A wire of resistance 4 $\Omega$ is stretched to twice its original ...

Question

A wire of resistance 4 $\Omega$ is stretched to twice its original length. What will be the resistance of the stretched wire?

Accepted Answer

New length $L' = 2L$. Volume is conserved: $A'L' = AL \Rightarrow A' = A/2$. $$R' = \frac{ho L'}{A'} = \frac{ho \cdot 2L}{A/2} = 4 \cdot \frac{ho L}{A} = 4R = 4 	imes 4 = 16\ \Omega$$

💬 Doubts & Comments

Abhi tak koi comment nahi

Apna comment likhein

💬 Doubts & Comments

Abhi tak koi comment nahi

Apna comment likhein

🔔 Email Updates Lein