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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