JEE Main 2024 — Current Electricity Question with Solution

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:

$$\frac{20\Omega }{10}=2\Omega$$

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 $$2\Omega$$.

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:

$$\frac{1}{R_{\text{parallel}}}=\frac{1}{R_1}+\frac{1}{R_2}$$

Given that $$R_1=R_2=2\Omega$$ (since the parts are identical), we have:

$$\frac{1}{R_{\text{parallel}}}=\frac{1}{2\Omega }+\frac{1}{2\Omega }=\frac{2}{2\Omega }$$

This simplifies to:

$$\frac{1}{R_{\text{parallel}}}=\frac{2}{2\Omega }=\frac{1}{\Omega }$$

From which it follows that:

$$R_{\text{parallel}}=1\Omega$$

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 $$1\Omega$$.

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:

$$R_{\text{series}}=5\times R_{\text{parallel}}=5\times 1\Omega =5\Omega$$

So, the equivalent resistance of the final combination is $$5\Omega$$.