Answer
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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.
Complete step by step answer:
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.
Additional Information:
The capacity of a material to resist flowing electric current is measured in resistivity. Insulators have a higher resistivity than conductors, as you would expect. For comparison, the resistivities of a few materials are mentioned below. Low-resistivity materials are excellent conductors of electricity.
Note:Electric resistivity is denoted by and is defined as the electrical resistance provided per unit length and unit cross-sectional area at a given temperature. Specific electrical resistance is another name for electrical resistance. The ohms. Metre is the SI unit for electrical resistivity.
Complete step by step answer:
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.
Additional Information:
The capacity of a material to resist flowing electric current is measured in resistivity. Insulators have a higher resistivity than conductors, as you would expect. For comparison, the resistivities of a few materials are mentioned below. Low-resistivity materials are excellent conductors of electricity.
Note:Electric resistivity is denoted by and is defined as the electrical resistance provided per unit length and unit cross-sectional area at a given temperature. Specific electrical resistance is another name for electrical resistance. The ohms. Metre is the SI unit for electrical resistivity.
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