Question

A wire of given material having length l and area of cross-section A has a resistance of 4 Ω. What would be the resistance of another wire of the same material having length l/2 and area of cross-section 2A? With solution

Answer 1

by the formula of resistance:
R=(σ) L/A
here σ is resistivity, L is length of wire and A is area of cross section

Given. σ1=σ2. ______eq 1

for first material R1=σ1 L/A.
we can write.. σ1=R1A/L
σ1=4A/L.

for second material σ2 = R2.4A/L

from eq1. 4A/L=R24A/L
R2=0.

hope you understand
have a nice day..

Answer 2

Given:

• Wire with length l and area A has resistance = 4 ohm.

To find:

• Resistance of a wire with length l/2 and area 2A?

Calculation:

First of all, let's write the general equation of resistance with respect to 1st wire :

$R = \rho \times \dfrac{l}{A} = 4 \: ohm$

Now, new wire has :

• Length = l/2
• Area = 2A
• Resistivity will be same as material is same.

So, resistance of new wire is :

$R_{2} = \rho \times \dfrac{length}{Area}$

$\implies R_{2} = \rho \times \dfrac{( \dfrac{l}{2}) }{2A}$

$\implies R_{2} = \rho \times \dfrac{l }{(2 \times 2)A}$

$\implies R_{2} = \dfrac{1}{4} \times (\rho \times \dfrac{l }{A} )$

$\implies R_{2} = \dfrac{1}{4} \times 4$

$\implies R_{2} = 1 \: ohm$

So, resistance of new wire is 1 ohm