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Question

# A $6\Omega$ resistance wire is doubled on itself. Calculate the new resistance of the wire.

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Solution

## Step 1: Given data and formula usedGiven data:Resistance of the wire is $R=6\Omega$Consider the length of the wire be $l$ and the area of the cross-section of the wire be $A$.Formula:The resistance of the wire is given by $R=\rho \frac{l}{A}$ Where $\rho$ is the resistivity of the material of the wire.When the wire is doubled on itself, the length of the wire becomes $l/2$ and the area of cross-section of the wire becomes $2A$.The new resistance of the wire becomes $R\text{'}=\rho \frac{l/2}{2A}$Step 2: Finding the formula of new resistance of the wireDivide $R\text{'}$ by $R$ to calculate the new resistance of the wire.$\frac{R\text{'}}{R}=\frac{\rho \frac{l/2}{2A}}{\rho \frac{l}{A}}\phantom{\rule{0ex}{0ex}}\frac{R\text{'}}{R}=\frac{l}{4A}×\frac{A}{l}\phantom{\rule{0ex}{0ex}}\frac{R\text{'}}{R}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}R\text{'}=\frac{R}{4}$Step 3: Calculating the new resistance of the wireSubstituting the value of resistance into the formula to calculate $R\text{'}$.$R\text{'}=\frac{6\Omega }{4}\phantom{\rule{0ex}{0ex}}R\text{'}=1.5\Omega$Thus, the new resistance of the wire is $1.5\Omega$.

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