Answer:
The current is 4 times the current of the full length wire.
Explanation:
The resistor value is directly proportional to its length:
[tex]R=\rho *\frac{L}{A}[/tex]
the current is given by:
[tex]I=\frac{V}{R}[/tex]
so if the wire was cut in the half:
[tex]R_n=\rho *\frac{L}{2*A}=\frac{1}{2}*\rho *\frac{L}{A}=\frac{1}{2}R[/tex]
Both ends of the wire were connected to the battery, so they are connected in parallel, the equivalent resistor is given by:
[tex]R_{eq}=\frac{1}{\frac{1}{R_n}+\frac{1}{R_n}}=\frac{1}{\frac{1}{\frac{R}{2}}+\frac{1}{\frac{R}{2}}}\\\\R_{eq}=\frac{R^2}{4R}=\frac{R}{4}[/tex]
So the current is:
[tex]I=\frac{V}{R_{eq}}\\I=4*\frac{V}{R}[/tex]
