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# A $6\,ohm$ resistance wire is doubled by folding. What is the new resistance? Verified
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Hint:When an electric current passes through a bulb or some other conductor, the conductor creates an obstruction, which is defined as electrical resistance and denoted by $R$. Electrical resistance exists in all materials, which is why conductors emit heat as current passes through them.

Electrical resistance is proportional to the conductor's length (L) and inversely proportional to its cross-sectional area (A). The following relationship explains it.
$R = \dfrac{{\rho L}}{A}$
Where $\rho$ is the resistivity of the material.
Given: Resistance $R = 6$ ohms
As we saw above
$R = \dfrac{{\rho L}}{A}$.................$(1)$
The wire is doubled by folding so,
Area (A)$= 2A$
Then the length will be (L)$= \dfrac{L}{2}$
The new resistance will be:
${R_{new}} = \dfrac{{\rho \dfrac{L}{2}}}{{2A}}$
$\Rightarrow {R_{new}} = \dfrac{{\rho L}}{{4A}}$
By equation $(1)$ we can simplify the above equation as
${R_{new}} = \dfrac{R}{4}$
$\therefore {R_{new}} = \dfrac{6}{4} = 1.5$ohms

So, the new resistance of the wire will be $1.5$ ohms.