Physics · Current Electricity

JEE Main 2024 — 5 April, Shift 2 — Question 58

A wire of resistance $20 \Omega$ is divided into 10 equal parts. 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 $\qquad$ $\Omega$ .

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

Each part has resistance $=2 \Omega$ '

2 parts are connected in parallel so, $\mathrm{R}=1 \Omega$

Now, there will be 5 parts each of resistance $1 \Omega$ , they are connected in series. $\mathrm{R}_{\mathrm{eq}}=5 \mathrm{R}, \mathrm{R}_{\mathrm{eq}}=5 \Omega$

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Exam: JEE Main 2024
Paper: 5 April, Shift 2
Subject: Physics
Chapter: Current Electricity
Topic: Combination of Resistors and cells, Wheatstone Bridge

Answer: 5

Step-by-step solution

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