Question

?a wire is stretched 30% what is the percentage change in resistance ​

Answer 1

Answer:

The percentage change in resistance of the wire is 69%.

Explanation:

Resistance of a wire can be determined by,

R = (ρl) ÷ A

Where R is its resistance, l is the length of the wire, A its cross sectional area and ρ its resistivity.

When the wire is stretched, its length and area changes but its volume and resistivity remains constant.

$l_{o}$ = 1.3l, and $A_{o}$ = $\frac{A}{1.3}$

So that;

$R_{o}$ = (ρ$l_{o}$) ÷ $A_{o}$ = (ρ × 1.3l) ÷ ($\frac{A}{1.3}$)

    = (1.3lρ) ÷ ($\frac{A}{1.3}$)

    = $(1.3)^{2}$ × [(ρl) ÷ A]

   = 1.69R               (∵ R = (ρl) ÷ A)

$R_{o}$ = 1.69R

Where $R_{o}$ is the new resistance, $l_{o}$ is the new length, and $A_{o}$ is the new area after stretching the wire.

The change in resistance of the wire = $R_{o}$ - R

                                      = 1.69R  - 1R

                                      = 0.69R

The percentage change in resistance = $\frac{0.69R}{R}$ × 100

                                                               = 0.69 × 100

                                                              = 69%

The percentage change in resistance of the wire is 69%.