Question

A uniform wire of resistance 50is is cut into five equal parts. These parts are now connected in

parallel. Then the Reff is ( )

a) 2 b) 12 c) 250 d) 625​

Answer 1

Answer:

option (a) 2 Ω

Explanation:

Given :

A uniform wire of resistance 50 Ω is cut into five equal parts.

To find :

the equivalent resistance when these parts are connected in parallel.

Solution :

Resistance of a wire is given by,

  $\longrightarrow \tt R=\dfrac{\rho l}{A}$

where

ρ denotes specific resistance/resistivity

l denotes length of the wire

A denotes the area of cross section of the wire

From this, we can note that resistance of the wire is directly proportional to the length of the wire.

 R ∝ l

The wire is cut into five equal parts.

length of each part, l' = l/5

Let R' be the resistance of each part.

 $\rm \dfrac{R}{R'}=\dfrac{l}{l'} \\\\ \rm \dfrac{50}{R'}=\dfrac{l}{\dfrac{l}{5}} \\\\ \rm \dfrac{50}{R'}=5 \\\\ \rm R'=\dfrac{50}{5} \\\\ \implies \rm R'=10 \Omega$

The resistance of each part = 10 Ω

• When resistors of resistances R₁ , R₂ , R₃ .... are connected in parallel, the equivalent resistance is given by

        1/R = 1/R₁ + 1/R₂ + 1/R₃ + ...

When 'n' identical resistors of resistance R Ω each are connected in parallel, the equivalent resistance is

  $\boxed{\longrightarrow \sf R_{eff}=\dfrac{R}{n}}$

When the 5 parts are connected in parallel, the equivalent resistance is

 $\sf R_{eff}=\dfrac{10}{5} \\\\ \sf R_{eff}=2 \Omega$

The required R(eff) is 2 Ω