A wire of resistance $20 \Omega$ is divided into 10 equal parts, resulting pairs. A combination of two parts are connected in parallel and so on. Now resulting pairs of parallel combination are connected in series. The equivalent resistance of final combination is _________ $\Omega$.

<p>Let's start by understanding the process of dividing the wire and recombining its parts to form the final configuration. Initially, we have a wire with a resistance of $20 \Omega$. This wire is divided into 10 equal parts, each part then has a resistance of: <p>$\frac{20 \Omega}{10} = 2 \Omega$</p> <p>Since each part has the same length and presumably the same material and cross-sectional area, then each part will have the same resistance of $2 \Omega$.</p> <p>When two parts are connected in parallel, the equivalent resistance, $R_{\text{parallel}}$, of this configuration can be calculated using the formula for two resistors in parallel:</p> <p>$\frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2}$</p> <p>Given that $R_1 = R_2 = 2 \Omega$ (since the parts are identical), we have:</p> <p>$$\frac{1}{R_{\text{parallel}}} = \frac{1}{2 \Omega} + \frac{1}{2 \Omega} = \frac{2}{2 \Omega}$$</p> <p>This simplifies to:</p> <p>$\frac{1}{R_{\text{parallel}}} = \frac{2}{2 \Omega} = \frac{1}{\Omega}$</p> <p>From which it follows that:</p> <p>$R_{\text{parallel}} = 1 \Omega$</p> <p>Now, since the original wire was divided into 10 equal parts, and pairs of these parts are connected in parallel, this results in $\frac{10}{2} = 5$ pairs. Each of these pairs has an equivalent resistance of $1 \Omega$.</p> <p>Finally, these pairs are all connected in series. The total resistance of resistors in series is simply the sum of their individual resistances. Therefore, the equivalent resistance of the final configuration, $R_{\text{series}}$, is:</p> <p>$R_{\text{series}} = 5 \times R_{\text{parallel}} = 5 \times 1 \Omega = 5 \Omega$</p> <p>So, the equivalent resistance of the final combination is $5 \Omega$. </p></p>