There are no changing magnetic fields in the problem, so we can apply Kirchhoff's Law around the loop.

$\begin{align*}120 \mbox{ V} - I R - 100 \mbox{ V} - I \cdot 1 \;\Omega = 0\end{align*}$
where we have modeled the battery as an ideal voltage source of $100 \mbox{ V}$ in series with a resistor of $1 \;\Omega$ . We are given that $I = +10 \mbox{ A}$ (the problem statement does not explicitly tell us in which direction the $10 \mbox{ A}$ current is flowing, but if the current is assumed to flow in counterclockwise direction, then the resistance $R$ turns out to be negative, which is absurd) and from this we can find the value of $R$ :
$\begin{align*}R = -\frac{100 \mbox{ V} - 120 \mbox{ V} + 10 \mbox{ A} \cdot 1 \;\Omega}{10 \mbox{ A}} = \boxed{1 \;\Omega}\en...$
Therefore, answer (C) is correct.

Solution to 1986 Problem 24