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A circuit has a current of 2 A. If the resistance in the circuit decreases to one-fourth of its original amount while the voltage remains constant, what will be the resulting current? 0. 5 A 2 A 4 A 8 A.

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If the resistance decreases to (1/4) of its original value while the voltage remained constant, the resulting current is 8A.

Option D) is the correct answer.

What is Ohm's Law?

Ohm’s law states that "the potential difference between two points is directly proportional to the current flowing through the resistance.

It is expressed as;

V = I × R

Where V is the voltage or potential difference, potential difference, I is the current and R is the resistance.

Given that the circuit has a current of 2A, this means;

V = 2A × R

Next, the resistance in the circuit decreases to one-fourth of its original amount while the voltage remains constant

V = I' × (1/4)R

So we can equate.

2A × R = I' × (1/4)R

2A × R = I' × R/4

2AR = I'R/4

8AR = I'R

I' = 8AR / R

I' = 8A

If the resistance decreases to (1/4) of its original value while the voltage remained constant, the resulting current is 8A.

Option D) is the correct answer.

Learn more about potential difference here: brainly.com/question/2364325

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