Question

if the diameter of wire is double then what will be it's resistance?​

Answer 1

Resistance of a given specimen can be calculated as,

R=p * l/A

Where,

R= Resistance

p= resistivity of the material

L=length of the specimen

A= cross section area of the specimen

Let, R1 is the initial resistance of the wire with radius of r1 having cross section area A1

So, R1=p*L/A1

=p*L/{π(r1)²}

Now let R2 be the new resistance of wire with r2 radius (where r2=2r1.) and A2 cross section area

So, R2=p*L/A2

=p*L/{π(r2)²}

So, change in resistance can be given as

R2/R1= [p*L/{π(r2)²}] / [p*L/{π(r1)²}]

=(r1)²/(r2)²

=(r1)²/(2r1)². ( Since r2 = 2r1)

=1/4

So by doubling the radius of wire , the resistance of the wire will become 1/4 time the previous resistance.

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Answer 2

Answer:

Explanation:

ANSWER:-

Given:-

• The diameter of the wire is doubled
• We need to find its resistance

Concept:-

We need to apply

$R = \dfrac{\rho \times l }{a}$

• R is the resistance
• Rho is resistivity
• A is the area.

Let's Answer!

Length of the wire is doubled , so :-

$R = \dfrac{ \times \rho \times l}{\dfrac{2a}{1}}$

$R = \dfrac{1 \times \rho \times l}{4a}$

But we know that:-

$R = \dfrac{\rho \times l }{a}$

So , the resistance will be 1/4 times than that of the original one.

Also, do not get confused in diameter as the diameter is the perimeter depicting the length of the wire only.