(i) Find the equivalent resistance between A and B.

(i) In figure, a parallel combination of resistors 6 Ω and 3 Ω are connected in series with another parallel combination of resistors 4 Ω and 12 Ω between the terminals A and B.

In the first part, let the equivalent resistance of parallel combination of resistors 3 Ω and 6 Ω be R₁.

$$\begin{aligned}\frac{1}{R_1} = \frac{1}{3} + \frac{1}{6} \\[0.5em] \Rightarrow \frac{1}{R_1} = \frac{2 + 1}{6} \\[0.5em] \Rightarrow \frac{1}{R_1} = \frac{3}{6} \\[0.5em] \Rightarrow \frac{1}{R_1} = \frac{1}{2} \\[0.5em] \Rightarrow R_1 = 2 Ω\end{aligned}$$

In the second part, 4 Ω and 12 Ω are in parallel connection. If the equivalent resistance is R₂ then

$$\begin{aligned}\frac{1}{R_2} = \frac{1}{4} + \frac{1}{12} \\[0.5em] \Rightarrow \frac{1}{R_2} = \frac{3 + 1}{12} \\[0.5em] \Rightarrow \frac{1}{R_2} = \frac{4}{12} \\[0.5em] \Rightarrow \frac{1}{R_2} = \frac{1}{3} \\[0.5em] \Rightarrow R_2 = 3 Ω\end{aligned}$$

In the third part R₁ and R₂ are in series. Therefore we get,

2 + 3 = 5 Ω

Hence, equivalent resistance between A and B = 5 Ω.

(ii) The resistivity of wire does not change with the change in the thickness of the wire.