A conducting wire of length 'l',


A conducting wire of length 'l', area of crosssection A and electric resistivity U 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 : 

  1. $\frac{1}{4} \frac{\mathrm{VA}}{\rho l}$

  2. $\frac{3}{4} \frac{\mathrm{VA}}{\rho l}$

  3. $\frac{1}{4} \frac{\rho l}{\mathrm{VA}}$

  4. $4 \frac{\mathrm{VA}}{\rho l}$

Correct Option: 1,


As per the question

Resistance $=\frac{\rho(2 l)}{(\mathrm{A} / 2)}=\frac{4 \rho l}{\mathrm{~A}}$

$\Rightarrow$ Current $=\frac{V}{R}=\frac{V A}{4 \rho l}$

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