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

Ohm's Law is \(V = IR\), where \(V\) is voltage, \(I\) is current, and \(R\) is resistance. Initially, \(V = I_1R_1\), with \(I_1 = 2\space A\).

Step2: Analyze the new - resistance situation

The new resistance \(R_2=\frac{1}{4}R_1\), and voltage \(V\) is constant. So \(V = I_2R_2\). 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

Substitute \(R_2\) into \(I_1R_1 = I_2R_2\), we get \(I_1R_1=I_2\times\frac{1}{4}R_1\). Cancel out \(R_1\) (since \(R_1
eq0\)) from both sides of the equation. Then \(I_2 = 4I_1\).

Step4: Calculate the new current

Given \(I_1 = 2\space A\), then \(I_2=4\times2\space A = 8\space A\).

Answer:

8 A