A wire of length and resistance 100 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 i

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Solution

Q.
A wire of length $‘ l ’$ and resistance 100 $Ω$ 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 [2024]

1 26 $Ω$

2 52 $Ω$

3 55 $Ω$

4 60 $Ω$

Ans.
(2)

As the wire having resistance $R = 100$ $Ω$ is divided into 10 equal parts, each part will have resistance $\frac{R}{10}$ .

Equivalent resistance of series combination of first 5 parts,

$R'_{eq} = R_{1} + R_{2} + R_{3} + R_{4} + R_{5}$

$= \frac{R}{10} + \frac{R}{10} + \frac{R}{10} + \frac{R}{10} + \frac{R}{10};$ $R_{e q}^{'} = \frac{R}{2}$

Equivalent resistance of parallel combination of next 5 parts,

$\frac{1}{R_{eq}} = \frac{1}{R_{6}} + \frac{1}{R_{7}} + \frac{1}{R_{8}} + \frac{1}{R_{9}} + \frac{1}{R_{10}} = \frac{10}{R} + \frac{10}{R} + \frac{10}{R} + \frac{10}{R} + \frac{10}{R} = \frac{50}{R}$

$R_{eq} = \frac{R}{50}$

Resistance of final series combination is, $R_{final} = R'_{eq} + R_{eq} = \frac{R}{2} + \frac{R}{50} = \frac{26 R}{50} = \frac{26 \times 100}{50} = 52 Ω$

$∴$ $R_{final} = 52$ $Ω$

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