Solution to 1986 Problem 24

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.

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