Alright, let's address this problem step-by-step using Ohm's Law. Ohm's Law indicates that:
[tex]\[ V = I \times R \][/tex]
where:
- [tex]\( V \)[/tex] is the voltage,
- [tex]\( I \)[/tex] is the current (or amperage),
- [tex]\( R \)[/tex] is the resistance.
We need to determine what happens to the voltage (V) if the resistance (R) is halved while maintaining the same amperage (I).
Step-by-Step Solution:
1. Identify the original conditions:
Let's denote the initial resistance as [tex]\( R \)[/tex] and the initial voltage as [tex]\( V \)[/tex]. To keep it simple, let's denote the initial current as [tex]\( I \)[/tex].
2. Express the initial voltage using Ohm's Law:
According to Ohm's Law, the initial voltage is:
[tex]\[ V = I \times R \][/tex]
3. Change the resistance:
The problem states that the resistance [tex]\( R \)[/tex] is halved. So, the new resistance [tex]\( R' \)[/tex] is:
[tex]\[ R' = \frac{R}{2} \][/tex]
4. Determine the new voltage with the halved resistance while keeping the current the same:
With the new resistance [tex]\( R' \)[/tex] and the same current [tex]\( I \)[/tex], the new voltage [tex]\( V' \)[/tex] can be calculated as:
[tex]\[ V' = I \times R' \][/tex]
Substituting the new resistance [tex]\( R' \)[/tex]:
[tex]\[ V' = I \times \left(\frac{R}{2}\right) \][/tex]
5. Simplify the new voltage equation:
[tex]\[ V' = \frac{I \times R}{2} \][/tex]
Notice that [tex]\( I \times R \)[/tex] is the original voltage [tex]\( V \)[/tex]:
[tex]\[ V' = \frac{V}{2} \][/tex]
Thus, when the resistance is halved and the current remains the same, the new voltage [tex]\( V' \)[/tex] is half of the original voltage [tex]\( V \)[/tex].
Conclusion:
Therefore, the correct answer is:
- The voltage is halved.
