$A$ 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
    $4 \frac{VA}{\rho l}$

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