A length of wire has a resistance of 18.0 Ω. When the wire is uniformly stretched (i.e., the total volume of the wire is unchanged during the stretching) to triple it original length, what resistance results?
A) 162 Ω
B) 54.0 Ω
C) 18.0 Ω
D) 6.00 Ω
E) 2.00 Ω

Question

Quizplus · Accepted Answer

The resistance of a wire is given by $R = \rho \frac{L}{A}$, where $R$ is the resistance, $\rho$ is the resistivity of the material, $L$ is the length of the wire, and $A$ is the cross-sectional area. When the wire is stretched to triple its original length, its length $L$ becomes 3 times larger, and its cross-sectional area $A$ becomes smaller to keep the volume constant. Since the volume $V = A \cdot L$ remains constant, if $L$ is tripled, $A$ must be reduced by a factor of 3 to keep $V$ constant. The new resistance $R'$ can be calculated as $R' = \rho \frac{3L}{A/3} = 9\rho \frac{L}{A} = 9R$. Given the original resistance is 18.0 Ω, the new resistance is $9 \times 18.0 \Omega = 162 \Omega$.

A Length of Wire Has a Resistance of 18