$A$ conducting wire of length $l$,area of cross-section $A$,and electric resistivity $\rho$ is connected between the terminals of a battery. $A$ potential difference $V$ is developed between its ends,causing an electric current. If the length of the wire of the same material is doubled and the area of cross-section is halved,the resultant current would be:

• A
  $\frac{1}{4} \frac{VA}{\rho l}$
• B
  $\frac{3}{4} \frac{VA}{\rho l}$
• C
  $\frac{1}{4} \frac{\rho l}{VA}$
• D
  $4 \frac{VA}{\rho l}$