If the voltage in a circuit is doubled and the resistance is halved, what happens to the intensity of the current?

To understand the impact of doubling the voltage while halving the resistance on the intensity of the current, we can refer to Ohm's Law, which states that the current (I) in a circuit is directly proportional to the voltage (V) and inversely proportional to the resistance (R). The formula is expressed as:

I = V / R.

When the voltage is doubled, it becomes 2V. When the resistance is halved, it becomes R/2. If we substitute these values into the equation, we get:

I' = (2V) / (R/2).

To simplify this expression, we can multiply the numerator and denominator:

I' = (2V) * (2/R) = 4V / R.

This shows that the new current (I') is four times the original current (I), indicating that when the voltage is doubled and the resistance is halved, the intensity of the current increases four times.

Therefore, the correct answer indicates that the intensity of the current increases four times due to the combined changes in voltage and resistance. This foundational understanding of Ohm's Law and the relationships between voltage, current, and resistance is essential for analyzing electrical circuits effectively.