In a series circuit, Ohm’s Law states \(I = \frac{V}{R}\), where \(V\) is voltage and \(R\) is resistance. If resistance is doubled (\(R \to 2R\)) and voltage remains constant:
\[
I_{\text{new}} = \frac{V}{2R} = \frac{1}{2} \cdot \frac{V}{R} = \frac{I}{2}
\]
Thus, the current halves, so option (2) is correct.