Question

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

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 new - resistance and current relationship

The new resistance $R_2=\frac{1}{4}R_1$, and voltage remains the same $V = I_2R_2$. 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

$I_1R_1 = I_2\times\frac{1}{4}R_1$. Cancel out $R_1$ (since $R_1
eq0$) on both sides of the equation. We get $I_2 = 4I_1$.

Step4: Calculate the new current

Substitute $I_1 = 2A$ into $I_2 = 4I_1$. So $I_2=4\times2A = 8A$.

Answer:

D. 8A