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?
A. 0.5 A
B. 2A
C. 4A
D. 8A
Answer (2)
To solve this question, we need to use Ohm's Law, which states that V = I × R , where V is the voltage, I is the current, and R is the resistance.
Original Situation:
Current, I = 2 A
Let the original resistance be R .
Using Ohm's Law, the voltage V = I × R = 2 A × R .
New Situation:
The resistance decreases to one-fourth of its original amount, so the new resistance is 4 R .
The voltage remains constant. Therefore, V = I ′ × 4 R , where I ′ is the new current.
Since the voltage stays constant, we can equate the two expressions for voltage:
2 R = I ′ × 4 R
Now, solve for I ′ :
I ′ = 4 R 2 R = 2 × 4 = 8 A
Thus, the resulting current when the resistance decreases to one-fourth while keeping the voltage constant is 8 A .
The correct option is:
8A
The resulting current, when the resistance decreases to one-fourth while the voltage remains constant, is 8 A. This is calculated using Ohm's Law, which states that voltage equals current times resistance. The correct option is 8A.
;
