \times R_{\text{Al}}}{R_{\text{Cu}} + R_{\text{Al}}}$$

Let's factor out the common term $$10^{-3}$$ to simplify calculation with fractions:

$$R_{AB} = \left( \frac{\frac{4.25}{3} \times \frac{6.5}{3}}{\frac{4.25}{3} + \frac{6.5}{3}} \right) \times 10^{-3}\ \Omega$$

$$R_{AB} = \left( \frac{\frac{27.625}{9}}{\frac{10.75}{3}} \right) \times 10^{-3}\ \Omega$$

$$R_{AB} = \left( \frac{27.625}{9} \times \frac{3}{10.75} \right) \times 10^{-3}\ \Omega$$

$$R_{AB} = \left( \frac{27.625}{3 \times 10.75} \right) \times 10^{-3}\ \Omega = \frac{27.625}{32.25} \times 10^{-3}\ \Omega$$

$$R_{AB} \approx 0.8566 \times 10^{-3}\ \Omega \approx 0.858\text{ m}\Omega$$

Final Answer: 0.858 m$$\Omega$$