Electric Current> A copper wire is stretched to increase it...

A copper wire is stretched to increase its length by .1% . what is the percentage change in the resistance of the wire. Pleas give me full details

Rajasanu boro , 9 Years ago

Grade 12

3 Answers

utpal raj

hii rajasanu

r=pl/A where p=resistivity of material , l=length of material, A = area of material

then you stretch length by 1% l1 = l+0.01l =1.01l

find r1=pl1/A and r2= pl/A and r1-r2 will be the change in resistance .

hope this will clear your doubt , please aproove

good luck

Last Activity: 9 Years ago

Kshitij Sharma

2%. because the formula is $R=\rho L/A$ . $\Rightarrow$ $AR=\rho L$ $\Rightarrow$ $\log (AR)=\log (\rho L)$ $\Rightarrow$ $\log A+\log R=\log \rho +\log L$

On partial differentiation, we get $\frac{dA}{A}+\frac{dR}{R}=\frac{d\rho }{\rho }+\frac{dL}{L}$ $.....(1)$ .since $\rho$ is const, $d\rho$ = 0 and d(volume)=d(LA)=0 $\Rightarrow$ LdA+AdL=0 $\Rightarrow$ $LdA=-AdL.....(2)$ .From(1) $\frac{AdL-LdA}{LA}=\frac{dR}{R}$ And from (2) $\frac{dR}{R}=\frac{2dL}{L}$ = 2 X 1 % =2 %

Last Activity: 9 Years ago

arYa kumaR

let Rinitial =R1 = R and Rfinal =R2
length= L1 =L and L2 = 1001L/1000
since wire is same, V1 =V2
A1.L1=A2.L2
L1/L2=A2/A1
Now R1/R2= ∫L1/A1 × ∫L2/A2
R2 = L1.L1 /L2.L2 × R
= 1002001/1000000 R
Change in R = R2-R1
= 2001/1000000 R
Change % = 2001/1000000 R × 1/R × 100
= 0.2001% = 0.2%[APPROX]

Last Activity: 8 Years ago