
If a resistance $R$ is melted and recast into a wire of length $\dfrac{2}{3}th$ of that original wire, what is the resistance of the new wire?
Answer
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Hint: The electric resistance is a property of a material that shows the opposition to a current flowing through that material. We will first find out the change in the cross-sectional area due to the increase in the length as the volume of the wire remains constant. Then, we will substitute the value of the variables in the formula for resistance of the wire to get the resistance of the new wire in terms of the resistance of the original wire.
Formula Used:
The formula for the resistance of a wire is $R = \rho \dfrac{l}{A}$.
Where $\rho $ is the resistivity of the wire,
$l$ is the length of the wire,
$A$ is the area cross-section of the wire.
Complete step by step answer:
The resistance of a conductor of length $l$ and area cross section $A$ is given by $R = \rho \dfrac{l}{A}$.
Where $\rho $ is the resistivity of the conductor which remains constant for particular temperature.
It is given that the length of the new wire $l' = \dfrac{2}{3}l$
The volume of the conductor is constant, $A'l' = Al$
$ \Rightarrow A' = \dfrac{{3A}}{2}$
The resistance of the new wire is $R' = \rho \dfrac{{l'}}{{A'}}$
Substitute all the required values in the above formula
$R' = \rho \dfrac{{\dfrac{2}{3}l}}{{\dfrac{3}{2}A}}$
$ \Rightarrow R' = \dfrac{4}{9}\left( {\rho \dfrac{l}{A}} \right)$
Or $R' = \dfrac{4}{9}R$
Hence, the resistance of the new wire is $\dfrac{4}{9}R$.
Additional information:
The temperature dependence of resistivity of a material is given by
${\rho _T} = {\rho _0}\left[ {1 + a\left( {T - {T_0}} \right)} \right]$
Where ${\rho _T}$ is the resistivity of the material at a temperature $T$
${\rho _0}$ is the resistivity of the material at a temperature ${T_0}$
$a$ is the temperature coefficient of resistivity.
The S.I. unit of resistance is ohm $\left( \Omega \right)$
The S.I. unit of resistivity is $\Omega \cdot m$.
Note:
The resistivity is independent of the length and area cross section of the conductor. It depends on nature and the temperature of the material. Hence, the resistivity of a material is constant for a particular temperature. The resistivity of metallic conductors increases with increase in temperature.
Formula Used:
The formula for the resistance of a wire is $R = \rho \dfrac{l}{A}$.
Where $\rho $ is the resistivity of the wire,
$l$ is the length of the wire,
$A$ is the area cross-section of the wire.
Complete step by step answer:
The resistance of a conductor of length $l$ and area cross section $A$ is given by $R = \rho \dfrac{l}{A}$.
Where $\rho $ is the resistivity of the conductor which remains constant for particular temperature.
It is given that the length of the new wire $l' = \dfrac{2}{3}l$
The volume of the conductor is constant, $A'l' = Al$
$ \Rightarrow A' = \dfrac{{3A}}{2}$
The resistance of the new wire is $R' = \rho \dfrac{{l'}}{{A'}}$
Substitute all the required values in the above formula
$R' = \rho \dfrac{{\dfrac{2}{3}l}}{{\dfrac{3}{2}A}}$
$ \Rightarrow R' = \dfrac{4}{9}\left( {\rho \dfrac{l}{A}} \right)$
Or $R' = \dfrac{4}{9}R$
Hence, the resistance of the new wire is $\dfrac{4}{9}R$.
Additional information:
The temperature dependence of resistivity of a material is given by
${\rho _T} = {\rho _0}\left[ {1 + a\left( {T - {T_0}} \right)} \right]$
Where ${\rho _T}$ is the resistivity of the material at a temperature $T$
${\rho _0}$ is the resistivity of the material at a temperature ${T_0}$
$a$ is the temperature coefficient of resistivity.
The S.I. unit of resistance is ohm $\left( \Omega \right)$
The S.I. unit of resistivity is $\Omega \cdot m$.
Note:
The resistivity is independent of the length and area cross section of the conductor. It depends on nature and the temperature of the material. Hence, the resistivity of a material is constant for a particular temperature. The resistivity of metallic conductors increases with increase in temperature.
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