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Question
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 :
$4{{VA} \over {\rho l}}$
${3 \over 4}{{VA} \over {\rho l}}$
${1 \over 4}{{VA} \over {\rho l}}$
${1 \over 4}{{\rho l} \over {VA}}$

Solution

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