A resistor is made out of a long wire having a length L. Each end of the wire is attached to a terminal of a battery providing a constant voltage V0. A current I flows through the wire. If the wire were cut in half, making two wires of length L/2, and both wires were attached to the battery (the end of both wires attached to one terminal, and the other ends attached to the other terminal), what would be the total current flowing through the two wires? A resistor is made out of a long wire having a length . Each end of the wire is attached to a terminal of a battery providing a constant voltage 0. A current flows through the wire. If the wire were cut in half, making two wires of length /2, and both wires were attached to the battery (the end of both wires attached to one terminal, and the other ends attached to the other terminal), what would be the total current flowing through the two wires? I 2I 4I I/4 I/2

If we had simply put in parallel with the wire one equal to it, the total current flowing through both resistors would double, due to the equivalent resistance of two resistors of equal value, in parallel, is just the half of one of them, so applying Ohm's Law, I will be double than I₀.

But in this case, everything happens as if we put in parallel two resistors of half the value (as the resistance is directly proportional to the length while Ohm's Law be valid), so the equivalent resistance becomes 1/4 of the original wire.

Applying Ohm's Law, we have I = (V / R/4) ⇒ I = 4 I₀