To analyze what happens to the voltage when the resistance in a circuit is halved while keeping the amperage (I) constant, we can use Ohm's Law, which states that \( V = IR \).
If we denote the original resistance as \( R \) and the original voltage as \( V \), we can express the relationship as:
\[ V = I \cdot R \]
When the resistance is halved, the new resistance \( R' \) becomes:
\[ R' = \frac{R}{2} \]
Now, substituting the new resistance into Ohm's Law:
\[ V' = I \cdot R' \]
This gives us:
\[ V' = I \cdot \left(\frac{R}{2}\right) = \frac{I \cdot R}{2} \]
Notice that this scenario keeps \( I \) constant, so we can say:
\[ V' = \frac{V}{2} \]
This means that if the resistance is halved and the amperage remains the same, the voltage is also halved.
Therefore, the correct response is:
The voltage is halved.