CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A wire of resistance $$1\ ohm$$ is stretched to double its length. What is the new resistance?


Solution

Given: a wire of resistance $$1\Omega$$is stretched  to double its length.
To find the new resistance
Solution:
Here, 
$$R=1\Omega, \text{Length}=L, \text{Cross sectional area, CSA}=A$$
The formula used is,
$$R=\rho\dfrac LA\\\implies \rho\dfrac LA=1..........(i)$$
The new length, $$L'=2L$$
And new CSA = $$A'$$
So we have, 
Volume of the wire= $$AL$$
Now after the act of stretching the volume remains constant.
Therefore, $$AL=A'L'$$
$$\implies AL=2LA'\\\implies A'=\dfrac A2$$
Now the new resistance of the wire becomes
$$R_{new}=\rho\dfrac {L'}{A'}\\\implies R_{new}=\rho\dfrac {2L}{\dfrac A2}\\\implies R_{new}=4\rho\dfrac LA$$
$$R_{new}=4\times 1$$ by using equation (i)
Therefore the new resistance is $$4\Omega$$

Physics

Suggest Corrections
thumbs-up
 
1


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image