FCC Exam Question: 6A35

Halving the cross-sectional area of a conductor will:

A. Double the resistance
B. Half the resistance
C. Not affect the resistance
D. None of the above
Correct Answer: A

Explanation: The resistance of a conductor is inversely proportional to its cross-sectional area. This fundamental relationship is described by the formula $R = \rho \frac{L}{A}$, where $R$ is resistance, $\rho$ (rho) is the material's resistivity, $L$ is the length, and $A$ is the cross-sectional area. If you halve the cross-sectional area ($A$ becomes $A/2$), the resistance will double. For example, if the original resistance is $R = \rho \frac{L}{A}$, halving the area makes the new resistance $R_{new} = \rho \frac{L}{(A/2)} = 2 \times \rho \frac{L}{A} = 2R$. A smaller pathway for current flow increases opposition. Therefore, option A is correct. Options B and C are incorrect because resistance is directly affected by changes in cross-sectional area, specifically in an inverse manner.

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Includes Elements 1, 3, 6, 7R, 8, and 9.