$A$ 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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