Kelvi

A wire of length $l$ and resistance $100\Omega$ 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:

medium

Current Electricity

2024

physics

$55\Omega$

$60\Omega$

$26\Omega$

$52\Omega$

Explanation

To solve this problem, we need to calculate the resistance of the final combination of the wire segments. $\\$ Given: $\\$ • Total resistance of the wire $R = 100 \Omega. \\$ • The wire is divided into 10 equal parts, so each part has a resistance $r. \\$ Calculate the resistance of each part: $\\ r = \frac{R}{10} = \frac{100 \Omega}{10} = 10 \Omega. \\$

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A wire of length l l l and resistance 100Ω 100\Omega 100Ω 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:

Explanation

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A wire of length $l$ and resistance $100\Omega$ 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: