A wire of resistance R has its length and cross section both doubled. Its resistance will become
A
$4 R$
B
$2 R$
C
$R$
D
$R/4$

Question

A wire of resistance R has its length and cross section both doubled. Its resistance will become

Examoogle · Accepted Answer

$R$

Explanation: The resistance of a wire is given by $R = \rho \frac{L}{A}$. When the length $L$ is doubled ($L' = 2L$) and the cross-sectional area $A$ is also doubled ($A' = 2A$), the new resistance $R'$ becomes $R' = \rho \frac{2L}{2A} = \rho \frac{L}{A} = R$.

Working Principle: The electrical resistance of a conductor is directly proportional to its length and inversely proportional to its cross-sectional area. By increasing both dimensions by the same factor (a scale factor of 2), the ratios cancel out, leaving the total resistance invariant.

Key Formulas: $R = \rho \frac{L}{A}$ — Ohm's law for resistance based on geometry; $R' = \rho \frac{nL}{nA} = R$ — Proof of invariance under proportional scaling

Key Points: Resistance is a property of the conductor's geometry and material resistivity ($\rho$).; Scaling factor 'n' applied to both L and A results in no change to resistance.; If only length were doubled, resistance would increase by a factor of 4 (assuming volume conservation).; If only area were doubled, resistance would decrease by a factor of 4.

Advantages: Uniform scaling preserves electrical properties.; Easier to analyze using geometric scaling factors.

Applications: Cable sizing in power distribution.; Design of heating elements and conductors.

Additional Info: Resistivity ($\rho$) depends only on the material type and temperature, not dimensions.. Option A (4R) would occur if only length were doubled while maintaining volume (which would force the area to decrease).

Answer: C is correct — Since the length and cross-sectional area are both scaled by the same factor, the ratio remains unchanged, resulting in the same resistance $R$.

Exam Tip: Always verify if the problem implies volume conservation. If volume is constant and you double length, area MUST decrease, leading to $4R$. If physical dimensions are explicitly stated as doubled (as in this case), use the direct formula ratios.

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