Physics, asked by neenirvik6370, 11 months ago

If the diameter of a resistance wire is halved then its resistance becomes

Answers

Answered by mokshbhatt97386
4

Answer:

4x

Explanation: yes its a bit confusing but if diameter is doubled then basically the area is 1/4 times

so area of cross section is inversely proportional to resistance so now resistance is 4 times.

Hope it was helpful

Answered by ArunSivaPrakash
1

Given: The resistance of the wire, R1

             the length of the wire, L.

             the diameter of the wire, d1

To Find: the resistance of the wire when its diameter is halved, R2

Solution:

To calculate R2, the formula used:

  • R = ρ x (L / A)

       here, R is the resistance of the wire

        ρ is the resistivity of the wire

        A is the area of cross-section of the wire

        L is the length of the wire

Applying the above formula:

For R₁:

R₁ = ρ x (L / A)

A = π x r²

here, r is the radius of the wire

and, radius = diameter / 2

or, r = d / 2

∴ A = π x (d/ 2)²

Putting the value of A in the formula for R:

R₁ = ρ x [L / π x (d/ 2)²]

  =  ρ x Lx 4 / π x d²                                 ⇒ 1

For R₂, when the diameter of the wire is halved:

R₂ = ρ x (L / A)

As diameter gets halved-

Area, A = π x [(d/2)/ 2]²

         A = π x (d/ 2x2)²

Putting the value of A in the formula for R₂:

R₂ =  ρ x [ L / π x (d/ 2x2)²]

    =  ρ x L / (π x d² / 4 x4)

    =  ρ x L / (π x d² / 16)

    =  ρ x L x 16 / π x d²                           ⇒ 2

On dividing the equations 1 and 2:

R₁ / R₂  =  (ρ x Lx 4 / π x d²)  / (ρ x L x 16 / π x d²)

R₁ / R₂  =  4 / 16

R₁ / R₂  = 1 / 4

Taking recirpocal on both the sides:

R₂ / R₁  = 4 / 1

R₂        = ( 4/1 ) x  R₁

R₂        =4 x  R₁

R₂        =4R₁

This means new resistance R₂  is four times its original value R₁.

Hence, when the diameter of a wire is halved its resistance becomes four times its original value.

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