VIR (voltage is equal to amperage multiplied by resistance)
I V/R (amperage is equal to voltage divided by resistance)
R V/I (resistance is equal to voltage divided by resistance)
The resistance in a circuit is halved. What happens to the voltage (if the
Amperage is kept the same)? (1 point)
The voltage is doubled.
The voltage is halved.
The voltage is increased by a factor of 4.
The voltage remains the same.

Verified Answer :

Alright, let's address this problem step-by-step using Ohm's Law. Ohm's Law indicates that:

\(\[ V = I \times R \]\)

where:
- \(\( V \)\) is the voltage,
- \(\( I \)\) is the current (or amperage),
- \(\( R \)\) 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 \(\( R \)\) and the initial voltage as \(\( V \)\). To keep it simple, let's denote the initial current as \(\( I \)\).

2. Express the initial voltage using Ohm's Law:
According to Ohm's Law, the initial voltage is:
\(\[ V = I \times R \]\)

3. Change the resistance:
The problem states that the resistance \(\( R \)\) is halved. So, the new resistance \(\( R' \)\) is:
\(\[ R' = \frac{R}{2} \]\)

4. Determine the new voltage with the halved resistance while keeping the current the same:
With the new resistance \(\( R' \)\) and the same current \(\( I \)\), the new voltage \(\( V' \)\) can be calculated as:
\(\[ V' = I \times R' \]\)

Substituting the new resistance \(\( R' \)\):
\(\[ V' = I \times \left(\frac{R}{2}\right) \]\)

5. Simplify the new voltage equation:
\(\[ V' = \frac{I \times R}{2} \]\)

Notice that \(\( I \times R \)\) is the original voltage \(\( V \)\):
\(\[ V' = \frac{V}{2} \]\)

Thus, when the resistance is halved and the current remains the same, the new voltage \(\( V' \)\) is half of the original voltage \(\( V \)\).

Conclusion:

Therefore, the correct answer is:
- The voltage is halved.