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 :
Solution
We know that<br><br>$R = \rho {l \over A}$<br><br>Now, new length : $l' = 2l$<br><br>new area of cross section : $A' = A/2$<br><br>$\therefore$ New resistance : $R' = \rho .{{2l} \over {A/2}}$<br><br>$\Rightarrow R' = 4{{\rho l} \over A}$<br><br>$\Rightarrow R' = 4R$<br><br>$\therefore$ Resultant current : $I = {V \over {4R}}$<br><br>$\Rightarrow$ $I = {1 \over 4}{{VA} \over {\rho l}}$
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Ohm's Law and Resistance
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