Physics, 23.07.2020 04:01 Derrielle6451

What will be the resistance Rstretched of the wire if it is stretched to twice its original length? Assume that the density and resistivity of the material do not change when the wire is stretched.

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Answer from: dassiekeaton

If the wire is stretched to twice its original length, the new resistance would be quadrupled.

The resistance (R) of a wire is given by:

$R=\rho\frac{L}{A} \\\\where\ L\ is\ length\ of\ wire, A\ is\ the\ cross\ sectional\ area\ and\ \rho\ \\is\ the\ resistivity$

Since the wire is stretched, the new length (L₁) is twice its original length, hence:

L₁ = 2L

An increase in length affects the area, the new area (A₁) is:

initial volume = volume after stretch

AL = A₁L₁

AL = A₁(2L)

A₁ = A/2

The resistance of the stretched wire (R₁) is:

$R_1=\rho\frac{L_1}{A_1} \\\\R_1=\rho\frac{2L}{A/2} \\\\R_1=4\rho\frac{L}{A}=4R$

Therefore if the wire is stretched to twice its original length, the new resistance would be quadrupled.

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Answer from: irlanda20

Hence, the resistance of the stretched wire will be four times that of the original wire.

Explanation:

The resistance of a material is expressed as R = ρL/A

Volume of the original material V = Area * Length = A*L

ρ is the resistivity of the material

R is the resistance

A is the cross sectional area

L is the length of the wire.

If the wire is stretched o twice its original length then new length of the wire L₂ = 2L. Note that an increase in the length of wire will affect its area but its volume and density will not change.

This means V = V₂

A*L = = A₂*L₂

A*L = A₂*(2 L)

A = 2 A₂

A₂ = A/2

The new resistance of the material Rf = ρL₂/A₂

R₂ = ρ(2 L)/(A/2)

Rf = 2ρL * 2/A

Rf = 4(ρL/A)

Since R = ρL/A

R₂ = 4R

Hence, the resistance of the stretched wire will be four times that of the original wire.

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