## Solution to 2001 Problem 85

 Recall the definition of resistance\begin{align*}R = \frac{\rho l}{A}\end{align*}where $l$ is the length of the wire, $A$ is the cross-sectional area of the wire, and $\rho$ is the resistivity of the wire material. Let $I > 0$ be the current that runs from the $8 \mbox{ V}$ terminal to the $1 \mbox{ V}$ through the Nichrome wire. Then by Ohm's Law the following relationship must hold:\begin{align*}I \frac{\rho 2 L}{A} + I \frac{L}{2 A} = 7 \mbox{ V}\end{align*}This implies that\begin{align*}I = \frac{28 \mbox{ V} \cdot A}{5 L}\end{align*}Therefore, the potential difference across the smaller wire is\begin{align*}I \frac{L}{2 A} = \frac{14 \mbox{ V} \cdot A}{5 L} \cdot \frac{L}{2 A} = \frac{7}{5} \mbox{ V} = 1.4 \mbox{ V}\...Therefore, the potential at the junction is\begin{align*}1 \mbox{ V} + 1.4 \mbox{ V} = \boxed{2.4 \mbox{ V}}\end{align*}Therefore, answer (A) is correct.
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