Question

# A given wire of residence $$1\Omega$$ is stretched to double its length. What will be its new resistance.

Solution

## Given resistance$$=1\Omega$$ and initial length =L and final length =2 L.Let Initial area$$A$$ and final area$${A}_{f}=\dfrac{A}{2}$$ Though the length and cross sectional area of the wire changes it volume will remain constant.$${V}_{i} ={V}_{f}$$Thus initial resistance$${R}_{i}=\rho \dfrac{L}{A}$$--------(1)  Final resistance $${R}_{f}=\rho\dfrac{2L}{\dfrac{A}{2}}$$--------(2)Computing 1 and 2 we get $${R}_{f}=4{R}_{i}$$$${R}_{f}=4 \Omega$$Physics

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