a circuit has a current of 2 a. if the resistance in the circuit decreases to one - fourth of its original…

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?\n0.5 a\n2 a\n4 a\n8 a

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?\n0.5 a\n2 a\n4 a\n8 a

Answer

Explanation:

Step1: Recall Ohm's Law

$V = IR$, where $V$ is voltage, $I$ is current and $R$ is resistance. Initially, $V = I_1R_1$, with $I_1 = 2A$.

Step2: Set up the new - resistance situation

The new resistance $R_2=\frac{1}{4}R_1$ and voltage remains constant $V = I_2R_2$. Since $V = I_1R_1$ and $V = I_2R_2$, we have $I_1R_1=I_2R_2$.

Step3: Substitute $R_2=\frac{1}{4}R_1$ into the equation

$I_1R_1 = I_2\times\frac{1}{4}R_1$. The $R_1$ terms cancel out on both sides of the equation. We get $I_1=\frac{1}{4}I_2$.

Step4: Solve for $I_2$

Given $I_1 = 2A$, then $2=\frac{1}{4}I_2$. Multiply both sides by 4 to find $I_2$. So $I_2=8A$.

Answer:

D. 8 A