A wire of length 'l' and resistance 100W is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

First, find the resistance of each of the 10 equal sections of the original wire.

$\mathrm{R\_\{\backslash text\{total\}\}}=100\Omega$ divided into 10 equal parts gives

$\mathrm{R\_\{\backslash text\{section\}\}}=\frac{100}{10}=10\Omega$

**Series part** – the first 5 sections are connected in series, so their resistances add:

$5\times 10=50\Omega$

**Parallel part** – the next 5 sections are connected in parallel. For n identical resistors, the equivalent resistance is the individual resistance divided by n:

$\mathrm{R\_\{\backslash text\{parallel\}\}}=\frac{10}{5}=2\Omega$

Finally, the series and parallel groups are connected in series, so their resistances add again:

$50+2=52\Omega$

Therefore, the resistance of the whole arrangement is **52 Ω**.